But isn’t your conclusion “1/2” for the visual case, in agreement with mine?
I didn’t quite parse the part about P(seen w|mw)P(mw). Are you saying that quantity is 0.5? Or, did you not mean to include the P(mw) factor, which itself is 0.25, making the product at least that small?
My assumption is that the two people waiting for me are randomly sticking their heads around the corner, and so I see a random one first. Is this inconsistent with what you’re trying to construct?
If not, P(seen w|mw) = 0.5. If you want to do the standing-in-a-line thing, it just rearranges the terms, but the answer is the same, namely P(seen w|mw) + P(seen w|wm) either is 0.5+0.5 or 0+1.
Okay let’s say we have two people who walk in in order: If we find the first one is female, then clearly it’s 50-50 that the second one is male as the only two possibilties are FM and FF.
Let’s say we examine the gender of one at random and find them to be female:
The we have 3 different choices FF, FM and MF, however the probabilty of selecting a female at random from FF is 1 and the probabilty selecting a female at random from FM or MF is 0.5. So it still works out as 50-50.
What you’ve assumed is that given that one selcted at randomis female then the three ordered pairs (FF, FM, MF) are equally likely, which is not the case.
Agreed. You’re twice as likely to hear a woman speak in the FF case as you are in the MF case.
It occured to me, after my first post, that this would be a good job for Bayes’ Rule. We want to know the probability that some earlier event happened (a man and a woman walked in) given that we know a later event happened (we heard a woman speak). Pasta already gave his analysis, but in case it helps, here’s mine:
The visitors were either two women (A[sub]1[/sub], with probability 1/4), a man and a woman (A[sub]2[/sub], with probability 2/4), or two men (A[sub]1[/sub], with probability 1/4).
The probability of hearing a woman if they were two women, P(B | A[sub]1[/sub]), is 1.
The probability of hearing a woman if they were one woman and one man, P(B | A[sub]2[/sub]), is 1/2.
The probability of hearing a woman if they were two men, P(B | A[sub]3[/sub]), is 0.
So the probability that they were a man and a woman if we heard a woman, P(A[sub]2[/sub] | B), is
(2/4)(1/2) / [(1/4)1 + (2/4)(1/2) + (1/4)*0]
= 1/2.
Whenever you can specify “the other person” at all, the probability is 1/2. In your case, there’s one person you heard, and one person you didn’t. It doesn’t matter that you can’t match the voice to the faces, because there’s still one you heard and one you didn’t. It’s actually pretty difficult to come up with a real-world situation where the 2/3 answer applies. Here’s one I came up with:
My high school (an all-male school) throws an appreciation dinner for parents of alumni, and my mother goes. At the dinner, she sits with another lady, and in the course of dinner conversation, the other lady mentions that she has two children. My mother knows that this lady has at least one son, since otherwise, she wouldn’t be at this dinner. What is the probability that both of her children are sons?
A key point to note, here: I can’t ask about the lady’s “other child”, since I haven’t specified a child. I could say “the child who went to that high school”, but that doesn’t work, since for all I (or my mom) know, maybe both of the kids went to that high school.
Now, if later in the conversation the other lady says “I’m sorry I’m not very alert; my toddler kept me awake all last night”, then the situation changes. Now, my mom does know that one and only one of the lady’s children went to that high school, and so we now can describe the two children as “the one who went to my high school” and “the one who didn’t”. The toddler has a 50% chance of being male or female.
I understand the B/G problem. In the audible office case, I could have heard a man, but I did not. My having heard a woman makes the “mw” and “wm” cases one-half as preferred a posteriori as “ww”. In other words, there is an assumption we 1/2’ers are taking, which is that either person was equally likely to speak, but it was a woman in this particular case.
In order to get the 2/3 answer, you need to know that one of the visitors is female and nothing else. If there’s any way to distinguish the visitor whose sex you learn from the other, the probability is 1/2.
Pasta–I confused the issue there, sorry about that…please see my response to Thudlow.
Thudlow-thanks for your additional analysis, it was very clearly expressed.
However, I’m still not getting the problem with this breakdown:
A woman speaks, so it can’t be MM.
A woman speaks, so it could be MF. (If it helps, think of them as standing in this arrangement, i.e. the man on the left and the woman to the right.)
A woman speaks, so it could be FM. (Woman is on the other side of him now.)
A woman speaks, so it could be FF.
