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Hi, you have probably heard of this classic math question about coin flipping, but in case you haven't I will tell it to you.
Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. Now how did the other coin land?
(Correct) answer: There is a 67% chance that the other coin landed tails. Why? Because after he told you one was heads, the remaining possibilities of the coin lands were heads-heads, heads-tails and tails-heads, therefore in two out of three cases the other coin is tails. If he told you that the first coin thrown was heads, then the chances would be 50-50, since there would be only two possibilities, heads-heads and heads-tails.
My opinion: WHAT? That is the stupidest thing I have ever heard. Two coins were thrown, coins don't have a memory, no matter what your friend tells you, the chances of other coin will always be 50-50. Let's look at this from a different perspective.
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side. Do we really think that if someone tells you something, the probabilities of coin flips change?
Let's create a fictional scene, where you use the (correct) math. You walk down the street, there is this guy who tells you he will give you $5 if you can throw two heads in a row. You think ok, 25% chance. Then he tells you it gets better, one of the coins you will throw is heads on both sides. Now you think wow, you throw that first and 50% chance you get second one heads as well. But then he says to you he won't tell do you throw the one sided coin first or second. Now you feel bummered since you only have 33% chance since the three possible ways the throwing can go are heads-heads, tails-heads and heads-tails.
Exactly, that does not make any sense what so ever. It does not matter when you or he, or anyone throws the one sided coin, you will always have 50% chance with other coin. Throwing two coins always has a 50% chance that both coins land same side, even if you know how one of the coins landed. FACT!
 
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Calgary25968 Posts
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities.
If he flipped a coin, told you it was heads, and then flipped the other coin, there would be a 50% chance one of them was tails. But that's not the same thing. In scenario 1, TT was an option was that later eliminated with information. In scenario 2, TT is never a possibility so it doesn't factor in.
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edit again - the last part of chill's post makes the most sense out of this lol
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It's the classic Montey's door scenario or what not. Learn it in every intro probability course. It's actually not intuitive if you haven't studied probability yet, so it's understandable how people get this wrong all the time.
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On July 16 2011 02:38 Chill wrote:
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities.
It eliminates one of the possibilites, but it also make the heads-heads possibility twise as possible as before making it 50-50.
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On July 16 2011 02:35 Sea_Food wrote:
Let's create a fictional scene, where you use the (correct) math. You walk down the street, there is this guy who tells you he will give you $5 if you can throw two heads in a row. You think ok, 25% chance. Then he tells you it gets better, one of the coins you will throw is heads on both sides. Now you think wow, you throw that first and 50% chance you get second one heads as well. But then he says to you he won't tell do you throw the one sided coin first or second. Now you feel bummered since you only have 33% chance since the three possible ways the throwing can go are heads-heads, tails-heads and heads-tails.
In this scenario, the outcomes you listed do not all have equal probability, hence the error in your calculations.
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On July 16 2011 02:40 lvatural wrote:
It's the classic Montey's door scenario or what not. Learn it in every intro probability course. It's actually not intuitive if you haven't studied probability yet, so it's understandable how people get this wrong all the time.
I have studied probability, and from a mathbook, I first saw this question, but I did not want to argue this with math teacher.
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On July 16 2011 02:43 Jumbled wrote:Show nested quote +On July 16 2011 02:35 Sea_Food wrote:
Let's create a fictional scene, where you use the (correct) math. You walk down the street, there is this guy who tells you he will give you $5 if you can throw two heads in a row. You think ok, 25% chance. Then he tells you it gets better, one of the coins you will throw is heads on both sides. Now you think wow, you throw that first and 50% chance you get second one heads as well. But then he says to you he won't tell do you throw the one sided coin first or second. Now you feel bummered since you only have 33% chance since the three possible ways the throwing can go are heads-heads, tails-heads and heads-tails.
In this scenario, the outcomes you listed do not all have equal probability, hence the error in your calculations.
Then why do all the outcomes in the orriginal story have equal probabilities?
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Calgary25968 Posts
On July 16 2011 02:42 Sea_Food wrote:Show nested quote +On July 16 2011 02:38 Chill wrote:
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities. It eliminates one of the possibilites, but it also make the heads-heads possibility twise as possible as before making it 50-50.
He told you the information after flipping the coins. It isn't twice as likely.
Think about it:
Scenario 1: I flip 4 coins. I tell you 3 of them are heads. Are you saying the chance of the last one being heads is the same as it being tails? Obviously not:
HHHH
THHH, HTHH, HHTH, HHHT
There are way more possibilities to have 1 tails and 3 heads than all 4 heads.
Scenario 2: I flip 3 coins. They are all heads. What is the chance of the fourth coin being heads? 50%. Coins don't have memories.
HHHH
HHHT
These are the only possibilities.
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On July 16 2011 02:39 rawb wrote:
edit again - the last part of chill's post makes the most sense out of this lol
-.-
The thing chill said in last part was also in the OP.
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Well, I don't really understand completely either, but I think the confusion results from one focusing on single coin instead of two coins which were thrown overall. He's not asking you the probability of 'a' coin landing tails, but what the result of the coin that is not head is.
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you have the question slightly wrong, as the coins aren't interdependent, it is still a 50% chance.
i believe the actual problem you are referring to involves 3 doors, two bad prizes, and one good prize.
you pick one door, from which you will receive the prize. the host then shows you one of other the doors with a bad prize, and asks you if you would like to switch your choice of door. you should, because:
initially you had a 33% chance of getting the good prize, meaning there is a 66% chance that the good prize is behind the other two doors. After the host shows you one of the other doors had a bad prize, the percentages remain the same, and your current door has a 33% chance, while the other one left has a 66% chance, meaning if you want the good prize, switch doors.
i hope this was clear to everyone.
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Calgary25968 Posts
On July 16 2011 02:46 Sea_Food wrote:Show nested quote +On July 16 2011 02:39 rawb wrote:
edit again - the last part of chill's post makes the most sense out of this lol -.- The thing chill said in last part was also in the OP.
I like how you're coming into this as if a famous example of probability is wrong and you are right. Maybe you should try to understand why you're wrong 
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Edit: misread his hypo. It's actually pretty confusing lol
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Calgary25968 Posts
On July 16 2011 02:46 Hesmyrr wrote:
Well, I don't really understand completely either, but I think the confusion results from one focusing on single coin instead of two coins which were thrown overall. He's not asking you the probability of 'a' coin landing tails, but what the result of the coin that is not head is.
Very good point. In Scenario 1, only one of the coins is heads. It doesn't matter which one. In Scenario 2, only the first coin can be heads. That's why Scenario 1 is more likely.
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You understand that one coin doesn't affect the outcome of the other coin, which is perfectly true. But the probability of 2/3 that you give IS affected by the result of the first coin.
The first coin doesn't affect the result of the second coin, but does affect the result when both coins are considered together, which is what you're doing.
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So COIN 1 have 50%/50% chance of getting head & tails.
COIN 2 also have 50%/50% chance of getting head & tails.
Thus
TT
TH
HT
HH
is equally probable. Now one coin landed heads, so visualizing the information:
(Bolded is the coin your friend said was heads.)
TT
TH
HT
HH
Two T, one H. I bolded the other coin with red to emphasize the forementioned result have not come from 'single' coin.
Edit: Just addressing this because I think this is more clearer.
Then why do all the outcomes in the orriginal story have equal probabilities?
TT 25%
TH 25%
HT 25%
HH 25%
The possibility of other coin being H - meaning 2 coins flipped heads - do not change from 25% to 50% because the coins indeed do flip TT 25% of the times. We are not just excluding the TT scenario because we have new information that at least one coin landed H.
So instead of thinking 2/3, one could also say there is 50% chance that coins will flip HT and 25% chance that coins will flip HH.
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I think i understand the mixup. if you have TH and HT as two separate options, shouldn't there also be two HH's? as your double-headed coin could land on either head.
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On July 16 2011 02:35 Sea_Food wrote:
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side.
Can someone try to explain me this?
Because acording to the correct math, if my friend tosses two coins 1 000 000 times, and tells me how one of the coins landed each time, I will say each time that the chacnes coins landed different sides is 67%, which means if im rigth about 677 777 times the coins did land different side.
Now if he didnt tell me anything, the chances would be the coins landed different side only about 500 000 times
On July 16 2011 02:47 Chill wrote:Show nested quote +On July 16 2011 02:46 Sea_Food wrote:On July 16 2011 02:39 rawb wrote:
edit again - the last part of chill's post makes the most sense out of this lol -.- The thing chill said in last part was also in the OP. I like how you're coming into this as if a famous example of probability is wrong and you are right. Maybe you should try to understand why you're wrong
If I didnt think that a famous example of probability is wrong and I are right, I would have not wrote this blog.
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Yeah, this is sort of like the classic Monty Hall problem, which has been repeated seemingly 32834720487 times...
I hope you get it; it's pretty interesting 
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They're two independent events that don't influence each other, the chances are still 50% either way.
This is just the same as the 'gambler's fallacy' in roulette, even if red comes up a hundred times in a row it's still just as likely to come up red the next time.
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Calgary25968 Posts
On July 16 2011 03:00 Sea_Food wrote:Show nested quote +On July 16 2011 02:35 Sea_Food wrote:
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side. Can someone try to explain me this?
The entire thread has explained this. The fact that either coin could be H gives you two chances to be right, whereas there is only one chance to be wrong (HH). The other chance of being wrong (TT) was eliminated by the statement "One coin is heads". 2 / 4 now becomes 2 / 3.
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On July 16 2011 03:07 Chill wrote:Show nested quote +On July 16 2011 03:00 Sea_Food wrote:On July 16 2011 02:35 Sea_Food wrote:
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side. Can someone try to explain me this? Your quoted statement is false.
EDIT:
Explain how it is wrong?
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Think about it this way maybe:
There are two balls in a bag, one green and one red. You pick one at random. So at the time that you picked there is pretty clearly a 50% chance that you pick the red ball.
Now, without looking at the ball you picked, your friend tells you that the green ball is still in the bag. Due to this additional information you now know that the chances of the ball you picked being red are 100%, even though at the time you picked the chance was only 50%.
The coin examples function basically the same way, but there are more possibilities.
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Coin 1 \ Coin 2 (Adding the title because it's important, first result refers to what coin 1 flipped as and second result refers to what coin 2 flipped as)
T T 25,0000
T H 25,0000
H T 25,0000
H H 25,0000
Above is the result you get from tossing two coins 1 000 000 times. Now you LITERALLY THROW AWAY THE 25,0000 results you got (TT) since it does not meet the specified parameter (at least one coin has flipped head).
Coin 1 / Coin 2
T H 25,0000
H T 25,0000
H H 25,0000
About 50,000 times one of the coin flipped heads and another flipped tails. And about 25,000 times both of the coins flipped head. Since we are not talking about how coin 1 or coin 2 flipped but how the coin that is not heads (it can be coin 1 OR coin 2) flipped, 500000/750000 = 2/3.
Try it yourself. Flip two coins 100 times, and you will understand what we mean.
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Chill explained it perfectly in the first reply. Actually, everything Chill posted is how you would normally show people why your friend is right.
You should take any intro to probability course at any level high school+. It's actually pretty interesting and should run you through the basics. A lot of people struggle through it since it really isn't intuitive unless you're willing to open your mind a bit to learn and accept this slightly crazier reality.
Your counterexample fictional scenario is wrong, though I'm not sure what you're talking about still.
The possible outcomes are either:
HT
HH
TH
HH
In other words, it doesn't matter which coin you throw first, only the results of the fair coin matter. And the fair coin is 50/50 for H/T. Not 33.
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On July 16 2011 03:11 Blisse wrote:
Your fictional scenario is wrong, though I'm not sure what you're talking about still.
The possible outcomes are either:
HT
HH
TH
HH
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On July 16 2011 02:55 Tatum26 wrote:
I think i understand the mixup. if you have TH and HT as two separate options, shouldn't there also be two HH's? as your double-headed coin could land on either head.
Nope. There is only 2 coins. HH means the first coin was a head and second was also a head. There can't be another scenario where there is another HH.
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United States10328 Posts
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3.
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Interesting discussion, but yeah along with most other people, Chill explained it pretty well in his first reply. Maybe I should try this on my younger cousins and see if they get it =P
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On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3.
This is best and simplest way of describing it.
