Restricted choice?

[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]

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]

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]

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.org/wiki/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]

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]

- 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]

- 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 :D

[cherdanno]

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 :D

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.edu/news/2004/june9/diaconis-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]

Firstly, there are more boys born than girls.

<raises eyebrow>

 

You sure about that?

[kenberg]

People are reading a lot into this question, so let me try. Recall the original question:

 

"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?"

 

Since the question came from his daughter, I suggest that the probability that all three of his children are boys is zero.

 

 

But if we take it as a typical textbook question the answer is 1/7 as Elianna said and for the reasons that she said, and 1/(2^n-1) with n kids, as Helene said. Restricted choice involves an ongoing collection of information and some assumptions about who would do what. For example if your opponent always plays the J from QJ tight then the probability that he holds the Q when he drops the J is about 1/2 and the probability that he holds the J when he drops the Q is zero (which is why one should play the quacks randomly, but is an assumption in the analysis that your opponent does so). In the question as posed, it is very reasonable to interpret it as "We choose a family from among all families that have exactly 3 kids, at least one of whom is a boy, what is the probability that all 3 are boys?". You can jazz it up with biological and social data. For example, some parents wish to have at least one boy and at least one girl. To take that to extremes, suppose all parents quit producing after 1 child of each sex and always have a third child after having two of the same sex. Then, given that the family has three children one of them being a boy, the birth sequence must have been ggb, or bbb, or bbg. The probability that all three are boys is perhaps 1/3. But further, bbg and ggb would then remain 3 child families, while bbb would not. So you might need to ask the age of the youngest child. Of course not all parents make such choices, but some do and it will influence the percentages in the real world.

 

This assumes that you wish to take the problem as a truly practical problem. Surely the intended interpretation is the one that leads to the answer 1/7. Whether census data would agree with this figure I do not know. My guess is that it would not agree because of some of the points raised.

 

 

Here is a thought about restricted choice that I have never seen mentioned: Playing 5 card majors you open 1H, you eventually play 4H, You hold AKT32, dummy holds 7654. You win the opening side suit lead in your hand and you lay down the ace of hearts. LHO plays a quack. Restricted choice makes it 2 to 1 that the other quack is on your right? Not so fast. Holding Q9 it is (fairly) safe for lho to drop the Q. If partner has the K he still gets it, if declarer has the K defender was (probably) not getting your Q anyway. If defender can induce declarer to go to the board planning on making a restricted choice play this may use up a dummy entry when the suit was 2-2 all along. Defender won't get a trump trick but he wasn't going to get one anyway, and there may be a tactical advantage. At any rate, the 2 to 1 odds only really get set if declarer does go to the board, leads a spot, and RHO follows low. Now it is known that LHO held either QJ tight or a stiff. Now it is 2 to 1 for the well-known reasons. But when the Q first fell, I would say the evidence is not quite in.

 

Youu can vary the question of course.

[EricK]

It isn't clear to me that the "natural" interpretation of the problem is "we choose a family at random from all the 3-children families which have at least one boy" rather than "we choose a random child from all the families which have 3 children and it turns out to be a boy"

[MattieShoe]

It isn't clear to me that the "natural" interpretation of the problem is "we choose a family at random from all the 3-children families which have at least one boy" rather than "we choose a random child from all the families which have 3 children and it turns out to be a boy"

Since we are given the answer (1/7), it's clear they meant "at least one of the three children is a boy" because other interpretations of the question would result in different answers. :-) English is sloppy enough that I don't think there is some hard-and-fast way to interpret the original statement -- everybody could have a differing and equally valid view. :-)

 

Regarding the trump one... I imagine the implicit odds there depend greatly on the quality of the players. For fish like me, I might throw the Q of QJ but I'd never throw the Q in Q9 in that situation, even though the logic makes sense :-)

[gwnn]

Firstly, there are more boys born than girls.

<raises eyebrow>

 

You sure about that?

yep definitely sure, but then boys are dying much quicker so in the 70+ camp there are more women than men. I also have the feeling that subsequent children are more likely to be of the gender of the first child (this probably just means that a given couple have a predisposition to give birth to one gender, in fact I don't know how the two effects could be differentiated) but I never saw solid statistical proof for this, I tried googling it and found nothing.

[blackshoe]

Can you provide a source? I'm pretty sure that I learned in school (though it was long ago) that the birth rate is about 52-48 girls to boys. Has it changed recently?