Boy or girl puzzle
January 14th, 2014 at 10:35:21 PM
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Quote: MathExtremistLet's ask it another way. If you go around to every parent you see in a park with a single child, and find out which of those parents have exactly two children, what is the probability of the child at home of the two-children parents being the same gender as the child in the park?
I see, you're insisting it's 1/3.
Well, you're wrong: it's still 50/50, for both problems. (The park and the OP that is, not your coin game, which is 1/3.)
Again, in your example, I wasn't going to be shown a coin unless it had come up tails. In all three worlds, that coin I saw had a 100% probability of being tails. I have been given no reason to think the children in the park couldn't have been the opposite sex. I can say after the fact there was a 100% probability they weren't - in fact, that's sort of my starting point - but I have to consider how likely it had been, within the parameters of the three possible worlds under consideration, and the fact that it was less likely in two of them makes those two worlds less likely. If the park allowed only little girls and their parents, then I would say that it was more likely the parents in question had a boy at home, but you said no such thing.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
January 14th, 2014 at 11:50:51 PM
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It's 1/2.
If BG = GB, then it's 1/2
If BG =/= GB, then we would have to add in an extra variable of BB because if BG =/= GB, then a "timeline" has occured. Which means B1B2 =/= B2B1.
If BG = GB, then it's 1/2
If BG =/= GB, then we would have to add in an extra variable of BB because if BG =/= GB, then a "timeline" has occured. Which means B1B2 =/= B2B1.
January 17th, 2014 at 6:13:25 AM
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OK, for those who cannot be convinced by pure mathematical proof, please resort to experimental proof.
Experiment A : launch a series of ordered pairs of coin-tosses (for example, a €1 coin and a €.50 coin). Select those where the €1 is tails. How many (on average) have the €0.50 exhibiting tails? Observed: about half of them.
Experiment B : launch a series of unordered pairs of coin-tosses. Select one coin in the pair at random. Set apart those where that coin is tails. How many (on average) have their counterpart exhibiting tails? Observed: about two thirds of them.
Please try it physically. You will be convinced that it is true. Then go back to the theoretical arguments to understand why it is so.
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It can never be repeated too often: probability is about the amount of information, not about the state of the world.
Experiment A : launch a series of ordered pairs of coin-tosses (for example, a €1 coin and a €.50 coin). Select those where the €1 is tails. How many (on average) have the €0.50 exhibiting tails? Observed: about half of them.
Experiment B : launch a series of unordered pairs of coin-tosses. Select one coin in the pair at random. Set apart those where that coin is tails. How many (on average) have their counterpart exhibiting tails? Observed: about two thirds of them.
Please try it physically. You will be convinced that it is true. Then go back to the theoretical arguments to understand why it is so.
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It can never be repeated too often: probability is about the amount of information, not about the state of the world.
Reperiet qui quaesiverit
January 17th, 2014 at 11:16:21 AM
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Quote: kubikulannExperiment B : launch a series of unordered pairs of coin-tosses. Select one coin in the pair at random. Set apart those where that coin is tails. How many (on average) have their counterpart exhibiting tails? Observed: about two thirds of them.
I dare you to write a simulation of this. If I were the type to enter into wagers via fora, I would bet you a pretty penny you'd find you were wrong.
(Or just to scribble it out: say you flip 20,000 pairs, you'll get, on average, 5k HH, 5k TT, 10k HT. Pick one at random from each, and you'll be left with all the TT, but only half the HT. I think you can take it from there.)
Information, by definition, is all about the state of the world.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
January 17th, 2014 at 1:08:54 PM
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Well, I'm sold.
I whipped up a quick program that tested it, as I understand it. It functions as per the following: Flip two coins. If they're both tails, toss it aside. (But record it for the sake of recordkeeping.) Were they both heads? If so, then HH. Otherwise, since we've already established that there was at least one head, and they're not both, one of them has to be tails-- HT.
Yes, my coding is awful. Ignore the two numbers beneath; those were there as an error check.
(And for funsies, I ran the same thing, except one of the values was locked to H. (or, well, .1): It comes out 50/50.)
I whipped up a quick program that tested it, as I understand it. It functions as per the following: Flip two coins. If they're both tails, toss it aside. (But record it for the sake of recordkeeping.) Were they both heads? If so, then HH. Otherwise, since we've already established that there was at least one head, and they're not both, one of them has to be tails-- HT.
Yes, my coding is awful. Ignore the two numbers beneath; those were there as an error check.
(And for funsies, I ran the same thing, except one of the values was locked to H. (or, well, .1): It comes out 50/50.)
January 17th, 2014 at 1:22:47 PM
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That's not the problem the OP described, nor the problem just described. He said pick one randomly, and if it's tails, keep that pair. You're always picking tails if it's possible to do so. That's the distinction people in this thread are missing.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
January 17th, 2014 at 1:31:55 PM
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Late to the party, but here was my take based on my reading of the problem. Also my take on where the discrepancy in approaches arises.
Consider four (two parent) families 1,2,3,4 that each have had two children BB, BG, GB, GG respectively (order of children as shown). Based on the premise of the problem these are equally likely. Let's associate parent 1 with child 1 and parent 2 with child 2. On Sunday, one parent and child from each family go out for a walk in the park, the other parent and child remain at home. You meet a parent with a male child, What is the probability that you met family 1? I am in the 1/3 camp. It can't be family 4, and the other 3 families are equally represented in the park.
This isn't different than others have said (and I liken it to MathExtremist's penny quarter description).