Btw, is there any other famous mathematical mindfucks? I am curious
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On July 16 2011 02:44 Sea_Food wrote:Show nested quote +On July 16 2011 02:43 Jumbled wrote:
In this scenario, the outcomes you listed do not all have equal probability, hence the error in your calculations. Then why do all the outcomes in the orriginal story have equal probabilities?
Because it's a different problem. In the first example, you have four outcomes with equal probability, one of which is excluded due to extra information you receive. In the second example, there are only three possible outcomes in the first place, and one of those is more likely than the others.
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yep, I have one that's probably more famous
You're on a gameshow, and there are three doors
Behind one of the doors there is a prize
You choose a door, and the game show host opens one of the other doors and shows there is nothing on the other side.
You can now switch doors, should you?
It's the same problem as the op in concept
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On July 16 2011 03:09 Hesmyrr wrote:
Coin 1 \ Coin 2 (Adding the title because it's important, first result refers to what coin 1 flipped as and second result refers to what coin 2 flipped as)
T T 25,0000
T H 25,0000
H T 25,0000
H H 25,0000
Above is the result you get from tossing two coins 1 000 000 times. Now you LITERALLY THROW AWAY THE 25,0000 results you got (TT) since it does not meet the specified parameter (at least one coin has flipped head).
No I dont, because if he tosses coins that many times, he will throw TT some times, in which case he cannot tell me one of the coins landed H.
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On July 16 2011 03:14 Hesmyrr wrote:Show nested quote +On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious
e^(i(pi)) = -1
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It's easy to find out whether it's 50% or 66%: test it a bunch of times. You'll get 66%. Once you agree that that's the right answer, it'll be easier to try to understand why.
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On July 16 2011 03:16 monkxly wrote:yep, I have one that's probably more famous You're on a gameshow, and there are three doors Behind one of the doors there is a prize You choose a door, and the game show host opens one of the other doors and shows there is nothing on the other side. You can now switch doors, should you? It's the same problem as the op in concept
Going straight with my intuition here, I should because now it is 1/2 instead of 1/3. Right? Actually yeah, that seems slightly more flimsy than the OP's example :O (or at least it feels more like it)
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On July 16 2011 03:19 Impervious wrote:Show nested quote +On July 16 2011 03:14 Hesmyrr wrote:On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious e^(i(pi)) = -1
What does "i" stands for?
btw, I understood the OP with what ]343[ wrote, thanks.
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On July 16 2011 03:20 Hesmyrr wrote:Show nested quote +On July 16 2011 03:16 monkxly wrote:yep, I have one that's probably more famous You're on a gameshow, and there are three doors Behind one of the doors there is a prize You choose a door, and the game show host opens one of the other doors and shows there is nothing on the other side. You can now switch doors, should you? It's the same problem as the op in concept Going straight with my intuition here, I should because now it is 1/2 instead of 1/3. Right? Actually yeah, that seems slightly more flimsy than the OP's example :O (or at least it feels more like it)
2/3 actually.
Why?
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On July 16 2011 03:21 Essbee wrote:Show nested quote +On July 16 2011 03:19 Impervious wrote:On July 16 2011 03:14 Hesmyrr wrote:On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious e^(i(pi)) = -1 What does "i" stands for? btw, I understood the OP with what ]343[ wrote, thanks.
i = sqrt(-1)
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You're too caught up in sequencing. Of course the result of the first coin has no bearing on the second. Stop thinking of the flipping happening in some order, just think of the overall results.
Two coins were flipped. One was heads. There are then three possible outcomes: HT, TH, HH. 67% that the unknown coin was heads. Not the second coin, the unknown coin.
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On July 16 2011 03:21 Impervious wrote:Show nested quote +On July 16 2011 03:20 Hesmyrr wrote:On July 16 2011 03:16 monkxly wrote:yep, I have one that's probably more famous You're on a gameshow, and there are three doors Behind one of the doors there is a prize You choose a door, and the game show host opens one of the other doors and shows there is nothing on the other side. You can now switch doors, should you? It's the same problem as the op in concept Going straight with my intuition here, I should because now it is 1/2 instead of 1/3. Right? Actually yeah, that seems slightly more flimsy than the OP's example :O (or at least it feels more like it) 2/3 actually. Why?
2/3? When you switch, right? Hmm, okay. I might not be understanding this then. Let me think about it...
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On July 16 2011 03:16 Sea_Food wrote:Show nested quote +On July 16 2011 03:09 Hesmyrr wrote:
Coin 1 \ Coin 2 (Adding the title because it's important, first result refers to what coin 1 flipped as and second result refers to what coin 2 flipped as)
T T 25,0000
T H 25,0000
H T 25,0000
H H 25,0000
Above is the result you get from tossing two coins 1 000 000 times. Now you LITERALLY THROW AWAY THE 25,0000 results you got (TT) since it does not meet the specified parameter (at least one coin has flipped head).
No I dont, because if he tosses coins that many times, he will throw TT some times, in which case he cannot tell me one of the coins landed H.
That is precisely why you are throwing them away. >_>
Chill is right.
This problem is poorly worded. The question shouldn't involve a friend at all, because you can make faulty arguments like "he wouldn't have told me that if...".
Here's how the problem should be worded.
You flip two coins, and you know that at least one of the coins turned up heads (but have no information as to which one, nor any other information). What is the probability that the other is heads?
The answer, for reasons explained above many times, is 1/3.
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Calgary25968 Posts
OP, please answer the following questions:
You have two magic coins. When tossed together, at least one of them must be heads. What's the probability of one coin landing on tails?
Compare that to:
You throw two coins together. One of them is heads. What is the probability of one coin landing on tails?
Compare that to:
Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. What is the probability of the other coin being tails?
They're all basically the same scenario with the same logic to get the answer. They're just worded differently, which is what I think is tripping you up.
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This is totally the Monty Hall problem with coins...and this would totally be a great application of Bayes' Theorem as well.
I can see a lot of people giving up on you if you don't simply research this very WELL explained problem that's all over the internet. But if you're in a probability class, Bayes theorem would give you an answer that is satisfying( imo most bayes theorem problems give interesting results to me anyway)
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Calgary25968 Posts
On July 16 2011 03:27 lolsixtynine wrote:Show nested quote +On July 16 2011 03:16 Sea_Food wrote:On July 16 2011 03:09 Hesmyrr wrote:
Coin 1 \ Coin 2 (Adding the title because it's important, first result refers to what coin 1 flipped as and second result refers to what coin 2 flipped as)
T T 25,0000
T H 25,0000
H T 25,0000
H H 25,0000
Above is the result you get from tossing two coins 1 000 000 times. Now you LITERALLY THROW AWAY THE 25,0000 results you got (TT) since it does not meet the specified parameter (at least one coin has flipped head).
No I dont, because if he tosses coins that many times, he will throw TT some times, in which case he cannot tell me one of the coins landed H. That is precisely why you are throwing them away. >_> Chill is right. This problem is poorly worded. The question shouldn't involve a friend at all, because you can make faulty arguments like "he wouldn't have told me that if...". Here's how the problem should be worded. You flip two coins, and you know that at least one of the coins turned up heads (but have no information as to which one, nor any other information). What is the probability that the other is heads? The answer, for reasons explained above many times, is 1/3.
The problem is not poorly worded. The OP is just frankly not getting it. You could word it so that it was easier to get the right answer, but that doesn't necessarily mean it's poorly worded. It's worded in a way that makes it inherently more difficult to get correct, by design.
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On July 16 2011 03:22 Impervious wrote:Show nested quote +On July 16 2011 03:21 Essbee wrote:On July 16 2011 03:19 Impervious wrote:On July 16 2011 03:14 Hesmyrr wrote:On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious e^(i(pi)) = -1 What does "i" stands for? btw, I understood the OP with what ]343[ wrote, thanks. i = sqrt(-1)
What the hell. I did it with a web calculator, I can't figure how this result comes out haha.
2.71828182845905 ^ (3.14159265358979 * sqrt(-1)) = -1
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On July 16 2011 03:24 Hesmyrr wrote:Show nested quote +On July 16 2011 03:21 Impervious wrote:On July 16 2011 03:20 Hesmyrr wrote:On July 16 2011 03:16 monkxly wrote:yep, I have one that's probably more famous You're on a gameshow, and there are three doors Behind one of the doors there is a prize You choose a door, and the game show host opens one of the other doors and shows there is nothing on the other side. You can now switch doors, should you? It's the same problem as the op in concept Going straight with my intuition here, I should because now it is 1/2 instead of 1/3. Right? Actually yeah, that seems slightly more flimsy than the OP's example :O (or at least it feels more like it) 2/3 actually. Why? 2/3? When you switch, right? Hmm, okay. I might not be understanding this then. Let me think about it...
You had a 1/3 chance of guessing the correct door. Therefore the probability of all other doors having the prize is 2/3. Since the host eliminated one door (which does not influence the chance that you originally picked the correct door), the remaining door must have a 2/3 probability of having the prize.
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On July 16 2011 03:29 Essbee wrote:Show nested quote +On July 16 2011 03:22 Impervious wrote:On July 16 2011 03:21 Essbee wrote:On July 16 2011 03:19 Impervious wrote:On July 16 2011 03:14 Hesmyrr wrote:On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious e^(i(pi)) = -1 What does "i" stands for? btw, I understood the OP with what ]343[ wrote, thanks. i = sqrt(-1) What the hell. I did it with a web calculator, I can't figure how this result comes out haha. 2.71828182845905 ^ (3.14159265358979 * sqrt(-1)) = -1
e^(pi*i)=-1 from Euler's identity, but don't derail this thread!
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On July 16 2011 03:28 Chill wrote:Show nested quote +On July 16 2011 03:27 lolsixtynine wrote:On July 16 2011 03:16 Sea_Food wrote:On July 16 2011 03:09 Hesmyrr wrote:
Coin 1 \ Coin 2 (Adding the title because it's important, first result refers to what coin 1 flipped as and second result refers to what coin 2 flipped as)
T T 25,0000
T H 25,0000
H T 25,0000
H H 25,0000
Above is the result you get from tossing two coins 1 000 000 times. Now you LITERALLY THROW AWAY THE 25,0000 results you got (TT) since it does not meet the specified parameter (at least one coin has flipped head).
No I dont, because if he tosses coins that many times, he will throw TT some times, in which case he cannot tell me one of the coins landed H. That is precisely why you are throwing them away. >_> Chill is right. This problem is poorly worded. The question shouldn't involve a friend at all, because you can make faulty arguments like "he wouldn't have told me that if...". Here's how the problem should be worded. You flip two coins, and you know that at least one of the coins turned up heads (but have no information as to which one, nor any other information). What is the probability that the other is heads? The answer, for reasons explained above many times, is 1/3. The problem is not poorly worded. The OP is just frankly not getting it. You could word it so that it was easier to get the right answer, but that doesn't necessarily mean it's poorly worded. It's worded in a way that makes it inherently more difficult to get correct, by design.
I don't think it's poorly worded if you assume your friend would say what he said in ALL cases where it was true.
To take an extreme example, imagine a friend that would only say what he said if there were exactly one head. Then the probability of it being tails would be 100%. The friend complicates things unnecessarily.
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On July 16 2011 03:31 Dave[9] wrote:Show nested quote +On July 16 2011 03:29 Essbee wrote:On July 16 2011 03:22 Impervious wrote:On July 16 2011 03:21 Essbee wrote:On July 16 2011 03:19 Impervious wrote:On July 16 2011 03:14 Hesmyrr wrote:On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious e^(i(pi)) = -1 What does "i" stands for? btw, I understood the OP with what ]343[ wrote, thanks. i = sqrt(-1) What the hell. I did it with a web calculator, I can't figure how this result comes out haha. 2.71828182845905 ^ (3.14159265358979 * sqrt(-1)) = -1 e^(pi*i)=-1 from Euler's identity, but don't derail this thread!
Alright alright, I just found it interesting
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On July 16 2011 03:32 lolsixtynine wrote:Show nested quote +On July 16 2011 03:28 Chill wrote:On July 16 2011 03:27 lolsixtynine wrote:On July 16 2011 03:16 Sea_Food wrote:On July 16 2011 03:09 Hesmyrr wrote:
Coin 1 \ Coin 2 (Adding the title because it's important, first result refers to what coin 1 flipped as and second result refers to what coin 2 flipped as)
T T 25,0000
T H 25,0000
H T 25,0000
H H 25,0000
Above is the result you get from tossing two coins 1 000 000 times. Now you LITERALLY THROW AWAY THE 25,0000 results you got (TT) since it does not meet the specified parameter (at least one coin has flipped head).
