[kgr]

Question from my daughter:
I have 3 children and one of them is a boy. What is the probability that they are all 3 boys?
The answer should be 1/7.
Is there an easy explanation for this? How should this be calculated?

Thanks,
Koen

[Elianna]

There are 8 possibilities:

BBB

BBG

BGB

GBB

BGG

GBG

GGB

GGG

You can eliminate the last possibility, so now there are only 7 choices. And you care only about one. So 1/7

[matmat]

kgr, on Sep 19 2009, 12:02 PM, said:

Question from my daughter:
I have 3 children and one of them is a boy. What is the probability that they are all 3 boys?
The answer should be 1/7.
Is there an easy explanation for this? How should this be calculated?

Thanks,
Koen

0, since you have a daughter.

[MattieShoe]

For the sake of clarity, maybe "at least one of them is a boy" would be better :-) English is sloppier than math, "One is a boy" could be interpreted to mean two are not.

But yeah, 8 possible permutations, all having equal odds of happening. One (GGG) is eliminated leaving 7. One of the remaining 7 is all boys (BBB) so 1/7th.

If you said "the oldest is a boy" or some such, the odds would be 1/4th... You'd be eliminating GGG, GGB, GBG, GBB, leaving BGG, BGB, BBG, BBB. :-)

[MattieShoe]

It gets really fun when you get into prior distributions....

You're at a gay bar consisting of 80% males and 20% females. You can determine a person's gender at a glance correctly three quarters of the time. You glance at somebody and think they're female. What are the odds that she's actually female?

The answer is 3/7ths or about 43%. :-)

Ain't math fun?

[EricK]

There is no unique answer to the question as you need to make some assumption as to how you came by the knowledge that (at least) one child is a boy.

Consider these scenarios:
1. You ask a person who has 3 children "is at least one a boy?" and he answers "yes"..
2. You ask a person who has 3 children to tell you the sex of one of his children. He says "male".

In the first scenario the probability that all his children are boys is 1/7 (assuming he was telling the truth!)

In the second scenario the probability that he has 3 boys is 1/4 (again assuming he isn't lying)

[kgr]

Thanks for the answers!
- Has this something to do with restircted choice?
- What if I have 20 children and at least one of them is a boy, what is the prob that they are all boys? (suppose you have too much children to write done all combinations)

[OleBerg]

kgr, on Sep 20 2009, 10:18 AM, said:

Thanks for the answers!
- Has this something to do with restircted choice?
- What if I have 20 children and at least one of them is a boy, what is the prob that they are all boys? (suppose you have too much children to write done all combinations)

A quick guess for the formula:

N = Number of children.
B = Number of children known to be boys.

1 divided by: 2(uplifted to N'th) - B

I am not sure, but I also believes that all this assumes that a child is exactly 50% likely to be of either gender.

I might be using the wrong words, English (and Maths) isn't my native language.

[MattieShoe]

OleBerg, on Sep 20 2009, 03:39 AM, said:

kgr, on Sep 20 2009, 10:18 AM, said:

Thanks for the answers!
- Has this something to do with restircted choice?
- What if I have 20 children and at least one of them is a boy, what is the prob that they are all boys? (suppose you have too much children to write done all combinations)

A quick guess for the formula:

N = Number of children.
B = Number of children known to be boys.

1 divided by: 2(uplifted to N'th) - B

I am not sure, but I also believes that all this assumes that a child is exactly 50% likely to be of either gender.

I might be using the wrong words, English (and Maths) isn't my native language.

Not the correct formula. For instance, take the same problem but say at least two are boys. That leaves
BBG
BGB
GBB
BBB

So the odds of all boys is 1/4, not 1/6 as your formula would suggest.

[EricK]

Assume you find out that a family with N children has at least B boys because you ask the question "do you have at least B boys?" and get the answer "yes".

Then, letting P be the probability that a random family with N children has at least B boys, the probability that this family has N boys is 1/(P*2^N)

P will equal to (C(N,B )+C(N,B+1)+...+C(N,N))/2^N where C is the standard Combination function
(http://en.wikipedia....iki/Combination)


On the other hand, if you find out the family has at least B boys because you see B children at random and they all turn out to be boys, then the probablity that they are all boys is simply 1/(2^(N-B ))

Note that this relates to restircted choice because the play of a bridge hand is more like the first scenario than the second - i.e. players don't play cards at random.