I present it this way to offer what I think the "1/2 camp" is saying
What I take them to be suggesting is that I could have met with equal possibility
-family 3 (parent 2 with B2)
-family 2 (parent 1 with B1)
-family 1 (parent 1 with B1)
-family 1 (parent 2 with B2)
i.e. that there are two ways to meet family 1.
But as one of the parent and child from each family is known to be at home, that to me, doesn't seem to be the case. There are 3 families in the park and only one is family 1.
Or have I changed the problem by having 4 famlies instead of 1 with four possibilities???
This isn't different than others have said (and I liken it to MathExtremist's penny quarter description).
I present it this way to offer what I think the "1/2 camp" is saying
What I take them to be suggesting is that I could have met with equal possibility
-family 3 (parent 2 with B2)
-family 2 (parent 1 with B1)
-family 1 (parent 1 with B1)
-family 1 (parent 2 with B2)
i.e. that there are two ways to meet family 1.
But as one of the parent and child from each family is known to be at home, that to me, doesn't seem to be the case. There are 3 families in the park and only one is family 1.
Or have I changed the problem by having 4 famlies instead of 1 with four possibilities???
January 17th, 2014 at 5:50:16 PM
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I think we're well past spoilers at this point.
What you're missing is that although the families are equally represented in the park, it could have equally likely been the other parent in the park, so the two parents from family 1 are no less likely, just because one of them (and you don't know which) could have stayed home. It was certain that there would be in that park a little boy from family 1, whereas from families 2 and 3 there may or may not have been one, so while meeting this boy totally excludes only 4, 1 becomes a bit more likely, since you knew 1 was going to bring a boy. Information doesn't work like a light switch.
Let's scale it again: families 2 and 3 each have six daughters and a son, whereas 1 has only boys. Which family do you think the boy you've met is most likely to be from now?
What you're missing is that although the families are equally represented in the park, it could have equally likely been the other parent in the park, so the two parents from family 1 are no less likely, just because one of them (and you don't know which) could have stayed home. It was certain that there would be in that park a little boy from family 1, whereas from families 2 and 3 there may or may not have been one, so while meeting this boy totally excludes only 4, 1 becomes a bit more likely, since you knew 1 was going to bring a boy. Information doesn't work like a light switch.
Let's scale it again: families 2 and 3 each have six daughters and a son, whereas 1 has only boys. Which family do you think the boy you've met is most likely to be from now?
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
January 18th, 2014 at 6:42:00 AM
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Caution! That is a different question.Quote: 24BingoLet's scale it again: families 2 and 3 each have six daughters and a son, whereas 1 has only boys. Which family do you think the boy you've met is most likely to be from now?
Reperiet qui quaesiverit
January 18th, 2014 at 11:22:06 AM
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It isn't, not really - or it is, but it suits its purpose. chevy's argument is predicated on the idea that since there's one child of each family in the park, the child is equally likely to have come from any of them. What practically everyone in this thread is missing is that the fact that the child is a boy is evidence as to which of the three families he comes from, as well as totally excluding the fourth. People are missing it because it's not a slam-dunk as with 4, so I've made it not quite a slam-dunk, but overwhelming enough that people can see it. In the problem I gave, that boy has a 7/9 chance of being from the all-boy family, for the exact same reason he has a 1/2 chance of being from the two-boy family in chevy's scenario.
I'm just waiting for some acknowledgment that your "experiment B" was as much a failure as might be "proving" two and two to be five.
I'm just waiting for some acknowledgment that your "experiment B" was as much a failure as might be "proving" two and two to be five.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
March 7th, 2014 at 10:19:50 AM
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Sorry that this is disjointed, i thought i was posting after an earlier point in the thread. i'm new to this and will learn how to quote-the post i am responding to. Anyhow you all should get the gist of what i am thinking and or asking.
I f you get 50% when you know it is the eldest son, the same can be calculated if you knew it was the youngest son ( you would rule out GG and BG in this case).
So why do you even need that info, its already there. Without being told, I know that the chances of a boy being either the eldest or the youngest out of two children is 100%. Both the eldest and the youngest son solutions end in the 50% not the 33.33% chance.
In other words how can you not have the information you need to get to 50%.
I f you get 50% when you know it is the eldest son, the same can be calculated if you knew it was the youngest son ( you would rule out GG and BG in this case).
So why do you even need that info, its already there. Without being told, I know that the chances of a boy being either the eldest or the youngest out of two children is 100%. Both the eldest and the youngest son solutions end in the 50% not the 33.33% chance.
In other words how can you not have the information you need to get to 50%.
March 9th, 2014 at 8:52:26 AM
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Perhaps this reasoning will help: consider two tossed coins with the outcomes hidden under cups. You are guaranteed at least one of the coins is "heads." I hope you will find it easily reasoned that the chance of both being "heads" is 1/3. (HH, HT, TH) A cup is removed and you are shown that one of the coins is "heads." Obviously, the probababiliy that both are "heads" is then 1/2. The difference is not knowing the result of either coin and knowing the result of one coin.
Edit: spelling
Edit: spelling
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
March 9th, 2014 at 9:14:31 AM
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Hi Doc, I read your response last night, posted to the "Mr. Smith" thread. I meant to post to that thread, but it sank below the recent posting list and I found this thread on a search. Yes, your answer is spot-on. I just thought I would present something based more on intuitive reasoning that rigorous proof.Quote: DocI posted the answer to that last night in the other thread you started on this same topic.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