No I dont, because if he tosses coins that many times, he will throw TT some times, in which case he cannot tell me one of the coins landed H. That is precisely why you are throwing them away. >_> Chill is right. This problem is poorly worded. The question shouldn't involve a friend at all, because you can make faulty arguments like "he wouldn't have told me that if...". Here's how the problem should be worded. You flip two coins, and you know that at least one of the coins turned up heads (but have no information as to which one, nor any other information). What is the probability that the other is heads? The answer, for reasons explained above many times, is 1/3. The problem is not poorly worded. The OP is just frankly not getting it. You could word it so that it was easier to get the right answer, but that doesn't necessarily mean it's poorly worded. It's worded in a way that makes it inherently more difficult to get correct, by design. I don't think it's poorly worded if you assume your friend would say what he said in ALL cases where it was true. To take an extreme example, imagine a friend that would only say what he said if there were exactly one head. Then the probability of it being tails would be 100%. The friend complicates things unnecessarily.
All this discussion reminds me of certain stupid math thread xD
Also, can anyone tell me what the quiz show problem is referred to as so I can google it? My brain is not just working today.
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I just feel bad for Chill because I've sen him try to explain math to fellow teamliquidians before and eventually give up because the OP simply doesn't understand. But there are tons of resources for the Monty Hall problem online...which gives you like hundreds of perspectives on the same problem...I think the OP just needs research a little bit.
P.S. You're not the only one who's had problems with this problem, even the esteemed Paul Erdos thought that the probability of switching the door (or flipping the coin) was 50%. Indeed it is very counter intuitive, but that's simply the way it is sometimes. Maybe if it helps....run a computer program to approximate the results. I bet there are people out there who have done monte carlo on this.
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On July 16 2011 03:39 Dave[9] wrote:
I just feel bad for Chill because I've sen him try to explain math to fellow teamliquidians before and eventually give up because the OP simply doesn't understand. But there are tons of resources for the Monty Hall problem online...which gives you like hundreds of perspectives on the same problem...I think the OP just needs research a little bit.
P.S. You're not the only one who's had problems with this problem, even the esteemed Paul Erdos thought that the probability of switching the door (or flipping the coin) was 50%. Indeed it is very counter intuitive, but that's simply the way it is sometimes. Maybe if it helps....run a computer program to approximate the results. I bet there are people out there who have done monte carlo on this.
There are. I read an article about this. If I can find it I'll link it here. Unsurprisingly, it was 66%.
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The easy way to make the Monty Hall problem obvious is just to take it to much larger proportions:
I have a standard deck of 52 cards. Without letting you see the relevant side of the cards, I tell you to pick out the Queen of Clubs. I look through all the remaining cards and throw out 50 of them, so that we now each have 1 card and I tell you that one of us has the Queen of Clubs.
Do you wish to trade?
If that doesn't make it really obvious, nothing will
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Calgary25968 Posts
On July 16 2011 04:12 psychopat wrote:The easy way to make the Monty Hall problem is just to take it to much larger proportions: I have a standard deck of 52 cards. Without letting you see the relevant side of the cards, I tell you to pick out the Queen of Clubs. I look through all the remaining cards and throw out 50 of them, so that we now each have 1 card and I tell you that one of us has the Queen of Clubs. Do you wish to trade? If that doesn't make it really obvious, nothing will
lol thats the best way ive ever seen it explained haha
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On July 16 2011 03:31 Dave[9] wrote:Show nested quote +On July 16 2011 03:29 Essbee wrote:On July 16 2011 03:22 Impervious wrote:On July 16 2011 03:21 Essbee wrote:On July 16 2011 03:19 Impervious wrote:On July 16 2011 03:14 Hesmyrr wrote:On July 16 2011 03:12 ]343[ wrote:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is best and simplest way of describing it. Btw, is there any other famous mathematical mindfucks? I am curious e^(i(pi)) = -1 What does "i" stands for? btw, I understood the OP with what ]343[ wrote, thanks. i = sqrt(-1) What the hell. I did it with a web calculator, I can't figure how this result comes out haha. 2.71828182845905 ^ (3.14159265358979 * sqrt(-1)) = -1 e^(pi*i)=-1 from Euler's identity, but don't derail this thread!
You can prove it through the direct use of exponential, sin, and cosine taylor series (which is what I did before I learned about Euler's formula). It's still a mathematical mindfuck, even if there are multiple ways of proving it.
+ Show Spoiler +
Also, Chill basically answered the OP about as well as anyone could.....
EDIT - cleared something up a bit.
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Thanks for the explanations. The wiki link (and illustrations) allowed me to get it straight away. Pretty simple and much more intuitive than coin flip.
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On July 16 2011 04:13 Chill wrote:Show nested quote +On July 16 2011 04:12 psychopat wrote:The easy way to make the Monty Hall problem is just to take it to much larger proportions: I have a standard deck of 52 cards. Without letting you see the relevant side of the cards, I tell you to pick out the Queen of Clubs. I look through all the remaining cards and throw out 50 of them, so that we now each have 1 card and I tell you that one of us has the Queen of Clubs. Do you wish to trade? If that doesn't make it really obvious, nothing will lol thats the best way ive ever seen it explained haha
yea thats how alot of people explain the game show one.
say there are a hundred doors. and you pick one. then the game show host eliminates 98 of them, should you switch?
you know another stupid argument we should have is whether .9999 repeating is the same thing as 1
(we really shouldnt)
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I think it's more obvious with a deck of cards example just because it's more of a "real life" application. People seem to instinctively realize that their odds of actually picking a specific card didn't change just because I threw out the rest of the deck, whereas I've had people continue arguing when I just used a larger sample of doors.
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On July 16 2011 03:39 Dave[9] wrote:
I just feel bad for Chill because I've sen him try to explain math to fellow teamliquidians before and eventually give up because the OP simply doesn't understand. But there are tons of resources for the Monty Hall problem online...which gives you like hundreds of perspectives on the same problem...I think the OP just needs research a little bit.
P.S. You're not the only one who's had problems with this problem, even the esteemed Paul Erdos thought that the probability of switching the door (or flipping the coin) was 50%. Indeed it is very counter intuitive, but that's simply the way it is sometimes. Maybe if it helps....run a computer program to approximate the results. I bet there are people out there who have done monte carlo on this.
Actually the reason I created this topic because I wanted to understand after hours of thinking about it my self.
On July 16 2011 03:27 Chill wrote:OP, please answer the following questions: Show nested quote +You have two magic coins. When tossed together, at least one of them must be heads. What's the probability of one coin landing on tails? Compare that to: Show nested quote +You throw two coins together. One of them is heads. What is the probability of one coin landing on tails? Compare that to: Show nested quote +Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. What is the probability of the other coin being tails? They're all basically the same scenario with the same logic to get the answer. They're just worded differently, which is what I think is tripping you up.
I have no idea what are you trying to say with this post.
And by the way to the guys telling me the door story, I know it and I accept the correct awnser, but I think its 100% different.
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On July 16 2011 03:07 Chill wrote:Show nested quote +On July 16 2011 03:00 Sea_Food wrote:On July 16 2011 02:35 Sea_Food wrote:
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side. Can someone try to explain me this? The entire thread has explained this. The fact that either coin could be H gives you two chances to be right, whereas there is only one chance to be wrong (HH). The other chance of being wrong (TT) was eliminated by the statement "One coin is heads". 2 / 4 now becomes 2 / 3.
So please someone awnser this awnser this:
Because acording to the correct math, if my friend tosses two coins 1 000 000 times, and tells me how one of the coins landed each time, I will say each time that the chacnes coins landed different sides is 67%, which means if im rigth about 677 777 times the coins did land different side.
Now if he didnt tell me anything, the chances would be the coins landed different side only about 500 000 times
Is that statement correct?
Because if it is correct then, cant I just say in the start about 677 777 times they will land different side?
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Try thinking of it this way:
Instead of your friend telling you that one coin landed heads and wanting you to guess the result of the other coin, he tells you that the coins did not both land on tails and wants you to guess what orientation both coins landed(HT and TH are the same).
So from the 4 possible outcomes:
HH
HT
TH
TT
you remove TT, and are lefit with
HH
HT
TH
From here each result is equally likely so there is a 66% chance of HT/TH and a 33% chance of HH
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Edit - see below. Meant to edit, not write a new post.
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Calgary25968 Posts
On July 16 2011 05:02 Sea_Food wrote:Show nested quote +On July 16 2011 03:39 Dave[9] wrote:
I just feel bad for Chill because I've sen him try to explain math to fellow teamliquidians before and eventually give up because the OP simply doesn't understand. But there are tons of resources for the Monty Hall problem online...which gives you like hundreds of perspectives on the same problem...I think the OP just needs research a little bit.
P.S. You're not the only one who's had problems with this problem, even the esteemed Paul Erdos thought that the probability of switching the door (or flipping the coin) was 50%. Indeed it is very counter intuitive, but that's simply the way it is sometimes. Maybe if it helps....run a computer program to approximate the results. I bet there are people out there who have done monte carlo on this. Actually the reason I created this topic because I wanted to understand after hours of thinking about it my self. Show nested quote +On July 16 2011 03:27 Chill wrote:OP, please answer the following questions: You have two magic coins. When tossed together, at least one of them must be heads. What's the probability of one coin landing on tails? Compare that to: You throw two coins together. One of them is heads. What is the probability of one coin landing on tails? Compare that to: Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. What is the probability of the other coin being tails? They're all basically the same scenario with the same logic to get the answer. They're just worded differently, which is what I think is tripping you up. I have no idea what are you trying to say with this post.
I directly asked you to answer the questions. I'm not saying anything. Answer the questions.
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Calgary25968 Posts
On July 16 2011 05:15 Sea_Food wrote:Show nested quote +On July 16 2011 03:07 Chill wrote:On July 16 2011 03:00 Sea_Food wrote:On July 16 2011 02:35 Sea_Food wrote:
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side. Can someone try to explain me this? The entire thread has explained this. The fact that either coin could be H gives you two chances to be right, whereas there is only one chance to be wrong (HH). The other chance of being wrong (TT) was eliminated by the statement "One coin is heads". 2 / 4 now becomes 2 / 3. So please someone awnser this awnser this: Show nested quote +Because acording to the correct math, if my friend tosses two coins 1 000 000 times, and tells me how one of the coins landed each time, I will say each time that the chacnes coins landed different sides is 67%, which means if im rigth about 677 777 times the coins did land different side.
Now if he didnt tell me anything, the chances would be the coins landed different side only about 500 000 times Is that statement correct? Because if it is correct then, cant I just say in the start about 677 777 times they will land different side?
Fully accept that you don't understand this question in the OP. Please stop making up analogies that don't make sense to try to prove yourself right.
In the situation you quoted, here's what would happen:
1. Your friend would flip the coin 1,000,000 times.
1a. 250,000 of these times, the coins would land TT and he would say "Oh that didn't count."
1b. If these did count, you could say "They are both different sides" and be right 500,000 times [50%].
2. 750,000 of these times, he would say "One of them is heads."
3. You say "They are both different sides" and are right 500,000 times [66.67%].
Do you see that you are right 500,000 times in both cases (1b and 3)? However, the percentage increases because 250,000 of the trials "didn't count".
4. You don't accept this and make some other non-applicable analogy.
5. I go get drunk. Be back later.
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On July 16 2011 05:02 Sea_Food wrote:Show nested quote +On July 16 2011 03:39 Dave[9] wrote:
I just feel bad for Chill because I've sen him try to explain math to fellow teamliquidians before and eventually give up because the OP simply doesn't understand. But there are tons of resources for the Monty Hall problem online...which gives you like hundreds of perspectives on the same problem...I think the OP just needs research a little bit.
P.S. You're not the only one who's had problems with this problem, even the esteemed Paul Erdos thought that the probability of switching the door (or flipping the coin) was 50%. Indeed it is very counter intuitive, but that's simply the way it is sometimes. Maybe if it helps....run a computer program to approximate the results. I bet there are people out there who have done monte carlo on this. Actually the reason I created this topic because I wanted to understand after hours of thinking about it my self. Show nested quote +On July 16 2011 03:27 Chill wrote:OP, please answer the following questions: You have two magic coins. When tossed together, at least one of them must be heads. What's the probability of one coin landing on tails? Compare that to: You throw two coins together. One of them is heads. What is the probability of one coin landing on tails? Compare that to: Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. What is the probability of the other coin being tails? They're all basically the same scenario with the same logic to get the answer. They're just worded differently, which is what I think is tripping you up. I have no idea what are you trying to say with this post. And by the way to the guys telling me the door story, I know it and I accept the correct awnser, but I think its 100% different.