[matmat]

EricK, on Sep 20 2009, 04:21 AM, said:

Note that this relates to restircted choice because the play of a bridge hand is more like the first scenario than the second - i.e. players don't play cards at random.

Have you seen me play?

[helene_t]

kgr, on Sep 20 2009, 09:18 AM, said:

- Has this something to do with restircted choice?

Yeah, if there are two honors out, at least one of which is with West, then the chance that the other is also with West is 33%. If you have two kids at least one of which is a boy then the chance that the other is also a boy is 33%. Same thing.

- What if I have 20 children and at least one of them is a boy, what is the prob that they are all boys? (suppose you have too much children to write done all combinations)

There are 2^20 combinations, one of which you have excluded (that they are all girls), that leaves 2^20-1. You are interested in one of them, so the answer is 1/(2^20-1), that is a little less than one in a million.

[Lobowolf]

The question of accounting for information depends, as has been suggested, by how the information has been acquired. Take the question, "Bob rolls 2 dice; at least one of them shows a 1. What is the probability that the other die shows a 1, as well?"

If Bob is trying to create the conditions for the premise, by rolling two dice behind a screen until he gets at least one 1, and then showing you a 1 and asking you the probability that the other one is a 1, as well, the probability is 1/11.

On the other hand, if Bob says, "Hey, let's try something!" and rolls two dice behind a screen, and without even looking just grabs the first one he touches, and brings it around the screen and it happens to be a 1, then he says, "Hey, what are the odds the other one is a 1, too?" then it's 1/6.

The parallel I'd draw in bridge was addressed in the book "For Experts Only," where you have to find a Q in one suit (suit "A"), and the author poses the fallacious idea of finessing the partner of the opening leader, because he's shorter in suit "B" (the suit led on opening lead). The author points out that if the lead were in the other hand, the non-opening-leader (now the opening leader) would have led suit "C," HIS longest suit, and now would it be correct to finesse the original opening leader, because he's shorter in suit C?

[MattieShoe]

helene_t, on Sep 20 2009, 08:02 AM, said:

kgr, on Sep 20 2009, 09:18 AM, said:

- Has this something to do with restircted choice?

Yeah, if there are two honors out, at least one of which is with West, then the chance that the other is also with West is 33%. If you have two kids at least one of which is a boy then the chance that the other is also a boy is 33%. Same thing.

- What if I have 20 children and at least one of them is a boy, what is the prob that they are all boys? (suppose you have too much children to write done all combinations)

There are 2^20 combinations, one of which you have excluded (that they are all girls), that leaves 2^20-1. You are interested in one of them, so the answer is 1/(2^20-1), that is a little less than one in a million.

Ah, but what if West opened that suit and East has only passed? :-) Someday I want to write a bridge playing program, but it seems like a tough proposition. Calculating odds based on partial knowledge gained through bidding, carding, and discards seems awfully tough.

[3for3]

One problem is the assumption of 50-50 ness.

Firstly, there are more boys born than girls.

Second, a person's likelihood of having the same sex of a second child is more than 50%, due to some factor I can't really recall.

Next time use coins

[cherdanno]

3for3, on Sep 20 2009, 12:10 PM, said:

One problem is the assumption of 50-50 ness.

Firstly, there are more boys born than girls.

Second, a person's likelihood of having the same sex of a second child is more than 50%, due to some factor I can't really recall.

Next time use coins

Uh-huh. Coin tossing doesn't work so well either, it has about a 51-49 bias to land the same way it starts out
http://news.stanford...iaconis-69.html

[PassedOut]

We had three sons. My Atlanta bridge partner (and great friend) had three daughters.

We could never agree on a trade because we couldn't bear to part with any of them (although I think his wife would have given us one of the girls outright).

One man I worked with (a Roman Catholic) had seven daughters when I last knew him, and desperately wanted a son. Don't know if he ever got one.

[Fluffy]

but if you had the knowledge that one of them is a boy because you saw one of them who happened to be a boy. Does it change the occurences of the 7 cases?

BGG = 1/3
BBG = 2/3
BGB = 2/3
BBB = 3/3
GBG = 1/3
GGB = 1/3
GBB = 2/3

So it would be 3/13 that all are boys.

(Not sure if my math is correct)

[helene_t]

No Gonzalo, that's different. If you only examined one of the kids and that kid turned our to be a boy, the there are only four combinations for the remaining two kids.

[blackshoe]

3for3, on Sep 20 2009, 01:10 PM, said:

Firstly, there are more boys born than girls.

<raises eyebrow>

You sure about that?