You see me out with one of my two children, who happens to be a boy.

What's the chances of the other child at home also being a boy?

The only info you have and are allowed to know is-

I have exactly two children, the boy with me and another child at home.

Assume a world wide boy/girl birth ratio of 50/50.

It's not a trick question, there's no play on words or any other shenanigans involved.

Quote:rudeboyoi25%. Can be BB GG BG GB. Only 1 out of 4 combos contain two boys.

Can't be GG, because you know one is a boy for sure.

But...

Anyways, any correlation can be ruled out for the purpose of this puzzle.

Sorry about that :)

We eliminate GG for obvious reasons.

BUT... we eliminate GB also, because putting the child brought with you first is as good an ordering as any, it being the only thing we have to differentiate the two.

This isn't the Three Prisoner problem. You've given us no information.

Quote:JeepsterWell rudeboyoi as you rounded down your answer I'll have to give it to the Wizard for giving the first correct answer.

Sorry about that :)

Lol I can't do anything right today apparently.

The only other info I gave you was I had two children.

Therefore you cannot eliminate GB

BG, GB, and BB are all valid and equal possibilities

If I have a dollar and a quarter and you have a quarter and a dollar, we have the same thing!

Quote:mwalz9BG and GB are the same thing!

Sorry but they are not.

When listing the probabilities the order of occurrence must be factored in.

It may not be intuitive, nevertheless it's important that it's done correctly.

An example, 4 x 2 coin tosses

HH, TT, HT, TH are the 4 possible outcomes

TH and HT are not the same thing, they are separate outcomes

With the assumption of a perfect 50/50 ratio, the known state is irrelevant-- there's no bearing on the sex of one child to the other so, for the purposes of the question, that child may as well not exist. What remains is a 50/50 chance of the unknown child being a boy or a girl.

Way I see it, the "trick" to this question is in the wording of "also being a boy", which seems to suggest the odds of BB occurring out of all possible permutations, rather than just out of the combinations where a B already exists. Assuming the two values were known, you'd have BB, BG, GB, GG. BG/GB are functionally identical in this question and can be merged, leaving BB, BG/GB, GG. Of these three, GG can be eliminated as one of the values is confirmed to be B. That leaves BB and BG/GB, or a 50% chance.

Phrasing it in another form:

I just flipped a (perfectly random 2-sided) coin that came up heads. What are the odds of the next flip coming up heads?

Quote:Venthus

I just flipped a (perfectly random 2-sided) coin that came up heads. What are the odds of the next flip coming up heads?

Obviously it's 50/50 for the NEXT flip to be heads

However after the events, if one of your flips was heads and the order was not known, it's only 1 in 3 that the other was also a head. Not 50/50.

If it is known that the first flip was a head then it's 50/50 that the other flip was also a head

This will prove the answer to be 1 in 3 in the case of the original puzzle.

Quote:Jeepster24Bingo, I never said the boy with me was born first.

The only other info I gave you was I had two children.

Therefore you cannot eliminate GB

BG, GB, and BB are all valid and equal possibilities

What's so special about birth order? I'm ordering them "near child, far child," just as I might order two dice "red die, black die." What do I care which die was manufactured first?

I see now what you're getting at:

But you've made a mistake by singling out your son as you did in your description of the problem.

Quote:24Bingo

"I've flipped a fair coin twice, and one of those flips was tails. With that information, what's the probability the other was tails?"

But you've made a mistake by singling out your son as you did in your description of the problem.

Why was it a mistake, it's just a scenario to set out the question.

It gave the info that one child is a boy and the sex of the other is unknown

It did not alter the answer, it's still 1 in 3 that the other is a boy.

I do like your example above though.

Let's change the question slightly and say...

You currently have one child and your wife is home pregnant with you're next child. The child I see you with is a boy. What's the chances that the unborn child is also a boy?

According to the question, the world rate for sex of babies is 50/50, so the answer to my question must be the chances that the other unborn child will be a boy is 50%

So, for the question posed by Jeepster, the only difference is the child has already been born. Why does that fact alone make it 17% less likely that the child is a boy. ???

The only difference between the two questions is the physical location of child #2, inside the mother or outside . Seems that shouldn't change the odds it's a boy

An hour later the kid is born. I tell you I have a son and another kid at home. Now the odds that other kid is a boy is 33% ?

Doesn't make sense, the only thing that changed is the physical location of the second kid .

Quote:michael99000

Let's change the question slightly and say...

You currently have one child and your wife is home pregnant with you're next child. The child I see you with is a boy. What's the chances that the unborn child is also a boy?

