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Coins tosses - Page 8

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Enhancer_
Profile Joined May 2011
Canada320 Posts
August 11 2011 04:58 GMT
#141
Poorly worded conditional probability problems are usually tough to understand, yes.
Cambium
Profile Blog Joined June 2004
United States16368 Posts
August 11 2011 05:24 GMT
#142
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.
When you want something, all the universe conspires in helping you to achieve it.
frogrubdown
Profile Blog Joined June 2011
1266 Posts
August 13 2011 23:18 GMT
#143
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.

TT
TH
HT
HH

2/4

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.

TT
TH
HT
HH

2/3

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.

TT
TH
HT
HH

2/3

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.

XX
XY
YX
YY

2/3

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.

TT
TH
HT
HH

2/4

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.

Show nested quote +
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.
frogrubdown
Profile Blog Joined June 2011
1266 Posts
Last Edited: 2011-08-14 01:40:07
August 14 2011 01:39 GMT
#144
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.

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.


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.

hypercube
Profile Joined April 2010
Hungary2735 Posts
Last Edited: 2011-08-14 02:29:16
August 14 2011 02:26 GMT
#145
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.
"Sending people in rockets to other planets is a waste of money better spent on sending rockets into people on this planet."
hypercube
Profile Joined April 2010
Hungary2735 Posts
Last Edited: 2011-08-14 02:55:46
August 14 2011 02:54 GMT
#146
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.

Show nested quote +
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.


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".
"Sending people in rockets to other planets is a waste of money better spent on sending rockets into people on this planet."
SpoR
Profile Blog Joined November 2010
United States1542 Posts
August 14 2011 13:00 GMT
#147
This thread pops up like every year I swear. Always ends up 20 pages long, can't you just search?
A man is what he thinks about all day long.
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