Mr Smith has two children, help me understand

Mr Smith has two children, one of them is a boy. What is the probability that his other child is a boy. The standard solution seems straight forward and makes perfect sense to me. There are four combinations of two children that are possible (BB, BG, GB, GG), knowing only that one is a boy rules out GG leaving 1/3 chance that the other child is a boy.

My problem is that I think that i can prove it is a 1/2 chance. I will argue that we know more information than we think. Please review and let me know where my logic fails.

First, if we knew Mr Smiths boy was the eldest, we could rule out GB and GG from the four combinations leaving it a 1/2 chance the other child is a boy.

Second, by the same method if we knew the boy was the youngest, we could rule out BG and GG from the four combinations leaving it a 1/2 chance that the other child is a boy.

Third, if Mr Smith has two children, one of them is a boy, the boy must be either the eldest (see first point), or the youngest (see second point), both of which leads to the 1/2 chance that the other child is a boy.

Please feel free to explain my error and which step or steps above are not accurate because I can not find it.

I think is has only been a month or so since the last time this was discussed.

If we assume that Mr. Smith is chosen at random from the pool of all men with exactly two children, where at least one is a boy, then the answer is 1/3.

Your argument for 1/2 doesn't hold water. If he has two boys then which of them is THE boy?

Thanks for responding, I have read two previous threads here and a few elsewhere online, but am still not understanding my error. I want to come to 1/3 because that is right, there are four ways to have two children. they must be distributed 1/4 BB, 1/2 BG/GB and 1/4 GG what i argued obviously cotradicts that. I just can't see the error.

You suggested that the error is because "if he has two boys then which is THE boy?" I am unclear on how that changes the problem. As long as THE boy is either the eldest or the youngest of the two, which i believe he must be, then we should be able to follow one or both options (described previously) to the same result. Otherwise it seems that I must consider THE boy to be neither the eldest nor the youngest (even though he is one of them) until told which he is? It makes no sense.

Four possible cases, BB BG GB GG, each assumed equally likely, with the first letter indicating the gender of the eldest.

Your solution consists of several steps that are essentially:

(1) Given that the eldest is a boy, what is the probability the youngest is a boy?
Ans: P(BB | (BB or BG) ) = P(BB) / ( P(BB)+P(BG) ) = .25/(.25 + .25) = .50

(2) Given that the youngest is a boy, what is the probability the eldest is a boy?
Ans: P(BB | (BB or GB) ) = P(BB) / ( P(BB)+P(GB) ) = .25/(.25 + .25) = .50

(3) Since a known boy must be either eldest or youngest, equally likely, combine these two:
P = .5*.50 + .5*.50 = .50

Unfortunately, this is not a correct solution. Perhaps you can see the difficulty this way: You have only been told that there is at least one boy, i.e., we have either BB, BG, or GB. That means that there is a 2/3 probability that the eldest is a boy and a 2/3 probability that the youngest is a boy. Since these probabilities don't add to 1, your combination in step (3), assuming that the equally-likely possibilities are 50% each, and multiplying that by the conditional probabilities from (1) and (2), just doesn't work.

The problem should be solved as:

P(BB | (BB or BG or GB) ) = P(BB) / ( P(BB)+P(BG)+P(GB) ) = .25 / (.25+.25+.25) = 1/3

Things are starting to clear up. Your response was very clear concise and makes a lot of sense. There are more ways that the boy can be the eldest , or the youngest for that matter. The equation at the bottom even made sense to me. Thanks.

I wonder if this is somehow related to the gamblers fallacy, I once threw 13 tails at a casino and everyone bet against me because they concluded that the chance of throwing the 14th was approx 1/16000 (14 in a row) minus 1/8000 to throw the 13, a whopping 1/8000 chance. They considered all the combinations possible thinking they did not have information to rule them out. I did the opposite by falsely ruling out combinations treating them as discrete unrelated events.

This is interesting. I was sure it was 50%, but then realized it gets weird when (at least) one is a boy.

I imagined rolling 2 dice, where odd numbers (1,3,5) are a Girl and even (2,4,6) are a boy. When two odds are rolled, its na (not available) because that's 2 girls. When at least one of the dice is even, its either a win or a fail, depending on if the other number is odd or even. The results:

11, na
12, fail
13, na
14, fail
15, na
16, fail
21, fail
22, win
23, fail
24, win
25, fail
26, win

You could do the same for 3x, 4x, 5x, and 6x.....but it'll just repeat as it did above.

Omit the three NA's, and you end up with 9 trials where at least one is a boy (even). Only three of the nine are wins (22,24,26) while the other 6 are fails (12,14,16,21,23,25).

Quote: 1call2many

Things are starting to clear up.

One more way you can look at it:

From your step (1) as I described it, there is a 50% probability of two boys if you know that the eldest is a boy. Also as I described, if you know that there is at least one boy, there is a 2/3 probability that the eldest is a boy. Combine these to get P = 2/3 * 50% = 1/3.

You can do the same thing starting with the step (2) and the 2/3 probability that the youngest is a boy, given that there is at least one boy.

I think 2 different problems are intermingled in this and another thread by the language used.
The 2 different problems are:

Problem 1
Smith has 2 children, at least one is a boy, what is the probability the 2nd is a boy.
The problem can also be stated If at least 1 is a boy, what is the probability of both boys.
Or simplified by removing the 'at least' and using the term 'one is a boy' which implies the 'at least'.
The answer is 1/3.

Problem 2
Smith has 2 children. You see one and is a boy. What is the probability that the other is a boy.
Or the eldest is a boy, what is the probability the other is a boy.
Or the smartest is a boy, what is the probability the other is a boy.
The answer is 1/2 in all cases.
There is no conditional probability in this problem as in problem 1. The identification of one of the children as a boy has nothing to do with the other.
They are independent events.
In problem 1 the 'at least' is the conditional which makes the 2 events dependent.

When the 1st problem is used without the 'at least' wording which is imlpies then
Problem 1: One is a boy with answer 1/3 for the other child
Problem 2: Eldest is a boy with answer 1/2 for the other child
And it seems absurd that when you specify the boy the prob changes from 1/3 to 1/2.
But the vague 'one is a boy' means 'at least one is a boy' which is the conditional.

I don't get how seeing is different from being told by Mr Smith, or identifying smartness either, these do not give information about the order of the birth which leaves us with the four two child combinations, three of which contain at least one boy.

I attempted to illustrate the difference in my Boy or girl puzzle thread post which discusses the same problem. I meant to post it to this more recent, active thread. I won't repost it here, least I incur the wrath of the Greenies!

Imagine a game with two dice.

Case one: you have thrown a six on one die. What is the probab that the other die will be a six? Answer: 1/6.
Case two: someone else has thrown the dice, you learn that "there is a six". What is the probability that both dice are sixes? Answer: 1/11.

But if you learn it by seeing one of the dice, then we are back to case one.
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When something appears paradoxical, try making a similar story with more strongly separable values. It often gives insight.

Quote: kubikulann

Case two: someone else has thrown the dice, you learn that "there is a six". What is the probability that both dice are sixes? Answer: 1/11.

There is a six: 1,6 2,6 3,6 4,6 5,6 6,1 6,2 6,3 6,4 6,5 6,6, 11 possible results
both sixes: given one is a six = 1/11