Because you are now putting a time line on the events, making the first born a boy.

That only leaves two possibilities. BG or BB, a 50/50 chance of the second being a boy.

When the order of events is not known there are 3 possibilities. BB, BG or GB, in two of these cases the other child is a girl, thus only a 1/3 chance of the other being a boy.

Quote:michael99000In other words, if I tell you I have a son and a pregnant wife, the odds of that second kid being a boy are 50/50.

An hour later the kid is born. I tell you I have a son and another kid at home. Now the odds that other kid is a boy is 33% ?

Doesn't make sense, the only thing that changed is the physical location of the second kid .

An hour later you tell someone else that you have two children, one of them a son. Not differentiating between them in any way, the odds that you have two sons rise from 25% to 33%. But tell that someone else your younger child is a son, it goes up to 50%.

But here's the thing - distinguish them in any way unrelated to gender, and it goes up to 50%. The child you've seen is a son; it goes up to 50%. The child earlier in the alphabet is a son; it goes up to 50%. One child has a birthmark and one doesn't, and the one without is a son; it goes up to 50%.

This is why I said "the westward coin." It could have been the faster-falling coin, or the shinier coin, or anything else, but it's just the same as saying "the first coin came up heads." What matters isn't chronology, but the fact that the two are distinct.

Let's think about it with the old familiar Monty Hall problem. A common way to illustrate that the change in probability isn't as ridiculous as it first appears is to go from three doors to ten, opening eight. Anyone can see then you should switch. But if the doors had been opened one at a time, with the host not knowing whether the car was behind each one, then it doesn't matter, because it was so improbable that you should have gotten there.

So let's put a similar scaling to this problem. "I've flipped a coin ten times, and got at least nine heads. With no other information, what's the chance I got ten?" 1/11. "You saw me flip heads nine times. I flipped the coin one more time, at some point buried in the nine flips, and you don't know the chronology, but which flips you saw were unrelated to their outcome. Which way was that hidden flip more likely to have come up?" They're equal.

Unless, of course, you'd never be seen in public with your daughter.

I'm not going near that one, most people will not accept the correct answer.

The following is assuming that you are not a Taliban, i.e. you might take your girl outside as well, unbiased. So we start with the following options:

B1B2 G1B3 B4G2 G3G4

The chances of meeting you are 12.5% for each of the children possible.

Now, by observation, we can rule out all the girls, that leaves us with :

B1 B2 B3 B4

So there is a 50% chances it is either boy 1or boy 2 form the two boy combination– you have two boys.

Or it might be a boy from either BG combination – 50% chance here as well.

The chance is 1/2.

Quote:CanyoneroLet me jump into the fray here:

The following is assuming that you are not a Taliban, i.e. you might take your girl outside as well, unbiased. So we start with the following options:

B1B2 G1B3 B4G2 G3G4

The chances of meeting you are 12.5% for each of the children possible.

Now, by observation, we can rule out all the girls, that leaves us with :

B1 B2 B3 B4

So there is a 50% chances it is either boy 1or boy 2 form the two boy combination– you have two boys.

Or it might be a boy from either BG combination – 50% chance here as well.

The chance is 1/2.

Sorry, I really have no idea of what you are saying.

Quote:JeepsterBecause you are now putting a time line on the events, making the first born a boy.

That only leaves two possibilities. BG or BB, a 50/50 chance of the second being a boy.

When the order of events is not know there are 3 possibilities. BB, BG or GB, in two of these cases the other child is a girl, thus only a 1/3 chance of the other being a boy.

ok, let's do all possibilities in a tree:

0

B G

BB BG GB GG

In total there are 4 possible outcomes of having 2 children. Now all we know that one of same is a boy. This eliminates only the solution GG. Leaving us with 3 possiblities (BB, BG, GB) which are equally likely. So the probability of the other also being a boy is 1/3.

edit:

I guess the problem that Canyonero has (and i also had for a short moment just now) is that he "creates" another solution where the order of the BB solution matters. Creating two variants B1B2 and B2B1 which doesn't exist. See table above. There is only one BB possibility and it doesn't matter that we don't know if we see the old or young boy of the boy-boy combo.

Quote:gameterror

edit:

I guess the problem that Canyonero has (and i also had for a short moment just now) is that he "creates" another solution where the order of the BB solution matters. Creating two variants B1B2 and B2B1 which doesn't exist. See table above. There is only one BB possibility and it doesn't matter that we don't know if we see the old or young boy of the boy-boy combo.

