today's fun with probability

[eub]

1) 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]

Comments

Re: do not taunt happy fun mathchick

[eub.livejournal.com]

3) I flip a loonie and a toonie (sp?). At least one comes up heads. What's P(both heads)?

4) I pick a random two-bit number. It's not 0. What's P(3)?

Re: do not taunt happy fun mathchick

[thatmathchick.livejournal.com]

For a moment I almost psyched myself out. I recently ended up discussing the problem:

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

[eub.livejournal.com]

I've lost track of what A and B bind to...

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?

[seanmcguire.livejournal.com]

I'm curious to know how both of you respond to this reply made more recently.

[eub.livejournal.com]

Hey hey hey, slummin' on LJ. How's tricks?

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.