Sometimes, it's easier to look at a question like this, and try to figure out the chance of it being wrong, rather than right. Then, since the only two options are "right" and "wrong", the probability of it being "right" is 100% - "wrong".
For the coins question:
You flip 2 coins, one is blue, the other is red. There are 4 possible outcomes, all equally likely. They are:
HH
HT
TH
TT
Now, you are told that one of the coins flipped heads. But you don't know which one. This information tells you that one of the outcomes is now impossible - TT cannot happen if one of the coins flipped heads. The new series of potential outcomes is:
HH
HT
TH
All of them are equally likely. Now, how often does both heads show up? Normally, without a constraint, that would be 25%. In this case, it is 33%.
Since the chance of a T showing up with the 2nd coin is is 100% - chance of 2 H showing up, and the chance of a 33% chance of 2 H showing up, the chance of the 2nd coin being T is 67%.
For the door question, same logic:
What is the chance that your pick of 3 doors is the wrong one? 2/3 times, you choose the wrong door, so it is 67%.
When the host opens a new door to show you it is empty, he does not somehow magically make it more or less likely that you picked the wrong door. The chance you picked one of the two empty doors is still the same.
The chance of the other door being wrong is 100% - the chance your initial pick was wrong = 33%, since there are only 2 doors.
Since the door you picked has a 67% chance of being wrong, that means it has a 33% chance of being the right door. And switching is a 67% chance of being right.
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On July 16 2011 03:39 Dave[9] wrote:
I just feel bad for Chill because I've sen him try to explain math to fellow teamliquidians before and eventually give up because the OP simply doesn't understand.
Yeah. The fact that Chill knows about conditional probability better than a few people in this blog combined makes it difficult to understand for some people, and that produces funny blogs. Thanks for the laughs guys.
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On July 16 2011 04:30 Orpheos wrote:
you know another stupid argument we should have is whether .9999 repeating is the same thing as 1
(we really shouldnt)
Oh god, not that argument. I don't see what's not to get. Just look at the freaking numbers.
0.999999...
1.000000...
Starting from the left, 0.9 repeating has a 0, while 1 has a 1. How can that not be enough for people, we learned this shit in kindergarten.
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The constraint your friend gives you isn't that "Coin A turned up heads". It's that "at least coin A OR coin B turned up heads". That still leaves the possibility that coin A is tails, and coin B is heads, which leaves you with the 3rd potential outcome.
Notice that the question also says "What's the probability of the other coin being tails?". It makes no mention of which is the first coin, and which is the second coin. It simply asks that, "given that you know all of the potential outcomes of two coins being flipped, and Tails+Tails isn't one of them, what are the odds of getting at least one tails out of those flips"? That would be 66.6667%.
Your confusions stems from the usage of the word 'other' in that question. You're thinking of the two coins in absolutes and always assuming the first coin is the one that came up heads. The question doesn't say that.
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I don't get it, instead of having a biased 3 hour thread discussion why don't you go flip 2 coins for 10 minutes, see that the results show that you're wrong and then approach this thread with that new knowledge in mind? "I don't understand this, it has to be wrong" is so much different than "I don't understand this, but I know it's right -- why?"
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a coin can also land on its border/edge, not being heads or tails.. how do you like that into your probability calculous
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On July 16 2011 06:11 josemb40 wrote:
a coin can also land on its border/edge, not being heads or tails.. how do you like that into your probability calculous
It's a hypothetical mathematical problem.
Also, I defy you to show me a video of someone flipping a coin in the air and legitimately landing it on a hard smooth surface. I think it's so improbable to not be worth calculating, like soooooooooooooo improbable.
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On July 16 2011 05:52 lolsixtynine wrote:Show nested quote +On July 16 2011 04:30 Orpheos wrote:
you know another stupid argument we should have is whether .9999 repeating is the same thing as 1
(we really shouldnt) Oh god, not that argument. I don't see what's not to get. Just look at the freaking numbers. 0.999999... 1.000000... Starting from the left, 0.9 repeating has a 0, while 1 has a 1. How can that not be enough for people, we learned this shit in kindergarten.
However. We know that if we multiply by X then divide by X we get the original number, right? Example: 10 * 2 = 20, 20 / 2 = 10.
Now take the number one. Divide it by 3, we get 0.33.., What happens when we multiply that by 3 again? 0.999..
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On July 16 2011 06:40 Adeny wrote:Show nested quote +On July 16 2011 05:52 lolsixtynine wrote:On July 16 2011 04:30 Orpheos wrote:
you know another stupid argument we should have is whether .9999 repeating is the same thing as 1
(we really shouldnt) Oh god, not that argument. I don't see what's not to get. Just look at the freaking numbers. 0.999999... 1.000000... Starting from the left, 0.9 repeating has a 0, while 1 has a 1. How can that not be enough for people, we learned this shit in kindergarten. However. We know that if we multiply by X then divide by X we get the original number, right? Example: 10 * 2 = 20, 20 / 2 = 10. Now take the number one. Divide it by 3, we get 0.33.., What happens when we multiply that by 3 again? 0.999..
The infinitely recurring 3 brings that multiple back to 1.
And here comes the off-topic train.
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On July 16 2011 02:38 Chill wrote:
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities.
If he flipped a coin, told you it was heads, and then flipped the other coin, there would be a 50% chance one of them was tails. But that's not the same thing. In scenario 1, TT was an option was that later eliminated with information. In scenario 2, TT is never a possibility so it doesn't factor in.
How about this:
Random:
HH -> H
HT -> T
TH -> T
TT -> H
50%
After knowing at least one was heads:
1 x HH -> H
0.5 x HT -> 0.5T
0.5 x TH -> 0.5T
TT
50%
I don't see a clear reason to prefer your distribution to mine. For example if he just looked at one of the coins in secret and told what he saw then the second distribution is the correct one (because for HT and TH he'd say tails 50% of the time).
edit: to clarify, everyone who got 2/3 was assuming that every time one of the coins comes up heads your friend will say so. If that's true you are correct to just count up the cases. However, this assumes that he has an a priori preference for heads, i.e for HT he'll always say one of them is heads and never that one of them is tails. I just don't feel like that assumption is justified, given the description of the problem.
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On July 16 2011 06:40 Adeny wrote:Show nested quote +On July 16 2011 05:52 lolsixtynine wrote:On July 16 2011 04:30 Orpheos wrote:
you know another stupid argument we should have is whether .9999 repeating is the same thing as 1
(we really shouldnt) Oh god, not that argument. I don't see what's not to get. Just look at the freaking numbers. 0.999999... 1.000000... Starting from the left, 0.9 repeating has a 0, while 1 has a 1. How can that not be enough for people, we learned this shit in kindergarten. However. We know that if we multiply by X then divide by X we get the original number, right? Example: 10 * 2 = 20, 20 / 2 = 10. Now take the number one. Divide it by 3, we get 0.33.., What happens when we multiply that by 3 again? 0.999..
I believe there is a 99.999...% chance that he was joking.
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On July 16 2011 05:52 lolsixtynine wrote:Show nested quote +On July 16 2011 04:30 Orpheos wrote:
you know another stupid argument we should have is whether .9999 repeating is the same thing as 1
(we really shouldnt) Oh god, not that argument. I don't see what's not to get. Just look at the freaking numbers. 0.999999... 1.000000... Starting from the left, 0.9 repeating has a 0, while 1 has a 1. How can that not be enough for people, we learned this shit in kindergarten.
I cannot believe you actually can't understand they are the same.
I feel so sorry for you. If Chill is drunk enough, he just might ban you for ignorance.
EDIT: On the off chance you're joking, well played.
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On July 16 2011 07:14 hypercube wrote:Show nested quote +On July 16 2011 02:38 Chill wrote:
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities.
If he flipped a coin, told you it was heads, and then flipped the other coin, there would be a 50% chance one of them was tails. But that's not the same thing. In scenario 1, TT was an option was that later eliminated with information. In scenario 2, TT is never a possibility so it doesn't factor in. How about this: Random: HH -> H HT -> T TH -> T TT -> H 50% After knowing at least one was heads: 1 x HH -> H 0.5 x HT -> 0.5T 0.5 x TH -> 0.5T TT50% I don't see a clear reason to prefer your distribution to mine. For example if he just looked at one of the coins in secret and told what he saw then the second distribution is the correct one (because for HT and TH he'd say tails 50% of the time). edit: to clarify, everyone who got 2/3 was assuming that every time one of the coins comes up heads your friend will say so. If that's true you are correct to just count up the cases. However, this assumes that he has an a priori preference for heads, i.e for HT he'll always say one of them is heads and never that one of them is tails. I just don't feel like that assumption is justified, given the description of the problem.
I'm having a hard time figuring out why you're giving double the probability of HH occurring there. That's an assumption nobody made at all. The only assumption made is that your friend flipped two coins and you know that ONE of them came up heads.
The most important piece of information to realize in the entire puzzle is that your friend doesn't specify WHICH SPECIFIC coin came up Heads. Only that one of them did. Out of the four potential possibilities, the clue he gives you only rules out ONE of them, not two of them.
Hypothetically, if he told you that the FIRST coin came up heads, then your outcome list would look like this (remember that the first letter represents the first coin, and the second letter represents the second coin):
HH
HT
TH
TT
And you'd have a 50:50 chance, of picking the right answer. But, that's NOT what he told you. There's a big distinction one needs to make between him telling you that a specific coin came up heads, and that either one of them did.
This was actually part of the original question:
Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. Now how did the other coin land?
(Correct) answer: There is a 67% chance that the other coin landed tails. Why? Because after he told you one was heads, the remaining possibilities of the coin lands were heads-heads, heads-tails and tails-heads, therefore in two out of three cases the other coin is tails. If he told you that the first coin thrown was heads, then the chances would be 50-50, since there would be only two possibilities, heads-heads and heads-tails.
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On July 16 2011 07:37 Bibdy wrote:Show nested quote +On July 16 2011 07:14 hypercube wrote:On July 16 2011 02:38 Chill wrote:
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities.
If he flipped a coin, told you it was heads, and then flipped the other coin, there would be a 50% chance one of them was tails. But that's not the same thing. In scenario 1, TT was an option was that later eliminated with information. In scenario 2, TT is never a possibility so it doesn't factor in. How about this: Random: HH -> H HT -> T TH -> T TT -> H 50% After knowing at least one was heads: 1 x HH -> H 0.5 x HT -> 0.5T 0.5 x TH -> 0.5T TT50% I don't see a clear reason to prefer your distribution to mine. For example if he just looked at one of the coins in secret and told what he saw then the second distribution is the correct one (because for HT and TH he'd say tails 50% of the time). edit: to clarify, everyone who got 2/3 was assuming that every time one of the coins comes up heads your friend will say so. If that's true you are correct to just count up the cases. However, this assumes that he has an a priori preference for heads, i.e for HT he'll always say one of them is heads and never that one of them is tails. I just don't feel like that assumption is justified, given the description of the problem. I'm having a hard time figuring out why you're giving double the probability of HH occurring there. That's an assumption nobody made at all. The only assumption made is that your friend flipped two coins and you know that ONE of them came up heads.
It's called discounting in poker. There's an equal chance of the throw coming up HH or HT. But HH leaves your friend with only one option, he can only say that the one of the coins is heads. HT leaves him with two options: he can either say that one of the coins is heads or (presumably) that one of them is tails.
The chances of the throw coming up HT and him saying one of the coins is heads is 0.25x0.5=0.125 25% for HT and 50% for choosing to point out heads instead of tails.
edit: for HH we have 0.25x1=0.25. 25% for HH and 100% for pointing out heads. 0.25 is twice as much as 0.125 hence the "double probability"
The most important piece of information to realize in the entire puzzle is that your friend doesn't specify WHICH SPECIFIC coin came up Heads. Only that one of them did. Out of the four potential possibilities, the clue he gives you only rules out ONE of them, not two of them.
It rules out one but it might also discount TH and HT by 50%, depending on our model of his behaviour.