No, there is only BB BG GB GG, the number I only put there as a point of reference (to little effect it seems).

In short, for a BB parent the chances of meeting them with a boy is 100%, while for either of the GB combinations it is only 50%. That is why it is not 1/3 but 1/2.

Quote:CanyoneroNo, there is only BB BG GB GG, the number I only put there as a point of reference (to little effect it seems).

ok, then i don't understand how you come to a different solution. 4 variants. each variant of birth has the same probability. now the fact that we see at least one boy eliminates the GG variant and only this one. so we are left with the 3 others BB, BG, GB. leaving the second kid we don't see as a B, G, G ...so 1/3.

Quote:Canyonero

In short, for a BB parent the chances of meeting them with a boy is 100%, while for either of the GB combinations it is only 50%. That is why it is not 1/3 but 1/2.

the fact that we would only see a boy in a GB or BG combination 50% of the time is no longer relevant...as we have already SEEN him.

Quote:michael99000Here's why I'm having trouble understanding why the chances the other child is a boy arent 50%.

Let's change the question slightly and say...

You currently have one child and your wife is home pregnant with you're next child. The child I see you with is a boy. What's the chances that the unborn child is also a boy?

According to the question, the world rate for sex of babies is 50/50, so the answer to my question must be the chances that the other unborn child will be a boy is 50%

So, for the question posed by Jeepster, the only difference is the child has already been born. Why does that fact alone make it 17% less likely that the child is a boy. ???

The only difference between the two questions is the physical location of child #2, inside the mother or outside . Seems that shouldn't change the odds it's a boy

You mostly beat me to this. Here is another way of putting it:

There is a "no zero" roulette game, the last time the wheel was spun it landed on red. WThat are the chances it will land on red the next spin?

The chances of the other child being a boy should be 50%. The sex of the child with the parent at the time should not matter.

I've always explained this one as switching from odds 1/3 to 1/2.Quote:JeepsterYou're opening a can of worms with the Monty Hall problem.

I'm not going near that one, most people will not accept the correct answer.

Quote:gameterrorok, then i don't understand how you come to a different solution. 4 variants. each variant of birth has the same probability. now the fact that we see at least one boy eliminates the GG variant and only this one. so we are left with the 3 others BB, BG, GB. leaving the second kid we don't see as a B, G, G ...so 1/3.

the fact that we would only see a boy in a GB or BG combination 50% of the time is no longer relevant...as we have already SEEN him.

It is very relevant.

You must not just eliminate the GG combination (= adjust to 0%) and ignore the adjustments to the other combinations.

If you meet a boy, it is twice as likely he came from the BB combination - it might be either of the two boys. You mustn't assume that all combinations BB GB BG have an equal likelyhood, since after already meeting a boy the odds are now clearly skewed towards the BB combination.

Can somebody with better explanatory skills back me up on this please. The Monty Hall reference totally applies btw...

Btw. the independent event approach also totally works and I have not seen any flaw in that logic. Imagine this: You meet a dad with only one child - a boy. That night he goes home and impregantes his wife. The next day you meet him again. What is the chance of the other child becoming a girl? - Exactly!

Quote:AZDuffmanYou mostly beat me to this. Here is another way of putting it:

There is a "no zero" roulette game, the last time the wheel was spun it landed on red. WThat are the chances it will land on red the next spin?

The chances of the other child being a boy should be 50%. The sex of the child with the parent at the time should not matter.

Yes, but this is a different question. If the question was : i already have one kid what is the chance the second is going to be a boy too ? the answer is 50% because the sex of the first kid has no influence on the second kid.

but the possible distribution of having 2 kids is BB, BG, GB, GG. all with the same probability of 25%. See tree above. Now I tell you that one of the kids is a boy. that eleminitates the GG variant. and leaves you with 3 possible distributions BB, BG, GB.... so 1/3 that it's two boys.

Quote:gameterror

edit:

I guess the problem that Canyonero has (and i also had for a short moment just now) is that he "creates" another solution where the order of the BB solution matters. Creating two variants B1B2 and B2B1 which doesn't exist. See table above. There is only one BB possibility and it doesn't matter that we don't know if we see the old or young boy of the boy-boy combo.

I think you are right.... after rereading your early statement that's exactly what i referenced in this edit.

The combination BB has two different boys which creates 4 scenarios for what we can see.

