today's fun with probability
Nov. 20th, 2002 07:33 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
eub1) You meet a man at the bowling alley [this is how it was posed, honest]. He has two children, bowling on lane 17. You see one girl there; the other child doesn't happen to be visible.
2) Man, bowling, two children. You ask if he has any daughters. He says yes.
In each case (any difference?), what's the probability that he has two daughters?
Assume male and female are mutually exclusive and 50-50 and independent between children. Make any further assumptions you need.
[edit: spoilers in comments, natch]
2) Man, bowling, two children. You ask if he has any daughters. He says yes.
In each case (any difference?), what's the probability that he has two daughters?
Assume male and female are mutually exclusive and 50-50 and independent between children. Make any further assumptions you need.
[edit: spoilers in comments, natch]
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do not taunt happy fun mathchick
Date: 2002-11-20 09:46 pm (UTC)There is a baby with the family bowling in the next lane. What is the probability that this baby is female?
(Under the assumption that families are not more likely to bring male babies to bowling alleys than female babies, fearing damage to a girl's delicate sensibilities.)
Your assumptions imply that gender of one child is completely independent of the gender of another child. 50/50 chance that the (other) child is a girl.
Re: do not taunt happy fun mathchick
Date: 2002-11-20 10:37 pm (UTC)4) I pick a random two-bit number. It's not 0. What's P(3)?
Re: do not taunt happy fun mathchick
Date: 2002-11-20 11:24 pm (UTC)Q: You roll 2 dice. What is the probability that you get at least one 6?
A: The answer is not 2/6- which would be derived from the notion that on each roll you have a 1/6 chance of rolling a 6. It is actually (2/6 - 1/36). If you simply add the probabilities, you double count the case where you roll 6 twice.
simpler example:
Q: You pick a 2 bit number. What is the probability that it has a 1 in it?
A: If you think of adding the probabilities, you would get 1/2 for the first digit and 1/2 for the second digit. Which would make it a 100% chance of at least one 1. HUH???
The possible choices are 10, 11, 00, 01. 3/4 possibilities have a 1 in them. Again, you take 1/2 + 1/2 - 1/4 = 3/4.
In general, for 2 independent events A and B, the problems can be stated:
I) Probability (A or B) = Probability (A) + Probability (B) - Probability (A and B)
II) Probability (A given B) = Probability (A).
The bowling, loonie - twoonie, and your 2 bit number examples are all problem II. The roll at least one 6 and my 2 bit number examples are problem I.
(Huh- apparently I'm also not sure how to spell the term for Canadian $2 coin.)
Re: do not taunt happy fun mathchick
Date: 2002-11-20 11:51 pm (UTC)In my (4) scenario, I'd want to talk about the events "2^0 bit is high" and "2^1 bit is high" for x uniform from {1, 2, 3}. Call those C and D. C and D are not independent, right? P(C) = P(D) = 2/3; P(C and D) = 1/3.
A and B, to be independent, would be those bit highnesses for x from {0, 1, 2, 3}, maybe? I don't remember exactly how this machinery works -- do we then talk about the independence (the lack thereof) of "A and not x=0" and "B and not x=0", or what?
no subject
Date: 2002-12-14 05:35 am (UTC)no subject
Date: 2002-12-14 03:42 pm (UTC)That reply is correct. Once you look at it in terms of four possibilities (which is where I was leading with the (2)-(3)-(4) sequence), you're going to get the right answer.
It's tempting, though, to think that one is a girl, the other is 50-50, so it's 50-50 that both are. The question is how to reeducate this intuition.
The reason that these two situation are different is that in the first, we can fix the gender of a child that is chosen at random from the two, whereas the second situation does nothing to specify the child.
Does this do it? Not for me; my tempting statement was in terms of "one" and "the other" anyway.
The key has to be that "the one is a girl" and "the other is a girl" are not independent, given that at least one is. If one is not a girl, then the other must be. This dependence means you can't break down the probability in the tempting manner. But why, the devil says, can't we talk "given that either the elder is, or the younger is", in either of which cases...? The math doesn't work, but I can't think of a nice intuitive appeal.
no subject
Date: 2002-11-21 07:51 am (UTC)assuming (in the first case) that he is equally likely to bring boys or girls with him to bowl,
the problems are equal, and it's a 50-50 chance. Seeing the girl, or being told he has a daughter, eliminates one child from consideration, so this reduces to the probability of an unknown child being female, which you have said is 50-50.
Now I'll go read the spoiler comments and see how far off I am. :-)
They're different!
Date: 2002-11-21 12:19 pm (UTC)In the first situation, there's a 50% chance of having two girls, and in the second, it's 1 in 3.
A good way to think about this is by writing out all of the possibilities and seeing which ones you can rule out. There are 4 equally likely child-pair possibilities:
A older and younger are girls
B older is boy and younger is girl
C older is girl and younger is boy
D older and younger are boys
In case 1, there are 4 things which could have occurred:
either we have pair A and we saw the older girl
we have pair A and we saw the younger girl
we have pair B and we saw the younger (only) girl
we have pair C and we saw the older (only) girl
Since each of these situations is equally likely, there's a 2 in 4 chance (50%) that both are girls.
In case 2, we can rule out possibility D (2 boys), which leaves us with A, B, and C to choose from. Out of those 3 possibilities, only one has two girls, so the chance that there are two girls is 1 in 3.
The reason that these two situation are different is that in the first, we can fix the gender of a child that is chosen at random from the two, whereas the second situation does nothing to specify the child.
If you'd like to experiment with the second situation, try flipping two coins, and count how many times there is at least one head as well as how many times there are two heads. If you divide the # of 2 heads by # of at least one head, you should get a ratio close to 1/3.
--A Random Probabilist
Re: They're different!
Date: 2002-12-06 04:07 am (UTC)Choose(2,2)/{ 2^2-Choose(2,0) } == 1/3