Hypothetically, if he told you that the FIRST coin came up heads, then your outcome list would look like this (remember that the first letter represents the first coin, and the second letter represents the second coin):
HH
HT
TH
TT
And you'd have a 50:50 chance, of picking the right answer. But, that's NOT what he told you. There's a big distinction one needs to make between him telling you that a specific coin came up heads, and that either one of them did.
Yeah, that's not what I mean.
This was actually part of the original question:
"Then your friend reveals that one of the coins he threw landed heads."
That's literally all we know about our friend. He tells the truth and he has the option to tell that one of the coins is heads. Is it a stretch to assume that he could say one of them is tails if HT or TH came up? I don't think it is.
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Coin 1 Coin 2
H1(.5) H(.5) = .25
H1(.5) T(.5) = .25
H2(.5) H(.5) = .25
H2(.5) T(.5) = .25
this is about as straight forward as it gets. H1 and H2 signify the two sides of the double-headed coin. as you can see, the probability of getting HH is 2/4, or 50%
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Okay, now you've both completely lost me. I don't see what's the point in modelling the guy's behaviour when its obviously a question that's meant to challenge your ability to consider how the information presented to you only rules out one possibility, nor can I figure out where the hell a double-headed coin comes from.
I'm just going to back away, slowly.
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On July 16 2011 08:29 Bibdy wrote:
Okay, now you've both completely lost me. I don't see what's the point in modelling the guy's behaviour when its obviously a question that's meant to challenge your ability to consider how the information presented to you only rules out one possibility, nor can I figure out where the hell a double-headed coin comes from.
I'm just going to back away, slowly.
If that's what you think there's no way I can convince you.
Won't bite on the double-headed coin.
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On July 16 2011 08:08 hypercube wrote:Show nested quote +On July 16 2011 07:37 Bibdy wrote:On July 16 2011 07:14 hypercube wrote:On July 16 2011 02:38 Chill wrote:
Random:
HH x
HT o
TH o
TT x
50%
After knowing at least one was heads:
HH x
HT o
TH o
TT
66.6%
O_o
I think you're just thinking about it wrong. His knowledge isn't affecting the outcome - the outcome was random but he gave you specific information about the result that eliminates one of the possibilities.
If he flipped a coin, told you it was heads, and then flipped the other coin, there would be a 50% chance one of them was tails. But that's not the same thing. In scenario 1, TT was an option was that later eliminated with information. In scenario 2, TT is never a possibility so it doesn't factor in. How about this: Random: HH -> H HT -> T TH -> T TT -> H 50% After knowing at least one was heads: 1 x HH -> H 0.5 x HT -> 0.5T 0.5 x TH -> 0.5T TT50% I don't see a clear reason to prefer your distribution to mine. For example if he just looked at one of the coins in secret and told what he saw then the second distribution is the correct one (because for HT and TH he'd say tails 50% of the time). edit: to clarify, everyone who got 2/3 was assuming that every time one of the coins comes up heads your friend will say so. If that's true you are correct to just count up the cases. However, this assumes that he has an a priori preference for heads, i.e for HT he'll always say one of them is heads and never that one of them is tails. I just don't feel like that assumption is justified, given the description of the problem. I'm having a hard time figuring out why you're giving double the probability of HH occurring there. That's an assumption nobody made at all. The only assumption made is that your friend flipped two coins and you know that ONE of them came up heads. It's called discounting in poker. There's an equal chance of the throw coming up HH or HT. But HH leaves your friend with only one option, he can only say that the one of the coins is heads. HT leaves him with two options: he can either say that one of the coins is heads or (presumably) that one of them is tails. The chances of the throw coming up HT and him saying one of the coins is heads is 0.25x0.5=0.125 25% for HT and 50% for choosing to point out heads instead of tails. edit: for HH we have 0.25x1=0.25. 25% for HH and 100% for pointing out heads. 0.25 is twice as much as 0.125 hence the "double probability" Show nested quote +The most important piece of information to realize in the entire puzzle is that your friend doesn't specify WHICH SPECIFIC coin came up Heads. Only that one of them did. Out of the four potential possibilities, the clue he gives you only rules out ONE of them, not two of them. It rules out one but it might also discount TH and HT by 50%, depending on our model of his behaviour. Show nested quote +Hypothetically, if he told you that the FIRST coin came up heads, then your outcome list would look like this (remember that the first letter represents the first coin, and the second letter represents the second coin):
HH
HT
TH
TT
And you'd have a 50:50 chance, of picking the right answer. But, that's NOT what he told you. There's a big distinction one needs to make between him telling you that a specific coin came up heads, and that either one of them did. Yeah, that's not what I mean. Show nested quote +This was actually part of the original question:
"Then your friend reveals that one of the coins he threw landed heads." That's literally all we know about our friend. He tells the truth and he has the option to tell that one of the coins is heads. Is it a stretch to assume that he could say one of them is tails if HT or TH came up? I don't think it is.
I'm not sure where you get the option for the friend from. The problem explains a very specific scenario that has occurred and has nothing to do with the friends behavior.
Here:
I have in the real world flipped two coins while posting at Teamliquid. One of the coins has landed heads. Now, what is the chance the other coin landed tails? This is reality, these coins are sitting in front of me right now. There is no other information. How do you determine the chance in this single scenario?
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He can't be this ignorant, he has to be playing devil's advocate at this point. The question he is discussing is losing all relevance to whats ACTUALLY being discussed, and its becoming another analogy that relates to a different scenario in order to make this logic make sense.
Math trolls. =/
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On July 16 2011 08:08 hypercube wrote:Yeah, that's not what I mean. Show nested quote +This was actually part of the original question:
"Then your friend reveals that one of the coins he threw landed heads." That's literally all we know about our friend. He tells the truth and he has the option to tell that one of the coins is heads. Is it a stretch to assume that he could say one of them is tails if HT or TH came up? I don't think it is.
Yes, it's a very big stretch. This is a math problem, not a real-life situation. The (implied) assumption in the original problem is that if at least one coin is heads, your friend will 100% of the time say that one coin is heads. Since we are considering a case where he does say one coin is heads....
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This is actually a problem with language and the framing of the problem. Your interpretation of the statement, "One of the coins is a head" determines the probability of the second coin being heads or tails.
If the statement being evaluated is, "Is at least one of these coins a head?" and the coin-flipper will always either answer, "One of the coins is a head," or, "It is not the case that one of the coins is a head," then the answer is that there is a 2/3 chance that one of the coins is a tail (I'd say it's misleading to say "the other" coin is a tail because a "first" or "original" coin was never specified).
If the statement the coin-flipper is answering is instead, "What side is showing on the first coin I examine?" then the answer is 50%.
For more information, see this article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
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United States10328 Posts
Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?)
On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3.
This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed.
EDIT: Compare to what the guy above me said: we don't pick ONE of the coins and say "oh this is heads." All we are doing is throwing out the case when both are tails.
We assume the coins are distinguishable, so H1T2 is different from H2T1. But H1H2 = H2H1... since both coins are heads.
In fact, to make it easier, we can impose an ordering on the coins being flipped. The first coin will come up H or T, the second coin will come up H or T, but not both (so no T1T2). Notice that the first coin always comes before the second coin, so there's no such thing as H2H1. So the possibilities are H1H2, H1T2, T1H2, and the probability both are heads is 1/3.
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On July 16 2011 11:05 crate wrote:Show nested quote +On July 16 2011 08:08 hypercube wrote:Yeah, that's not what I mean. This was actually part of the original question:
"Then your friend reveals that one of the coins he threw landed heads." That's literally all we know about our friend. He tells the truth and he has the option to tell that one of the coins is heads. Is it a stretch to assume that he could say one of them is tails if HT or TH came up? I don't think it is. Yes, it's a very big stretch. This is a math problem, not a real-life situation. The (implied) assumption in the original problem is that if at least one coin is heads, your friend will 100% of the time say that one coin is heads. Since we are considering a case where he does say one coin is heads....
Except that assumption isn't in the problem, which makes it ambiguous. It's not the only possible assumption and different assumptions will lead to different results. Read the Analysis of ambiguity section in Roketha's link, it discusses exactly this question.
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On July 16 2011 11:45 ]343[ wrote:Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?) Show nested quote +On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed.
That's a huge assumption. What you're doing is that you are rewriting the problem, introducing new information and then go on to prove that the answer is 2/3. It is in your problem. But it's a different problem, so it doesn't help us.
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United States10328 Posts
On July 16 2011 11:57 hypercube wrote:Show nested quote +On July 16 2011 11:45 ]343[ wrote:Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?) On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed. That's a huge assumption. What you're doing is that you are rewriting the problem, introducing new information and then go on to prove that the answer is 2/3. It is in your problem. But it's a different problem, so it doesn't help us.
I'm pretty sure "at least one of the coins turns up heads" is mathematically precise. It means that both are not tails.
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Calgary25968 Posts
lol this thread is a fucking train wreck. Jesus christ.
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On July 16 2011 12:28 ]343[ wrote:Show nested quote +On July 16 2011 11:57 hypercube wrote:On July 16 2011 11:45 ]343[ wrote:Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?) On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed. That's a huge assumption. What you're doing is that you are rewriting the problem, introducing new information and then go on to prove that the answer is 2/3. It is in your problem. But it's a different problem, so it doesn't help us. I'm pretty sure "at least one of the coins turns up heads" is mathematically precise. It means that both are not tails.
It's also not from the OP.
"Then your friend reveals that one of the coins he threw landed heads."
This is the problem statement. The interpretation that he could have said "one of the coins landed tails" if he got TH or HT is reasonable. Others agree with me, read the boy or girl paradox article on wikipedia if you don't believe me.
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I feel bad for Chill....he is trying so hard to explain it to him lol!!!!!
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On July 16 2011 14:25 hypercube wrote:Show nested quote +On July 16 2011 12:28 ]343[ wrote:On July 16 2011 11:57 hypercube wrote:On July 16 2011 11:45 ]343[ wrote:Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?) On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed. That's a huge assumption. What you're doing is that you are rewriting the problem, introducing new information and then go on to prove that the answer is 2/3. It is in your problem. But it's a different problem, so it doesn't help us. I'm pretty sure "at least one of the coins turns up heads" is mathematically precise. It means that both are not tails. It's also not from the OP. "Then your friend reveals that one of the coins he threw landed heads." This is the problem statement. The interpretation that he could have said "one of the coins landed tails" if he got TH or HT is reasonable. Others agree with me, read the boy or girl paradox article on wikipedia if you don't believe me.
Thanks for that, I knew something wasn't sitting right with me.
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United States10774 Posts
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If my friend tells you that you don't know the laws of probability, what's the chance that you'll still be beating my GSTL fantasy team by 25 points?
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Ok I didn't read the whole thread but I take it that your main objection is the "unintuitive" nature of counting
HT
TH
as separate cases?
Well think of it this way: if you flip two coins the chances of getting exactly one head is twice that of getting two heads (if you don't buy this I suggest you try flipping some coins lol)
Ok now we only have two cases:
1: One head one tail (probability 2/3)
2: Two heads (probability 1/3)
(why? because the probability of case 1 must be twice that of case 2, and the sum must equal 1. Or 2x+x = 1 and solve for x)
Now you see that case one has probability 2/3
Or maybe another way, suppose I color one coin blue and one coin red (same as imposing an "ordering"). Your friend tells you that at least one coin has heads. Well then I think you would agree that we should consider two cases: blue head red tails, blue tails red heads. Now if the friend could only say whether the blue coin was heads, well then he could only speak half the time (just like monty hall problem).
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On July 16 2011 14:01 Chill wrote:
lol this thread is a fucking train wreck. Jesus christ.
We can't convince them all :<
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On July 16 2011 05:30 Chill wrote:Show nested quote +On July 16 2011 05:15 Sea_Food wrote:On July 16 2011 03:07 Chill wrote:On July 16 2011 03:00 Sea_Food wrote:On July 16 2011 02:35 Sea_Food wrote:
We all know that if two coins are thrown, there is only 50% chance that both land same side up. But according to this, after he tells you one coin landed X, there is 67% chance that the other coin landed different side. Can someone try to explain me this? The entire thread has explained this. The fact that either coin could be H gives you two chances to be right, whereas there is only one chance to be wrong (HH). The other chance of being wrong (TT) was eliminated by the statement "One coin is heads". 2 / 4 now becomes 2 / 3. So please someone awnser this awnser this: Because acording to the correct math, if my friend tosses two coins 1 000 000 times, and tells me how one of the coins landed each time, I will say each time that the chacnes coins landed different sides is 67%, which means if im rigth about 677 777 times the coins did land different side.