B1B2 -> can be either B1 seen, B2 hidden or B1 hidden and B2 seen.

and then the two BG GB variants which in fact have no hidden boys...as we see the boys of those variants.

now - and i can't wrap my head around it - we have to figure out if it really matters which boy we see of the BB combination. your pregnancy example strongly indicates it doesn't :-)

i'm now also in the 50/50 camp :-)

I disagree. Here's a table of possibilities:Quote:Jeepster24Bingo, I never said the boy with me was born first.

The only other info I gave you was I had two children.

Therefore you cannot eliminate GB

BG, GB, and BB are all valid and equal possibilities

Child With You | Child At Home |
---|---|

Boy | Boy |

Boy | Girl |

Girl | Boy |

Girl | Girl |

Here's the question:

Keep in mind that the child that you brought with you is completely irrelevant according to this wording. You may as well have said, You see me out with my Subaru which happens to be Red, what are the chances that my child at home is a boy? So..Quote:PuzzleYou see me out with one of my two children, who happens to be a boy.

What's the chances of the other child at home also being a boy?

The only info you have and are allowed to know is-

I have exactly two children, the boy with me and another child at home.

Assume a world wide boy/girl birth ratio of 50/50.

Because the child that you brought with you is a boy, the last two rows are impossible, you must choose from what remains and that makes the child at home a 50/50 chance of being a boy.

Child #1 | Child #2 | Probability |
---|---|---|

Boy | Boy | 0.25 |

Boy | Girl | 0.25 |

Girl | Boy | 0.25 |

Girl | Girl | 0.25 |

Total | 1 |

Since we know that at least one of them is a boy, we can eliminate Girl-Girl as a possibility, leaving the following:

Child #1 | Child #2 | Probability |
---|---|---|

Boy | Boy | 0.25 |

Boy | Girl | 0.25 |

Girl | Boy | 0.25 |

Total | 0.75 |

However, the probabilities of the remaining possibilities only sum to 0.75, so we need to scale each probability by dividing them by 0.75. This is the same thing done in craps to calculate the probability of making or missing a point; we don't care about rolls of 2, 3, 11, 12, etc., and therefore we remove those possibilities from the calculation and scale the probability of the only outcomes that matter such that they sum to 1. So after doing this, we are left with:

Child #1 | Child #2 | Probability |
---|---|---|

Boy | Boy | 0.333333... |

Boy | Girl | 0.333333... |

Girl | Boy | 0.333333... |

Total | 1 |

And since the only remaining possibility that allows for the other child to be a boy has a probability of 1/3, that is the answer.

Read the puzzle carefully. The boy that is with the parent is completely irrelevant.Quote:JBAnd since the only remaining possibility that allows for the other child to be a boy has a probability of 1/3, that is the answer.

Child #1 | Child #2 | Child with me | Child at home |
---|---|---|---|

Boy1 | Boy2 | Boy1 | Boy2 |

Boy1 | Boy2 | Boy2 | Boy1 |

Boy | Girl | Boy | Girl |

Girl | Boy | Boy | Girl |

Girl1 | Girl2 | N/A | |

Girl1 | Girl2 | N/A |

50/50

Quote:JBYes, I see now that the semantics of the original wording are ambiguous. (First, there is no indication that the child with him is a boy, next there is.) So the answer is either 1/2 or 1/3 depending on how the question is interpreted.

Agreed. And as currently written, it's 1/2. Hope we didn't kill Monty Hall with this discussion.

Funny when you think of it as cards from a finite set, it seems easier to see it's not the same and something would be amiss. Draw two cards, one is red, even money if the other is red.

Quote:JeepsterWhy was it a mistake, it's just a scenario to set out the question.

It gave the info that one child is a boy and the sex of the other is unknown

It did not alter the answer, it's still 1 in 3 that the other is a boy.

Now that I've sobered up, I think I can explain this better: that's not the only information you gave me. You gave me the information that the child you were out with was a boy. Assuming you wouldn't keep a girl locked up, I've essentially taken a random sample and found a boy, which would be impossible if you had two girls, but would be certain if you had two boys, and possible but uncertain if you had one of each. Because it would be less likely to have happened in that world, the fact that it has happened makes that world less likely.

Here's the question you meant to ask: "Given I have at least one boy, what's the chance I have two?" And we get the chance you'd have at least one boy if you had two, 1, times the probability you'd have two boys, 1/4, over the probability you'd have at least one, 3/4, yielding 1/3. But that's wrong.

It's wrong because here's the question you actually asked: "Given this child was a boy, what's the chance I have two?" So what we get is the chance a randomly selected child if yours would be a boy if you had two, 1, times the probability you'd have two, 1/4, divided by the probability a random child would be a boy, 1/2, and that gives 1/2.

