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Sigh, believe it or not, its not the same.
Let us suppose there were 3 genders. Male Female and Gaymale.
If the first child is male, the second can still be Male, Female or Gaymale
In the doors question there are 3 doors. When you open one it eliminates that possibility.
Another analogy:
You have a bag of red and blue marbels, let us say that their number is unlimited. You pick 1 marble and it is red. You pick marble 2, 50% red, 50% blue.
If on the other hand you had a bad of red marbles containing 1 red marble and 1 blue marble. You pick 1 marble and its red. You pick marble 2, 100% chance blue.
That's the difference between gender and door riddles.
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On November 22 2004 06:22 KarlSberg~ wrote:
You have the right to be wrong but just in case you want to know the result of your own riddle.
There are 8 possible situations (all of which are as likely to happen).
BB and the 1st boy answers
BB and the 2nd boy answers
BG and the boy answers
BG and the girl answers
GB and the girl answers
GB and the boy answers
GG and the 1st girl answers
GG and the 2nd girl answers
Of these 8, we can get rid of 4 (because we know a girl answered)
the remainings are:
BG and the girl answers
GB and the girl answers
GG and the 1st girl answers
GG and the second girl answers
I've already explained this several times. Your problem is you act like those 8 possibilities are equally likely because you are working from the point of view of trying to make G1/G2 options equally likely in all cases. The point is that G1 at door from GG happens a lot less frequently than G1 at door from GB.
Let me try to explain with Fam's coinflip problem (altho I think he might have mangled the example previously, thus confusing things).
Let's say I flip 100 coins, and I know the first 99 are tails. Then I ask, what is the probability the 100th coin is hands? Obviously the correct answer is %50 since events are independent. This is however NOT what's going on in the Boy/Girl problem. A similar question would be to say, "my first child is a girl, what is the probability my second child is a boy?"
From a coinflip perspective, this is what IS happening. I flip a 100 coins and I observe that within all the tosses, there are at least 99 tails (I don't know which 99 happen to be the tails).
Now let's look at the possibilities (with the current amount of information we have).
All of these are equally likely.
HTTTTT ...... TTT
THTTTT ...... TTT
TTHTTT ...... TTT
...
TTTTTT ...... HTT
TTTTTT ...... THT
TTTTTT ...... TTH
TTTTTT ...... TTT
What is key here, is notice that I don't count the last case 100 times. The chance of getting all tails is the same chance as getting a head in the first flip or a head in the 20th flip or a head in the 87th flip.
Now I ask, having removed 99 tails (notice in the last case I have a choice of whichever 99 I wish) what are the chances that the remaining coin flip is a tail?
So what the 50/50 people are trying to argue, is that from the TTTTT ... TTT case, I could have selected the first 99 flips, the last 99 flips, or any other combination. That is fine. It's just that each of these combinations occur 1 in 100 times when the TTTTT ..... TTT option occurs which itself is 1 in 101 times. However, selecting the last 99 flips happens %100 of the time the HTTTT ... TTT option occurs, which itself also occurs 1 in 101 times. This is why it doesn't make sense to pretend like having selected the 99 last flips from TTTTT .... TTT happens as often as selecting the 99 last flips from HTTTTTT .... TTT.
This is what the Boy/Girl problem is equivalent too.
Here's an easier coinflip situation which hopefully makes things clearer. Have a friend flip a coin 10 times but don't look at the results. Now your friend says, "there are at least 5 tails and I am going to remove them". You now have to make a bet on H/T for your choice of one of the remaining 5 coins. Notice that there are lots of possible ways to get 5 heads and 5 tails or 4 heads and 6 tails. There are much fewer possible ways of getting 1 head and 9 tails or 0 heads and 10 tails. What removing the 5 tails has done, is just improved the odds that whatever is left is heads. So you'd obviously be better off betting that something left over is heads.