Now if he didnt tell me anything, the chances would be the coins landed different side only about 500 000 times Is that statement correct? Because if it is correct then, cant I just say in the start about 677 777 times they will land different side? Fully accept that you don't understand this question in the OP. Please stop making up analogies that don't make sense to try to prove yourself right. In the situation you quoted, here's what would happen: 1. Your friend would flip the coin 1,000,000 times. 1a. 250,000 of these times, the coins would land TT and he would say "Oh that didn't count." 1b. If these did count, you could say "They are both different sides" and be right 500,000 times [50%]. 2. 750,000 of these times, he would say "One of them is heads." 3. You say "They are both different sides" and are right 500,000 times [66.67%]. Do you see that you are right 500,000 times in both cases (1b and 3)? However, the percentage increases because 250,000 of the trials "didn't count". 4. You don't accept this and make some other non-applicable analogy. 5. I go get drunk. Be back later.
This is exactly my problem from understanding. 1b was the analogy I was searching for.
So does he say 1 tails, or 1 heads, I say they landed different side 67% chance 1 000 000times, but only 500 000 times did they land different side?
If I have a two out of three chance to guess correctly, how can I only guess correct one out of two times?
Or did you mean if we also count when he flips two tails and say one is tails I have 50% chance goess correctly? Because I would not understand why I would only have 50% chance then.
PS. Im trying to prove myself wrong.
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Calgary25968 Posts
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On July 16 2011 11:34 RoKetha wrote:This is actually a problem with language and the framing of the problem. Your interpretation of the statement, "One of the coins is a head" determines the probability of the second coin being heads or tails. If the statement being evaluated is, "Is at least one of these coins a head?" and the coin-flipper will always either answer, "One of the coins is a head," or, "It is not the case that one of the coins is a head," then the answer is that there is a 2/3 chance that one of the coins is a tail (I'd say it's misleading to say "the other" coin is a tail because a "first" or "original" coin was never specified). If the statement the coin-flipper is answering is instead, "What side is showing on the first coin I examine?" then the answer is 50%. For more information, see this article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
most important post of this thread
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Canada7170 Posts
This reminds me of when I was trying to convince people that 'trucker' didn't rhyme with 'diner'.
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On July 16 2011 22:54 mikeymoo wrote:
This reminds me of when I was trying to convince people that 'trucker' didn't rhyme with 'diner'.
Any word rhymes with any word if the speaker tries hard enough. There is no real definition of the word rhyme.
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Lol lets start arguing over what a rhyme is.
Chill said it best when he said its proven and you should be trying to understand why.
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Calgary25968 Posts
On July 17 2011 00:37 Sea_Food wrote:Show nested quote +On July 16 2011 22:54 mikeymoo wrote:
This reminds me of when I was trying to convince people that 'trucker' didn't rhyme with 'diner'. Any word rhymes with any word if the speaker tries hard enough. There is no real definition of the word rhyme.
Record a voice clip of you rhyming "Coin" with "Toss". PM me it and I will unlock this blog.
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Ok, now that I got my topic unlocked I would like to leave a final post (famous not last words.)
I will probably not understand any time soon how this coin flip math actually works, I'm just that stupid. However, I would appreciate if someone who does understand replies this with an answer, so I can someday hopefully understand on my own after more thought.
I am going to type claims, and you will just have to type which are the claims that are incorrect. You can explain why if you want.
1. Whenever two coins are tossed there is a 1/2 chance they both land different side up.
2. Whenever two coins are tossed, and you are told one of them landed on heads, you know there is a 2/3 chance the two coins landed different side up.
3. Whenever two coins are tossed, and you are told one of them landed on tails, you know there is a 2/3 chance the two coins landed different side up.
4. Whenever two coins are tossed, and you are told one of them landed on X, you know there is a 2/3 chance the two coins landed different side up.
5. When someone says he is going to toss two coins, but he promises he will tell you on what side one of them landed, you can before the coins are tossed already tell that there is 2/3 chance they will land different side up.
6. Two coins are tossed 1 000 000 times. Every time two tails land you are told one landed tails. Every time two heads land you are told one landed heads. When its 1 tails, one heads, you are told 50% times one landed heads, other 50% times one landed tails. No matter what, every time you are told which side up one of the coins landed, you will say there is 2/3 chance the two coins landed different side up. The rules on what base are you told what coin landed which side are not told to you.
7. Same as 6, but the rules are told to you.
Predictions:
+ Show Spoiler +Unsure about 4, but 5, 6 and 7. But then again im the guy who dosnt even understand 2 or 3.
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i think your problem is that you are too stubborn. you do understand why it is 2/3 chance, but you dont like the fact that your intuition led you to believe it is 1/2, so you keep adding weird rules to try to make it 1/2. if your weird hypothetical does in fact make it 1/2 its because you changed the situation too much so it doesnt really apply.
the point is that while the chance of heads or tails is 1/2, the additional information you get from flipping two and him telling you that one of them is heads, gives you ADDITIONAL INFORMATION. so you can say with more certainty that one thing happened over another.
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clearly:
![[image loading]](http://upload.wikimedia.org/math/4/7/e/47e0d646891053c1f3b41fa8c81b6ebb.png)
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On July 17 2011 02:05 Chill wrote:Show nested quote +On July 17 2011 00:37 Sea_Food wrote:On July 16 2011 22:54 mikeymoo wrote:
This reminds me of when I was trying to convince people that 'trucker' didn't rhyme with 'diner'. Any word rhymes with any word if the speaker tries hard enough. There is no real definition of the word rhyme. Record a voice clip of you rhyming "Coin" with "Toss". PM me it and I will unlock this blog.
Any chance on the voice clip being posted?
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One coin is heads:
HH, TH, HT, so 2/3 of the time the second coin is heads.
First coin is heads:
HH, HT, so 1/2 of the time the second coin is heads.
Your problem is that you don't consider that the second coin could have been the heads when you say "one coin is heads".
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On July 19 2011 04:14 Orpheos wrote:
i think your problem is that you are too stubborn. you do understand why it is 2/3 chance.
No, actually I dont.
I do know all the reasoning of people saying its 2/3. Math has always been my strongest subject, but no matter how much I think of the problem, I always find all the reasoning false.
On July 19 2011 04:51 Cassel_Castle wrote:
One coin is heads:
HH, TH, HT, so 2/3 of the time the second coin is heads.
First coin is heads:
HH, HT, so 1/2 of the time the second coin is heads.
Your problem is that you don't consider that the second coin could have been the heads when you say "one coin is heads".
I do consider it, but my mind just says that, when I know first coin is heads its 50%, when i know second coin is heads its 50%. When i know if its either first or second, I calculate it 50% + 50% / 2 = 50%, reasoning behind it being its either 1st or 2nd, the coin hes talking bout.
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On July 16 2011 11:34 RoKetha wrote:This is actually a problem with language and the framing of the problem. Your interpretation of the statement, "One of the coins is a head" determines the probability of the second coin being heads or tails. If the statement being evaluated is, "Is at least one of these coins a head?" and the coin-flipper will always either answer, "One of the coins is a head," or, "It is not the case that one of the coins is a head," then the answer is that there is a 2/3 chance that one of the coins is a tail (I'd say it's misleading to say "the other" coin is a tail because a "first" or "original" coin was never specified). If the statement the coin-flipper is answering is instead, "What side is showing on the first coin I examine?" then the answer is 50%. For more information, see this article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
This has been quoted before, but Roketha gets it. Chill and everyone else are doing a good job of stating the correct explanations, but they're horrible at actually explaining it because they are not properly addressing OP's misunderstanding. They're just hammering him with more re-phrasings of the same explanation without explaining WHY both sides are talking about different things. When I first saw this thread I was like "wtf it's obviously 50%" until I realized that the problem wasn't saying what I thought it was.
The real problem is the wording of the problem being a little too open to interpretation, as Roketha said.
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On July 19 2011 05:38 Vod.kaholic wrote:Show nested quote +On July 16 2011 11:34 RoKetha wrote:This is actually a problem with language and the framing of the problem. Your interpretation of the statement, "One of the coins is a head" determines the probability of the second coin being heads or tails. If the statement being evaluated is, "Is at least one of these coins a head?" and the coin-flipper will always either answer, "One of the coins is a head," or, "It is not the case that one of the coins is a head," then the answer is that there is a 2/3 chance that one of the coins is a tail (I'd say it's misleading to say "the other" coin is a tail because a "first" or "original" coin was never specified). If the statement the coin-flipper is answering is instead, "What side is showing on the first coin I examine?" then the answer is 50%. For more information, see this article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox This has been quoted before, but Roketha gets it. Chill and everyone else are doing a good job of stating the correct explanations, but they'd be horrible teachers because they are not addressing OP's misunderstanding. They're just hammering him with more re-phrasings of the same explanation without explaining WHY both sides are talking about different things. The real problem is the wording of the problem being a little too open to interpretation, as Roketha said.
This is actually not the case.
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On July 19 2011 05:41 Sea_Food wrote:Show nested quote +On July 19 2011 05:38 Vod.kaholic wrote:On July 16 2011 11:34 RoKetha wrote:This is actually a problem with language and the framing of the problem. Your interpretation of the statement, "One of the coins is a head" determines the probability of the second coin being heads or tails. If the statement being evaluated is, "Is at least one of these coins a head?" and the coin-flipper will always either answer, "One of the coins is a head," or, "It is not the case that one of the coins is a head," then the answer is that there is a 2/3 chance that one of the coins is a tail (I'd say it's misleading to say "the other" coin is a tail because a "first" or "original" coin was never specified). If the statement the coin-flipper is answering is instead, "What side is showing on the first coin I examine?" then the answer is 50%. For more information, see this article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox This has been quoted before, but Roketha gets it. Chill and everyone else are doing a good job of stating the correct explanations, but they'd be horrible teachers because they are not addressing OP's misunderstanding. They're just hammering him with more re-phrasings of the same explanation without explaining WHY both sides are talking about different things. The real problem is the wording of the problem being a little too open to interpretation, as Roketha said. This is actually not the case.
EDIT: nevermind, I didn't read after this post, I didn't see your post-reopening statement. My bad.
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On July 19 2011 05:37 Sea_Food wrote:Show nested quote +On July 19 2011 04:14 Orpheos wrote:
i think your problem is that you are too stubborn. you do understand why it is 2/3 chance. No, actually I dont. I do know all the reasoning of people saying its 2/3. Math has always been my strongest subject, but no matter how much I think of the problem, I always find all the reasoning false. Show nested quote +On July 19 2011 04:51 Cassel_Castle wrote:
One coin is heads:
HH, TH, HT, so 2/3 of the time the second coin is heads.
First coin is heads:
HH, HT, so 1/2 of the time the second coin is heads.
Your problem is that you don't consider that the second coin could have been the heads when you say "one coin is heads". I do consider it, but my mind just says that, when I know first coin is heads its 50%, when i know second coin is heads its 50%. When i know if its either first or second, I calculate it 50% + 50% / 2 = 50%, reasoning behind it being its either 1st or 2nd, the coin hes talking bout.
Man, if you are going to guess at a formula, at least think of the extreme cases to see if it's reasonable. Say you have one normal coin and one coin with tails on both sides. Then the chances of heads are 50% and 0% respectively (assuming a fair coin). According to your logic, if I told you that one coin was heads, you would conclude that the probability of both being heads would be (50% + 0%)/2 = 25% which is obviously wrong.
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your Country52797 Posts
On July 19 2011 05:37 Sea_Food wrote:Show nested quote +On July 19 2011 04:14 Orpheos wrote:
i think your problem is that you are too stubborn. you do understand why it is 2/3 chance. No, actually I dont. I do know all the reasoning of people saying its 2/3. Math has always been my strongest subject, but no matter how much I think of the problem, I always find all the reasoning false. Show nested quote +On July 19 2011 04:51 Cassel_Castle wrote:
One coin is heads:
HH, TH, HT, so 2/3 of the time the second coin is heads.
First coin is heads:
HH, HT, so 1/2 of the time the second coin is heads.