Actually, if I know your second child is just born, the answer is 50%. I f I know nothing, the answer is 33.33%Quote:michael99000In other words, if I tell you I have a son and a pregnant wife, the odds of that second kid being a boy are 50/50.

An hour later the kid is born. I tell you I have a son and another kid at home. Now the odds that other kid is a boy is 33% ?

Doesn't make sense, the only thing that changed is the physical location of the second kid .

Similarly, if you tell me "this boy is my elder child", the answer is 50%.

Probability is not about describing the real world. It is about describing information. In the original problem, I have less information about which child happened to wander with its father than in the 'pregnant' or 'elder' situations. That is why probabilities change. In the latter, the lack of info is just about the sex of one child (the absent one, that is forcibly absent, either because it is just born or because it is the younger one). In the original problem, the lack of info is also about which child I have been presented.

Quote:kubikulannActually, if I know your second child is just born, the answer is 50%. I f I know nothing, the answer is 33.33%

Similarly, if you tell me "this boy is my elder child", the answer is 50%.

Probability is not about describing the real world. It is about describing information. In the original problem, I have less information about which child happened to wander with its father than in the 'pregnant' or 'elder' situations. That is why probabilities change. In the latter, the lack of info is just about the sex of one child (the absent one, that is forcibly absent, either because it is just born or because it is the younger one). In the original problem, the lack of info is also about which child I have been presented.

And let me guess - the long run doesn't matter in a real casino, right?

What's so special about birth order? Let's say someone throws two dice, and you're told one of the numbers was a six. There's then a 1/11 chance that they're both sixes. But say you see one die, and it's a six. There's a 1/6 chance that they're both sixes. What's the difference? In the first case, you didn't know which die was the six, so the five cases where only die A was a six and those where only die B was were equally probable. "But," you say, "I don't know which die I saw!" Sure you do. You saw the die the saw. You didn't see the die you didn't see. Only the possibilities where the die you saw was a six are still possible; you don't have to consider that it might have been the other die. If we call that die "die A," we can see that only those six are possible. What if it's die B? Then we've made a mistake in naming them. But you can't do that if you've just been told that one was a six, because then if you pick the wrong name for the die that was a six, you may have made a mistake in naming them, or may have made a mistake in the actual die. Ultimately, if you don't know which die you saw, but have no reason to think one more likely than the other, you can say that all the one-die probabilities have gone down by half, for the 50% chance that that was the die you saw.

In the same way, I saw the child I saw, and didn't see the child I didn't see. It doesn't matter which was born first, because the possibility that the child I saw is a girl and the one I didn't see was a boy is eliminated, whether in birth order that's GB or BG. I don't know which, but they're equally likely, so if you prefer, I halve the probability of each.

(I'm starting to wonder how many ways I can come up with to explain this.)

Quote:24BingoIn the same way, I saw the child I saw, and didn't see the child I didn't see. It doesn't matter which was born first, because the possibility that the child I saw is a girl and the one I didn't see was a boy is eliminated, whether in birth order that's GB or BG. I don't know which, but they're equally likely, so if you prefer, I halve the probability of each.

(I'm starting to wonder how many ways I can come up with to explain this.)

Imagine a coin flip game.

Version 1: Your friend is flipping a penny and a quarter. He flips both coins, and keeps flipping if they both show up tails. That way he's guaranteed to have at least one head. Once he arrives at a valid outcome, which you can't see, your friend slides forward a coin showing heads. If only one of the penny and quarter is heads, that's the one he slides forward. If both the penny and quarter are heads, he picks one at random. Then he says "This coin is a head. I have flipped another coin. What are the chances of both being heads?" It's easy enough to see that the odds of both coins being heads are 1/3 under the circumstances: of the four possible outcomes in flipping 2 heads, TT has been eliminated leaving (with equal probability) HH, HT, and TH.

In Version 1, the two coins were differentiable -- you knew beforehand how to identify them. In Version 2 they're not. Version 2 is exactly the same but with two pennies instead. How could the odds possibly have changed?

(I think you're agreeing with me, but I'm not sure.)

Quote:24BingoOf course, the difference there is that your friend would under no circumstances have shown you a coin that came up tails, whereas there's no reason to think I'd be unlikely to see this man with his daughter if he has one.

(I think you're agreeing with me, but I'm not sure.)

But you didn't see the man with his daughter. You saw him with his son. It's not possible for the man in the story to have two girls, just as it's not possible in the case where you see him with his daughter for him to have two boys.

Let'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?