Clearly, out of the three possibilites, two involve the other person being a man. So I would conclude 2/3. What is the flaw in my reasoning here? Let’s put aside Bayes’ rule for a moment–that’s not because I doubt the method, I’m just trying to get at it from a conceptual, verbal level.
Thudlow, you say “We want to know the probability that some earlier event happened (a man and a woman walked in) given that we know a later event happened (we heard a woman speak).” But indeed, we knew (MM, MF, FM, FF), that this probability was 1/2 even before knowing that a woman spoke. So I’m not sure that Bayes’ rule is being applied to the crux of this question…the question is more closely considered as “if there are two people, and one is a woman, what is the chance that the other is a man?” This is the boy/girl paradox…what I’ve attempted to do by taking away the element of sight, and by creating a non-exclusive distinction, is to recreate this in a different form. When you say “eldest/youngest”, then it’s impossible for both participants to be either. But in this case, the gender isn’t mutually exclusive, and therefore hearing a woman doesn’t tell you anything more than that there is a woman.
MM
MW
WM
WW
If one is a woman, 2 out of the three remaining possible outcomes include a man.
If you were told only “It is not the case that the people waiting are both men”, then 2/3 is right.
But, you have learned more than just that by hearing a woman. You now know both:
(a) that it is not the case that the people waiting are both men, and
(b) that a random person chosen to speak from the pair was a woman.
ultrafilter, you’re right. my intention with the hearing example is to PREVENT that differentiation from taking place. hearing that there is a woman only tells you that there is at least one woman. so maybe it’s a problem with my wording, but again the goal of this was to be tricky. completely ambiguous is bad, but i’m okay with tricky
I told you that you hear a woman’s voice in order to establish that one of them was a woman as a GIVEN. (This was an inadequate way to do that.)
You cleverly retrofitted the problem to adjust for the possibility of that.
This strikes me as a flaw with my wording more than a flaw with my reasoning. Let’s say that just before turning the corner, the receptionist heads you off and says “One of them is a woman.” Would you still come up with 1/2 for that, or would you now see that as a 2/3 situation?
It’s the same flaw as this: I roll a die. It can either come up six, or not. Therefore the probability that I will roll six is 50%.
The possibilities that you have considered are not equally likely.
I always think of Bayes as sort of reverse probability. You hear a woman speak, and that in itself makes the FF case a bit more likely. Imagine that the only possibilities are MMMMMMMMMF and FFFFFFFFFF. You hear a female voice. Don’t you now think that FFFFFFFFFF is more likely? Not 50%, much more probable than that.
This makes it much more complicated, because now we have to deal with the motivation of the receptionist. Phrased that way, I would assume that what the receptionist meant to communicate was that exactly one of them was a woman, and that if they both were, the receptionist would have just said so.
Instead, you could have it like this:
Receptionist: Two people are waiting for you.
You: Was one of them a woman?
Receptionist: Yes.
Or, if you want to be completely thorough about it, you can ask “Was at least one of them a woman?”.
The receptionist stopping you and saying that “at least one is a woman” prevents you from learning the outcome of the random voice process, so 2/3 is correct in that scenario. This would indeed be identical to the boy/girl problem.
Your original wording allows the listener to probe the sample further, which you did not intend. However, if you just want to mess with your friends, you could keep the scenarios the same and hope that they improperly assume it is identical to the boy/girl problem. (That is, I’m assuming they are familiar with the boy/girl problem, since you aren’t messing with them using that problem. Thus, you could present these two scenarios with the right vocal inflections to make them think that the scenarios should give different answers, when in fact they both yield 1/2.)
That makes sense. But does my variation with the receptionist telling you that one is a woman correct for that? I think it does, since instead of a speaker being selected “randomly”, the receptionist makes a judgement upon seeing them, in which case the receptionist could be seeing two women, a woman and man, or a man and a woman.
You’re expecting two visitors at your office. There are two meeting rooms, a blue room and a pink room. The pink room has several posters promoting the woman’s place in corporate America. You, being the sexist pig that you are, tell your receptionist that if there are “any chicks”, they should go to the pink room with the posters, but if it’s just the good old boys, then we’ll go to the blue room and light up a few stogies. When they arrive, the receptionist calls you and tells you to go to the pink room. You’re all pissed off because you hate women, but you realize that the meeting hasn’t completely gone to sh*t. There could still be a man left. What are the chances that you’ll salvage the meeting by finding a man in the pink room?