The KEY is not knowing which 5 of the 10 coins happens to be tails. Just as we don't know that the girl answering the door is an older sister or younger sister. Don't make the mistake of believing that just because G1 from GG is just as likely as G2 from GG, that each of these is equally as likely as G1 from GB. The total GG probability has to occur just as often as the GB and BG probabilities combined. So G1 from GG will only get a fraction of the allowed 1/3 GG while G1 from GB gets all of the allowed 1/3 GB.
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i like this karlsberg guy
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Moltke, the problem is that saying the first child is male is NOT the same as saying as one of the children is male.
Fam, no need to be an asshole about this. It's a hard logical problem.
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Back to this again?
Famouzze reply to this one, it's from around page 12 and I think nails the problem simply. Refute it so I can see where you are misunderstanding.
On November 21 2004 23:46 RoTaNiMoD wrote:
I haven't read all the heated debate regarding this, but Famouzze's riddle is worded wrong. Thus it really is 1/2 boy instead of 2/3.
Famouzze the problem with your argument is that, despite BG and GB being twice as likely as GG, GG needs to be counted twice as a possibility. BG and GB each have only one girl that could answer the door, whereas GG has G1 (first girl) and G2 (second girl) that could answer the door. So really the other child is a boy the two times it is BG or GB, and a girl the two times it is G1 or G2. So 2/4 of the times, or 50%, the other child is a boy!
----YES, the answer is SUPPOSED to be 2/3 in the properly worded riddle, I have seen it and fully understand that answer. That being said, this one IS worded wrong and does not have a boy twice as likely as a girl.
[edit: Karlsberg posted this almost exactly just a few posts ago, he's smart. listen to him]
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On November 21 2004 20:42 Famouzze wrote:
ok i will just post one more, this is the second hardest riddle i know of and has been the subject of much debate among my friends (some of them still won't accept the truth of the right answer)
Let's say you're friends with a woman, and you know she has two children, but you don't know the gender of either child. Then one day you go over to her house and a girl answers the door and says my mom will be right down. So you now know one of her children is a girl, what are the chances that the other child is a boy?
facts:
2 children.
unknown genders.
possible genders: male / female
[THIS IS WHERE PEOPLE ARE GETTING CONFUSED]
a son is going to be called "x". a daughter is going to be called "y"
so at this point the woman could have a:
xy (she gave birth to a son, and then she gave birth to a daughter)
xx(she gave birth to a son, and then another son)
yy(she gave birth to a daughter, and then another daughter)
yx(she gave birth to a daughter, and then a son)
some people are claiming yx and xy are the same thing. they are not the same thing.
pretend the birth of the children are quarters being flipped. with 2 quarters, respectively, the total sides are:
heads, tails, tails, heads
the different possible outcomes include:
heads, heads
tails, tails
heads, tails
tails, heads
heads, tails and tails, heads are two different things because they are traits that exist in 2 different quarters. it is the exact same with the children.
famouzze disprove anything I have said thus far. use logic please.
ok to continue on. you find out 1 child is a girl. so you now know she gave birth to:
y*
looking at the total possibilities, the choices are:
yy
and
yx
so 1 out of 2 times it is yy. 1 out of 2 times it is yx. thats 50/50
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<3 travis, karlsberg, catyoul
<3 simple logic
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short version:
2 items exist. the items can possibly be "x" or "y"
you find out the first item is "x"
what is the 2nd item?
ok well checking the list of possibilities, the 2nd item can be "x" or "y"
so 50% of those are x. 50% of those are y.
therefore 50% of the time the 2nd item is x. 50% of the time the 2nd item is y.
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famouzze is corbalt?
if so it all makes sense now.
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On November 22 2004 07:23 gg2w wrote:
Moltke, the problem is that saying the first child is male is NOT the same as saying as one of the children is male.
Fam, no need to be an asshole about this. It's a hard logical problem.
I understand what you are saying:
GB 25% G1 Answers 100%
BG 25% G2 Answers 100%
GG 25% G1 Answers 50% G2 Answers 50%
BB 25% 0%
These are the probabilities before you know that a girl answers the door. The second set are based on the probabilities after you know a girl answes the door.
If you want to look at the entire problem BEFORE you know a girl answers a door, it comes down to this. You must confess that there was a 50% chance that in GB and BG that the boy would have answered the door.