Your problem is that you don't consider that the second coin could have been the heads when you say "one coin is heads". I do consider it, but my mind just says that, when I know first coin is heads its 50%, when i know second coin is heads its 50%. When i know if its either first or second, I calculate it 50% + 50% / 2 = 50%, reasoning behind it being its either 1st or 2nd, the coin hes talking bout.
This exactly. Coins are independent from each other. If one of them is heads (garunteed) then the other one still has a 50% chance.
Or, it's actually either coin that could be it, in which case we have HT, TH, and HH. TH and HT are redundant, so its HT(TH) and HH, either of which has a 50% chance.
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Stop thinking of the chance of the individual coins.
The problem gives you 3 options. Don't focus on which coin is which, you just have the following 3 possibilities:
Coin A OR B----Coin B OR A
H----------------------- H
H ---------------------- T
T----------------------- H
HT and TH are distinct, and the question is what percentage of the possibilities contain Tails. So, 2/3, or 66%, are tails.
EDIT: I'm just re-wording what chill said without adding much, because IDK what else there is to add, we can only gift-wrap the explanation differently now that I know it's not a misunderstanding.
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Everyone is assuming that the cases HH, HT and TH are equally likely. That's true right after the coin throw but not necessarily after the friend's statement.
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A = HH
B = HT
C = TH
D = At least one coin is heads
P(A|D) = P(A&D)/P(D) (bayes)
=0.25/0.75
=1/3
P(B|D) = P(B&D)/P(D)
=0.25/0.75
=1/3
and same for C. P(A&D)=P(A), P(B&D)=P(B), etc. because A, B, and C all imply D.
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On July 19 2011 03:30 Sea_Food wrote:I am going to type claims, and you will just have to type which are the claims that are incorrect. You can explain why if you want. 1. Whenever two coins are tossed there is a 1/2 chance they both land different side up. 2. Whenever two coins are tossed, and you are told one of them landed on heads, you know there is a 2/3 chance the two coins landed different side up. 3. Whenever two coins are tossed, and you are told one of them landed on tails, you know there is a 2/3 chance the two coins landed different side up. 4. Whenever two coins are tossed, and you are told one of them landed on X, you know there is a 2/3 chance the two coins landed different side up. 5. When someone says he is going to toss two coins, but he promises he will tell you on what side one of them landed, you can before the coins are tossed already tell that there is 2/3 chance they will land different side up. 6. Two coins are tossed 1 000 000 times. Every time two tails land you are told one landed tails. Every time two heads land you are told one landed heads. When its 1 tails, one heads, you are told 50% times one landed heads, other 50% times one landed tails. No matter what, every time you are told which side up one of the coins landed, you will say there is 2/3 chance the two coins landed different side up. The rules on what base are you told what coin landed which side are not told to you. 7. Same as 6, but the rules are told to you. Predictions: + Show Spoiler +Unsure about 4, but 5, 6 and 7. But then again im the guy who dosnt even understand 2 or 3.
Anyone?
Just tell me the incorrect statements please.
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Calgary25968 Posts
I'll answer these when I get back. I deleted my original response because I wasn't satisfied with it.
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Yeah, I learned about this in statistics, only the example was with boys and girls instead of heads and tails.
Definitely very interesting!
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There are multiple ways to interpret even the OP question:
On July 16 2011 02:35 Sea_Food wrote:
Your friend tosses two coins, then asks you to guess how the coins landed. You reply that you cannot know. Then your friend reveals that one of the coins he threw landed heads. Now how did the other coin land?
Your friend tosses two coins and lay them out on the table, covered (by ESPORTS!):
![[image loading]](http://i.imgur.com/xb1T0.png)
He asks you to guess how the coins landed. You reply that you cannot know.
Then your friend reveals that one of the coins he threw landed heads by uncovering it:
![[image loading]](http://i.imgur.com/fvpdK.png)
or
He points to the covered coin and asks how the other coin landed?
If your friend actually "reveals" which coin is heads, and asks only what the "other" coin is, then the answer is that the other coin is equally likely to be heads or tails, since any coin, considered individually, has those odds (assuming perfect balance and not landing on its edge, etc).
(This is one interpretation, and based on the original phrasing, one could argue that it's the most logical interpretation. I'm working on the other interpretation, the one that involves the "unintuitive" 1/3 heads, 2/3 tails, odds...)
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Wait, the OP actually managed to rhyme "coin" and "toss" together? O.O
On to the topic, yes all seven sentences are valid. Someone point it out if I am wrong though.
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Calgary25968 Posts
1. Whenever two coins are tossed there is a 1/2 chance they both land different side up. True.
TT
TH
HT
HH
2/4
2. Whenever two coins are tossed, and you are told one of them landed on heads, you know there is a 2/3 chance the two coins landed different side up. True.
TT
TH
HT
HH
2/3
3. Whenever two coins are tossed, and you are told one of them landed on tails, you know there is a 2/3 chance the two coins landed different side up. True.
TT
TH
HT
HH
2/3
4. Whenever two coins are tossed, and you are told one of them landed on X, you know there is a 2/3 chance the two coins landed different side up. True.
XX
XY
YX
YY
2/3
5. When someone says he is going to toss two coins, but he promises he will tell you on what side one of them landed, you can before the coins are tossed already tell that there is 2/3 chance they will land different side up. False.
All options are still possible. If it lands TT, he will say "One of them is tails." If it lands HH, he will say "One of them is heads." Because he gets to choose, we can't eliminate any options.
TT
TH
HT
HH
2/4
6. Two coins are tossed 1 000 000 times. Every time two tails land you are told one landed tails. Every time two heads land you are told one landed heads. When its 1 tails, one heads, you are told 50% times one landed heads, other 50% times one landed tails. No matter what, every time you are told which side up one of the coins landed, you will say there is 2/3 chance the two coins landed different side up. The rules on what base are you told what coin landed which side are not told to you. True. This is actually the same as question 3 and 4 but you've worded it strangely.
7. Same as 6, but the rules are told to you. True. Because the rules are the same as question 3 and 4 (fair and random), knowing the rules doesn't change anything.
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Calgary25968 Posts
On August 11 2011 05:56 Hesmyrr wrote:
Wait, the OP actually managed to rhyme "coin" and "toss" together? O.O
On to the topic, yes all seven sentences are valid. Someone point it out if I am wrong though.
No, it was terrible but I opened it because he put in the effort.
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Calgary25968 Posts
To summarize: You are still trying to prove yourself right by coming up with weird scenarios (#5, #6, #7). You can't use probability to predict the future of a truly random event (#5). Despite what your friend says he is going to do, a coin flip is still 50%.
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OP, when someone says he flipped two coins and one of the two was heads (whether or not he tells you which one turned up heads), and asks you how the other coin landed, one interpretation of the question is to consider that "other" coin separately, in which case any coin considered separately has even odds of landing heads or tails, as mentioned many times before (including my own attempt at explaining it above).
However, the other interpretation of the question is that he's asking what the chances are that the other coin landed heads (therefore, since he already said he has one heads, he would end up with two heads), and what are the chances that the other coin landed tails (therefore he has a heads and a tails).
---
Let's play three card monte, but instead of a red queen, we'll use a heads up coin, and we'll use more ESPORTS to cover the coins.
I put three coins on a table, one heads, two tails:
+ Show Spoiler +
I cover them up and rearrange them.
+ Show Spoiler +
Can we all agree that if I point at a covered coin, that it no longer has an equal chance of being heads or tails? You saw one heads and two tails and all I did was cover them up, so even though they're coins, they're not being reflipped, so if you pick one of the three at random, you should have a 1/3 chance of picking the heads and a 2/3 chance of picking the tails.
This is the key to the whole thing. In the above situation, you're no longer calculating whether a coin has landed heads or tails. You're picking from a fixed quantity of heads and tails. This may seem contrived, but let's keep going.
On another table, I have three coins, all heads:
+ Show Spoiler +
I then put one of my three covered coins next to each of the heads, forming three rows of two coins each:
+ Show Spoiler +
Even if I don't know what each covered coin is, I know one row has two heads, and the other two rows have a heads and a tails each.
I then add a row of two coins, both tails:
+ Show Spoiler +
The four rows represent all the possible ways two coins flipped together could end up: HH, HT, TH, TT. Yes, since I had all heads on the left to start, if I flipped over all the coins right now, there would be a "repeat" row, like this:
+ Show Spoiler +
But that doesn't mean you can discount that there's still four separate possibilities for how two coins could land, even with the "repeat" row.
The above is a reverse of how to calculate the 1/3 heads, 2/3 tails odds. You start with the four possible ways two coins could land together, eliminate the TT (no pun intended) since you know at least one of the coins is heads, and see what the odds are that both coins are heads or that the other coin is tails.
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On August 11 2011 06:39 Chill wrote:
To summarize: You are still trying to prove yourself right by coming up with weird scenarios (#5, #6, #7). You can't use probability to predict the future of a truly random event (#5). Despite what your friend says he is going to do, a coin flip is still 50%.
Isn't the OP on the "50%" camp and not on the 1/3, 2/3 side?
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Calgary25968 Posts
On August 11 2011 06:47 aNDRoM wrote:Show nested quote +On August 11 2011 06:39 Chill wrote:
To summarize: You are still trying to prove yourself right by coming up with weird scenarios (#5, #6, #7). You can't use probability to predict the future of a truly random event (#5). Despite what your friend says he is going to do, a coin flip is still 50%. Isn't the OP on the "50%" camp and not on the 1/3, 2/3 side?
Yes. And he's invented scenario #5 to show how you can affect the future with your words. That's my point. Scenario #5, where your friend says "Okay in the future I'll do this" is different than the rest because he hasn't affected anything by saying that. So that case is 1/2. Everything else is 2/3.
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On August 11 2011 06:51 Chill wrote:Show nested quote +On August 11 2011 06:47 aNDRoM wrote:On August 11 2011 06:39 Chill wrote:
To summarize: You are still trying to prove yourself right by coming up with weird scenarios (#5, #6, #7). You can't use probability to predict the future of a truly random event (#5). Despite what your friend says he is going to do, a coin flip is still 50%. Isn't the OP on the "50%" camp and not on the 1/3, 2/3 side? Yes. And he's invented scenario #5 to show how you can affect the future with your words. That's my point. Scenario #5, where your friend says "Okay in the future I'll do this" is different than the rest because he hasn't affected anything by saying that. So that case is 1/2. Everything else is 2/3.
Sorry, got it.
(I think the question is vague enough for interpreting it the way he wants to understand it, which is an issue with a lot of math probability questions, as shown in the boy/girl wikipedia entry.)
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Hey, how about thinking about it like this:
You friend flips two coins, and asks, "What is the probability of the each of the possible results?"
Answer:
HH = 1/4
HT = 1/4
TH = 1/4
TT = 1/4
Now the guy says "I have checked the coins, and I can tell you that the result is not TT. Now, what is the probability of the coin results?"
Answer:
HH = 1/3
HT = 1/3
TH = 1/3
TT = 0
Now the guy points to the first coin and asks, "What is the probability that this coin is heads?"
Answer:
HH + HT = 2/3
This is just rewording the same thing again, but maybe this progression will make more sense to you.
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Ok, this answer makes sense to me in a way but there is still something bothering me.
Lets say I flip two coins.
I check the first coin, and see it is heads. I now know that the first coin landed heads.
That means that there is a 50% chance that the second coin landed heads, and a 50% chance that it landed tails. So if I guess, I have a 50/50 chance of getting it right.
If I then tell my friend that at least one of the two coins landed heads (a true statement), according to the results of this thread, he guesses tails at 66% accuracy.
However, there is no way that his incomplete information gives him better odds than me. In fact, I am certain I must be going about this the wrong way, but just can't see it.
So please tell me what I'm doing wrong.
edit: nvm, I get it. lol I was dumb for a moment there 
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Calgary25968 Posts
He tells you one of the coins is heads, not the first one specifically.
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Wait, I messed myself up again.
So if the situation is this:
I flip 2 coins 100 times. In each case, I look at the first one and guess that the other coin will be whichever the first one wasn't. Because I know the first coin is X, then not X will have 50% likely and I will be correct approx. 50 times.
Also in each case, I tell my friend that at least one coin is X, so he guesses that the other coin will be not X. By the logic of the thread, he should be correct at a 66% rate, but he will in fact be correct approx. 50 times (same as me, we have the same guess).
Explain to me why this is, and why it does not show that the percentage should in fact be 50, not 66?
Actually, I guess it is just that the way you guys are doing it, out of the 100 cases we would only play those that landed HT, TH, HH, and we wouldn't play TT at all. By playing all four options and switching between H or T for X we ruin it.
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Calgary25968 Posts
There's 7 pages explaining that...
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You got it with your last paragraph for the second scenario. Even though you flip two coins 100 times, the game you're playing with your friend would only include 75 of those times, because the other 25 flips would be when you get "not X" for both coins.
For the first scenario, you're not excluding any of the 100 times the coins are flipped. Once again, assuming perfect coins:
Coin One is X on 50 flips. You guess "not X" on those flips. Coin 2 will be X on 25 of those 50 flips and "not X" on the other 25 flips. You'll be correct 50% of those 50 flips.
Coin One is "not X" on the other 50 flips. You guess X on those flips. Coin 2 will be "not X" on 25 of those 50 flips and X on the other 25 flips. You'll be correct 50% of those 50 flips.
So you'll be 50% correct on all 100 flips. Using the result of the first coin to guess the opposite result for the second coin is as arbitrary a way of choosing heads or tails as flipping a coin... Wait.
Edit: So Chill, do you know if the OP has come back and been convinced by any of the new explanations?
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Poorly worded conditional probability problems are usually tough to understand, yes.
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On July 16 2011 14:25 hypercube wrote:Show nested quote +On July 16 2011 12:28 ]343[ wrote:On July 16 2011 11:57 hypercube wrote:On July 16 2011 11:45 ]343[ wrote:Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?) On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed. That's a huge assumption. What you're doing is that you are rewriting the problem, introducing new information and then go on to prove that the answer is 2/3. It is in your problem. But it's a different problem, so it doesn't help us. I'm pretty sure "at least one of the coins turns up heads" is mathematically precise. It means that both are not tails. It's also not from the OP. "Then your friend reveals that one of the coins he threw landed heads." This is the problem statement. The interpretation that he could have said "one of the coins landed tails" if he got TH or HT is reasonable. Others agree with me, read the boy or girl paradox article on wikipedia if you don't believe me.
I read your post about thirty times, and wavered between whether I should agree with you.
My conclusion: no, this problem has nothing to do with ambiguous protocol.
The problem states one specific instance with a defined protocol, and asked for the probability of the other coin for this particular instance. It seems to me that you have taken game theory, so in game theory terms, the state space is well defined.
This is different from the hypothetical future where your friend would be throwing coins. He already threw the coins. Also, your claim that he could reveal a T is also false because the only ambiguity lies within the case of what would happen if both coins land T. The friend would never reveal T if the coins land HT or TH.
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On August 11 2011 06:29 Chill wrote:Show nested quote +1. Whenever two coins are tossed there is a 1/2 chance they both land different side up. True. TTTHHTHH2/4 Show nested quote +2. Whenever two coins are tossed, and you are told one of them landed on heads, you know there is a 2/3 chance the two coins landed different side up. True. TTTHHTHH2/3 Show nested quote +3. Whenever two coins are tossed, and you are told one of them landed on tails, you know there is a 2/3 chance the two coins landed different side up. True. TTTHHTHH2/3 Show nested quote +4. Whenever two coins are tossed, and you are told one of them landed on X, you know there is a 2/3 chance the two coins landed different side up. True. XXXYYXYY2/3 Show nested quote +5. When someone says he is going to toss two coins, but he promises he will tell you on what side one of them landed, you can before the coins are tossed already tell that there is 2/3 chance they will land different side up. False. All options are still possible. If it lands TT, he will say "One of them is tails." If it lands HH, he will say "One of them is heads." Because he gets to choose, we can't eliminate any options. TTTHHTHH2/4 Show nested quote +6. Two coins are tossed 1 000 000 times. Every time two tails land you are told one landed tails. Every time two heads land you are told one landed heads. When its 1 tails, one heads, you are told 50% times one landed heads, other 50% times one landed tails. No matter what, every time you are told which side up one of the coins landed, you will say there is 2/3 chance the two coins landed different side up. The rules on what base are you told what coin landed which side are not told to you. True. This is actually the same as question 3 and 4 but you've worded it strangely. True. Because the rules are the same as question 3 and 4 (fair and random), knowing the rules doesn't change anything.
All seems to be good here except for your interpretation of 6, which has stranger rules than I think you realize.
Suppose you are told that one of the coins is heads. This will happen 50% of the time. 25% of the time the coins will land HH, every time of which you will be told that one coin landed heads. So 25% of all cases will be HH in which you are told one landed heads.
50% of the time (overall, not just in cases told heads) it will land HT or TH, and for each you will be told 50% of the time that one landed heads. So, 25% of all cases will be a HT or TH in which you are told that one of the coins landed heads.
Therefore, exactly half of all cases in which you are told that one of the coins is heads will be cases in which the other coin is heads. There are twice as many rolls in which the two land differently, but these are reported as one landing heads half as often as when both land heads, evening the two situations out.
So in scenario 6, being told one coin landed heads (or tails) doesn't actually give you any additional information. Since you aren't told the rules, it might still be rational for you to bet as though it did give you the usual additional information, but that seems tangential to the point of the thread.
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On August 14 2011 08:18 frogrubdown wrote:Show nested quote +On August 11 2011 06:29 Chill wrote:1. Whenever two coins are tossed there is a 1/2 chance they both land different side up. True. TTTHHTHH2/4 2. Whenever two coins are tossed, and you are told one of them landed on heads, you know there is a 2/3 chance the two coins landed different side up. True. TTTHHTHH2/3 3. Whenever two coins are tossed, and you are told one of them landed on tails, you know there is a 2/3 chance the two coins landed different side up. True. TTTHHTHH2/3 4. Whenever two coins are tossed, and you are told one of them landed on X, you know there is a 2/3 chance the two coins landed different side up. True. XXXYYXYY2/3 5. When someone says he is going to toss two coins, but he promises he will tell you on what side one of them landed, you can before the coins are tossed already tell that there is 2/3 chance they will land different side up. False. All options are still possible. If it lands TT, he will say "One of them is tails." If it lands HH, he will say "One of them is heads." Because he gets to choose, we can't eliminate any options. TTTHHTHH2/4 6. Two coins are tossed 1 000 000 times. Every time two tails land you are told one landed tails. Every time two heads land you are told one landed heads. When its 1 tails, one heads, you are told 50% times one landed heads, other 50% times one landed tails. No matter what, every time you are told which side up one of the coins landed, you will say there is 2/3 chance the two coins landed different side up. The rules on what base are you told what coin landed which side are not told to you. True. This is actually the same as question 3 and 4 but you've worded it strangely. 7. Same as 6, but the rules are told to you. True. Because the rules are the same as question 3 and 4 (fair and random), knowing the rules doesn't change anything. All seems to be good here except for your interpretation of 6, which has stranger rules than I think you realize. Suppose you are told that one of the coins is heads. This will happen 50% of the time. 25% of the time the coins will land HH, every time of which you will be told that one coin landed heads. So 25% of all cases will be HH in which you are told one landed heads. 50% of the time (overall, not just in cases told heads) it will land HT or TH, and for each you will be told 50% of the time that one landed heads. So, 25% of all cases will be a HT or TH in which you are told that one of the coins landed heads. Therefore, exactly half of all cases in which you are told that one of the coins is heads will be cases in which the other coin is heads. There are twice as many rolls in which the two land differently, but these are reported as one landing heads half as often as when both land heads, evening the two situations out. So in scenario 6, being told one coin landed heads (or tails) doesn't actually give you any additional information. Since you aren't told the rules, it might still be rational for you to bet as though it did give you the usual additional information, but that seems tangential to the point of the thread.
Not sure why I said all of the others are fine. The problem with 6 affects 7 in a fairly self evident way. Also, I think I can make my point clearer with some help from Reverend Bayes. Let A=You are told that there is at least one heads. The priors are as follows.
P(A)=0.5 (we are given this in scenario 6)
P(HH)=P(HT)=P(TH)=P(TT)=0.25 (that's how coins work)
P(A|HH)=1 (we are given this)
P(A|HT)=P(A|TH)=0.5 (same)
P(A|TT)=0 (same)
By Bayes' Theorem:
P(HH|A)=P(A|HH)*P(HH)/P(A)=1*0.25/0.5=0.5
P(HT|A)=P(A|HT)*P(HT)/P(A)=0.5*0.25/0.5=0.25
P(TH|A)=P(A|TH)*P(TH)/P(A)=0.5*0.25/0.5=0.25
Therefore, P(HH|A)=P(HT|A) + P(TH|A). That is, the probability that the other coin is heads given that you are told one coin is heads is equal to the probability that the other coins is tails given that you are told one is heads.
Please, no one comment on this point without reading what it is a response to and recognizing that the hypothetical is a slightly different one than the one in the OP.
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On August 11 2011 14:24 Cambium wrote:Show nested quote +On July 16 2011 14:25 hypercube wrote:On July 16 2011 12:28 ]343[ wrote:On July 16 2011 11:57 hypercube wrote:On July 16 2011 11:45 ]343[ wrote:Hi dudes, I'd like to quote myself from earlier in the thread (in case you didn't read it?) On July 16 2011 03:12 ]343[ wrote:
OP: unfortunately your "counterexample" is different from the original problem. Instead, it would go like this:
You have two fair coins, each with a heads and a tails side. You aren't allowed to flip 2 tails: that is, whenever you flip two tails, flip both again. The probability you will get 2 heads now is indeed 1/3. This is what it means when we're saying "at least one of the coins turns up heads": the possibility TT is thrown out. Nothing else has changed. That's a huge assumption. What you're doing is that you are rewriting the problem, introducing new information and then go on to prove that the answer is 2/3. It is in your problem. But it's a different problem, so it doesn't help us. I'm pretty sure "at least one of the coins turns up heads" is mathematically precise. It means that both are not tails. It's also not from the OP. "Then your friend reveals that one of the coins he threw landed heads." This is the problem statement. The interpretation that he could have said "one of the coins landed tails" if he got TH or HT is reasonable. Others agree with me, read the boy or girl paradox article on wikipedia if you don't believe me. I read your post about thirty times, and wavered between whether I should agree with you. My conclusion: no, this problem has nothing to do with ambiguous protocol. The problem states one specific instance with a defined protocol, and asked for the probability of the other coin for this particular instance. It seems to me that you have taken game theory, so in game theory terms, the state space is well defined. This is different from the hypothetical future where your friend would be throwing coins. He already threw the coins. Also, your claim that he could reveal a T is also false because the only ambiguity lies within the case of what would happen if both coins land T. The friend would never reveal T if the coins land HT or TH.
I don't really want to go into too much detail, but I want to point out that your friend could hustle you by offering his $2 against your $3 in a bet to guess the second coin. Naively this bet has a positive expectation of (2/3)*2-(1/3)*3=1/3 dollars per bet.
But in reality, even using fair coins, your friend can manipulate the frequency of heads to exactly 50%. He could do it in a variety of ways, including by alternating saying "one of the coins is heads" and "one of the coins is tails" or just by not offering the bet at all after looking at the coins. In this case your expectation would be negative, of course.
edit: You could of course disallow all of these actions under the terms of the bet. My point is that you would have to be very specific to avoid getting hustled.
In the end some people think the OP implies that the friend will allways say "one of the coins is heads" when it is. Maybe. I don't think there's much point in debating it, as long as it's clear where the disagreement lies. Which it is in your case.
It's not a disagreement of logic or math, just interpretation.
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On August 11 2011 06:29 Chill wrote:Show nested quote +6. Two coins are tossed 1 000 000 times. Every time two tails land you are told one landed tails. Every time two heads land you are told one landed heads. When its 1 tails, one heads, you are told 50% times one landed heads, other 50% times one landed tails. No matter what, every time you are told which side up one of the coins landed, you will say there is 2/3 chance the two coins landed different side up. The rules on what base are you told what coin landed which side are not told to you. True. This is actually the same as question 3 and 4 but you've worded it strangely. True. Because the rules are the same as question 3 and 4 (fair and random), knowing the rules doesn't change anything.
You are mistaken. If you accept the bolded part (in the first spoilered quote) as true the correct answer becomes 1/2. Most people who argue for 2/3 don't accept it and implicitly or explicitly assume that for one heads one tails you are always told "that one of the coins he threw landed heads".
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This thread pops up like every year I swear. Always ends up 20 pages long, can't you just search?
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EPT![[image loading]](http://i.imgur.com/jgPw3.png)