GB G1 Answers 50% B2 Answers 50%
BG B1 Answers 50% G2 Answers 50%
GG G1 Answers 50% G2 Answers 50%
BB B1 Answers 50% B2 Answers 50%
Therefore the sum probability of a girl answering the door is 50%. Let us look at the scenarios where these are possible:
GB- 50
BG- 50
GG- 100
Therefore GG does have double the weight.
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rota you only slept like 4 or 5 hours :-(
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Yeah but on a sidenote firepay is letting me deposit again so I can play pokerstars more!
In like the 10 hands I played last night I lost $50 then doubled up, then doubled someone else up, then won an ok pot then ended even.
Like an idiot I just started throwing coins though, just to prove it to myself. I would throw two quarters, check one and then check if the opposite was heads or tails. It felt ridiculous to actually do, but yes the other one was heads or tails 50% of the time regardless of the first.
CHEW ON EXPERIMENTAL EVIDENCE FAM!
[edit: OMG I'M PLAYING WITH REKRUL HWAAAAAAAAAAA]
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Catyoul
France2377 Posts
<3 you too rota ^^
On November 22 2004 06:25 Famouzze wrote:
one thing i noticed this whole thread is the stubborn 50/50 people will only debate by ignoring what the 2/3 people say and then laying down their own train of thjought that the 2/3 people will then respond to. if u really want to discuss it and validate ur opinion i challenge you to really read what i wrote and write a response to it.
I suggest you apply this to yourself. You obviously have not read the discussion after you left, where the mistake in your solution (or in the wording of your problem) has been clearly explained. Rotanimod restated his very clear post above, Karlsberg too, travis too, let me say it in yet another different way just in case (page 8) :
On November 21 2004 22:46 Catyoul wrote:
Actually your mistake is in my bolded statement : "chosen randomly". If you take the two flips/children as a set, the very fact that you saw a girl in the first place should enter your probability calculus. Indeed :
GG, probability 1/4, getting a girl to open the door 1
GB/BG, probability 1/2, getting a girl to open the door 1/2
BB, probability 1/4, getting a girl to open the door 0
With your own words, if you look at the two events together, the probability actually changes. Without knowing the sex of any of the 2 children, there is a 1/4 probability of a girl opening you first with GG, and a 1/2*1/2=1/4 with BG/GB. So, eventually there is an equal chance for GG or BG/GB.
same thing, page 10, but more detailed :
On November 21 2004 23:21 Catyoul wrote:Show nested quote +On November 21 2004 23:00 radar14 wrote:
Hear me out. Given that one of the children is a girl, you have now limited the sample size to families with a girl and a boy, and families with a girl and a girl.
Do you agree with this? I see you coming. Ok, there we go for a basic probability course then. There are basically two ways of considering this problem : 1. the 2 children are 2 separate events. Those events are independent (well, I don't know if it's 100% true in biology, I guess it's pretty close at least but anyway we're working with the 50/50 model of getting a boy and a girl). 2. the 2 children are 1 single event. Choice 1 is the easy way, but Famouzze and you seem to have chosen way 2, up to you, no problem. But do it correctly then. Now, let's sum up what kind of possibilities we have with both of these choices. In each case, the GG scenario has a probability of 1/4 (25%), GB/BG of 1/2 (50%) and BB of 1/4 (25%), that's pretty obvious. 1. Being that the 2 children are 2 separate events, ok I see a girl first. The probability for the other child is completely unrelated, and unchanged, 50%. 2. Ok, so we have one single event we are measuring, and there are different cases with different probabilities. I will put into ( ) the partial (is that the technical term in English for this ?) probability of everything that should be taken into account and that is random. 1- GG (1/4) and we see a girl first (1). Total probability : 1/4 * 1 = 1/4 2- GG (1/4) and we see a boy first (0). TP : 1/4 * 0 = 0 3- GB/BG (1/2) and we see a girl first (1/2). TP : 1/2 * 1/2 = 1/4 4- GB/BG (1/2) and we see a boy first (1/2). TP : 1/2 * 1/2 = 1/4 5- BB (1/4) and we see a girl first (0). TP : 1/4 * 0 = 0 6- BB (1/4) and we see a boy first (1). TP : 1/4 * 1 = 1/4 Any problems with that ? I guess not, since it's still pretty obvious. The sum of the probabilities of these cases is of course 1, since we have covered all cases. Now let's look at what we want : what happens if we get a girl first ? Only in cases 1- and 3- is this possible, and they both have the same probability. So the GB/BG -> girl and GG have the same probability. Hence the 50% chance that the other child is a boy. Notice that I'm not objecting to the fact there are 2* more girl/boy than girl/girl in real world. Just that every other time you will get a boy to open the door in that case, and if you take the 2 children as one single event you just can't discard the probability for that.
and just to be sure, a misunderstanding cleared up with radar14, that yet again explains the point (page 13) :
On November 22 2004 00:10 Catyoul wrote:Show nested quote +On November 21 2004 23:25 radar14 wrote:
Catyoul,
the order in which you see the kids are not the order in which they are born. you don't see the girl "first" You simply know there is a girl, period. Ok, back. Actually in my post I was answering Famouzze's problem. The wording is critical here and yours is different. Let's reformulate it into a question to see the information we get : Your question : "Is there a girl ?" If the answer is yes, then move over, the probability of a boy is 2/3, no questions asked you're right. Famouzze : "Show me one of your 2 children" If you see a girl, the probability of the second one being a boy is still 1/2, refer to my previous posts for that. I think we can agree on that radial right ? Now on to read and comment baal's post 
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Catyoul
France2377 Posts
On November 22 2004 07:20 gg2w wrote:
From a coinflip perspective, this is what IS happening. I flip a 100 coins and I observe that within all the tosses, there are at least 99 tails (I don't know which 99 happen to be the tails).
This is what the Boy/Girl problem is equivalent too.
No that's not equivalent to his wording of the problem at least. Refer to my previous answer to radar14 :
Ok, back. Actually in my post I was answering Famouzze's problem. The wording is critical here and yours is different. Let's reformulate it into a question to see the information we get :
Your question :
"Is there a girl ?" If the answer is yes, then move over, the probability of the other one being a boy is 2/3, no questions asked you're right.
Famouzze :
"Show me one of your 2 children"
If you see a girl, the probability of the second one being a boy is still 1/2, refer to my previous posts for that.
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Odds of having a girl/girl.
1/2 * 1/2 = 25%
Odds of girl/girl with a girl answering the door = 25%
Odds of having a boy/girl.
1 * 1/2 = 50%
Odds of having a boy/girl with a girl answering the door = 25%
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haha famouzze thinks we spent 33 pages discussing his problem
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I'll concede. The relief of not trying to rack my brain over this far outweighs the agony of being wrong.
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Catyoul
France2377 Posts
On November 22 2004 09:09 radar14 wrote:
haha famouzze thinks we spent 33 pages discussing his problem
hehe
not surprising since he didn't read it, he can't imagine the marvelous riddles hidden inside
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I have not read through the entire thread. But i will shed some light on the question posed by radar14 in around the 10th page.
There are a ton of applications to coding theory with problems like these. Let's suppose there are 3 people and each are wearing a blue or a red hat. You can't see your own hat, but you can see other people's hats.
To win the game, at least one person out of the 3 must guess the color of his hat correctly and nobody may guess wrong.
You can win this game 75% of the time. The strategy is, if you see 2 hats of the same color, guess the opposite. possible combinations of hats are as follows:
RRR
RRB
RBR
BRR
BBR
BRB
RBB
BBB
The only way you lose is if the combination is RRR or BBB. So, this game can be won 75% of the time (6 ouf of 8).
This same type of logic can be applied to radars problem.
As for the childbirth question, ill shed some light on that too.....
The answer is 50%. It is NOT ISOMORPHIC TO THE DOOR PROBLEM. Baal, you CANNOT use the same logic for each question.
Every math major ive seen in this thread has taken that stance as well.
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EPT
