A Numbers Guy Quiz on Probability, from (you guessed it!)
The Numbers Guy at the
Wall Street Journal. The post gives an eight-question quiz derived from problems in Leonard Mlodinow's new book,
The Drunkard's Walk: How Randomness Rules Our Lives. I claim that the people who designed the cover of that book got the idea for the cover from the title of my blog; it has dice on it, and since a lot of people's conception of God is invisible, I'll claim it has God on it as well. (But I really can't complain, since my title's cotaken from Einstein anyway. Yes,
cotaken, as in I took the complement of what Einstein said; there are no arrows getting reversed involved here.) Carl Bialik, who writes that blog, will be interviewing Mlodinow next week. Mlodinow apparently
has a PhD (from the title of his thesis,
The Large N Expansion in Quantum Mechanics, I'd have to guess it's in theoretical physics) and is also
a screenwriter, which probably makes for a good story in itself; if I remember I'll link to the interview when it comes out.
The hardest of the eight questions, I think, is this one:
You know that a certain family has two children, and you remember that at least one is a girl with a very unusual name (that, say, one in a million females share), but you can’t recall whether both children are girls. What is the probability that the family has two girls — to the nearest percentage point?
I won't give an answer, and I ask that you don't either -- but think about it; the answer is surprising. (Bialik says he'll be giving the answers next week.)
30 comments:
Link again next week when he posts answers?
Don't we have to assume something about whether this family independently draws names for their children from the distribution of all children's names?
It seems too foolish to ask this as a math question with out giving guidance for assumptions in such a psych dominated topic. Being given a name like `emily' would be a lame move by parents.
In the inaugural Carnival of Mathematics, Heath Raftery had a post which dealt with essentially the same issue but in a different guise. That post generated a lot of comments, many of which were trying to explain to Heath why he was wrong (he wasn't). If I remember correctly, at some point in the discussion he brought up this problem of the family with the two kids (worded somewhat differently) in order to give another example of the same phenomenon.
50%
The answer I got doesn't seem very surprsing to me, so I'll look forward to being surprised next week.
Forget about names. Just write:
"You know that a certain family has two children, among them a girl which is blind, but you can't recall . . ."
Anonymous,
It certainly is not 50% !
Sure is 1/3 = 0.333333...
Infact these are the possibilities:
MF - FM - FF -> 1/3
ERRATA:
(IMHO) the answer is 0.66
but these are the possibilities:
1) FF -> I've seen the first of two sisters
2) FF -> I've seen the second of two sisters
3) FM -> I've seen the only sister
so -> 2/3 -> 0.66
riemann: Using your argument, option 3 should actually read,
3) FM -> I have seen the only sister who is the older one.
to which you will need to add a fourth option:
4) MF -> I have seen the only sister who is younger.
Then, we get 2/4 = 50%
No. I told the "unique" sister not the "older" sister.
(I'll give a look at your nice blog)
Bye
riemann: That's exactly what I said:
"The only sister (which is the same as the unique sister) but then now I need to consider the two obvious cases: younger and older!"
I don't understand your remark, the question is not about the age but about the possibility the family has two girls.
riemann: Vishal is interpreting "FM" as "older F, younger M" rather than "one F, one M".
OK! F = Female, M = Male. Sorry!
Dr Armstrong: Thanks for helping me clarify that point!
riemann: I read that problem in the following way. The sample space is the set S = {MM, MF, FF, FM}, which takes (in fact, should take) into account if a sibling is younger or older. Now, the statement "given that I saw one female, what is the probability that both are females" is equivalent to computing the probability P({FF}/A), where A = {MF, FF, FM}, i.e. we are asking the question, given that event A has occurred, what is the probability of the occurrence of event {FF}? And, from here, it is easy to calculate the required "conditional probability'.
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I am not saying you are wrong! I think you have constructed the sample space in a different manner, and hence your final answer will be different. In fact, having different sample spaces for the same problem may lead to different answers as is evident in the famous Bertrand's paradox.
riemann: And, I wish to add one more (subtle?) point. We usually construct the sample space before making any "observation" (which in this problem is the same as observing that I saw one female child).
I think in your solution, you are constructing the sample space after making that observation!
Hmmm, so far the values 1/3, 2/3, and 1/2 have been ventured as the answer. I think it's worth mentioning that in the Number Guy quiz, items 2 and 3 are very similar (#3 is the one Isabel quotes). If it looks like your model of the problem would fit #2, then I think it's safe to say that your answer will be wrong for #3. (That is, if you accept that those are indeed two distinct questions. The way #3 is worded, I think it's somewhat ambiguous, although I've seen other versions of this problem stated where it is unambiguous what the interpretation should be.)
Let me think another bit... I have not yet seen the other two questions.
Hmmm... my solution is indeed for problem 2. But, I really wonder how the clause/condition "girl with a very unusual name" in problem 3 changes anything at all! Of course, I could be missing something, in which case, it would be great if Isabel threw some light on that strange condition!
Anyway you cannot consider the possibility MM anymore 'cause your observation has excluded it! I think you were talking about "a priori possibility" which is not the case.
So conditional probability takes you to the right answer...
riemann: I see what you are saying. But, my earlier point had to do with first constructing the sample space, i.e. determining the set of of all possible outcomes before any observation is made. In my solution, the sample space I provided was {MM, MF, FF, FM}. But, I couldn't determine the sample space in your solution. If you could provide one, then I think our disagreement will be over. And, I do agree we could have different answers depending on what sample spaces we choose.
(My solution is really for problem #2. Problem #3, I admit I find it confusing.)
are you sure? #2 reads:
2. You know that a certain family has two children, and that at least one is a girl. But you can’t recall whether both are girls. What is the probability that the family has two girls — to the nearest percentage point?
So in this case isn't the sample space = {MF,FM,FF}?
riemann: if we take {FF, FM, MF} as our sample space, then the probability that both the children are girls is of course 1/3, as you will agree.
However, if we choose {FF, FM, MF, MM} as our sample space, then we are computing the conditional probability P({FF}/{FF, FM, MF}), which is again 1/3. So, our answers will be the same! It's just that I am calculating the probability using a different route!
I understand! So what's your answer for #3?
#3 is just crap! :) Unless someone points out some obvious thing that I have been missing all along!
Vishal: the technical term is "red herring".
Dr Armstrong: Touché! :D
Infact the strange name is a non sense. Not really important for the answer. Also the age! But if you try to think plainly the answer is 1/2 = 50%. Infact: given you have seen a girl what is the probability for the other child to be a girl? Nothing else that 1/2. But, what's the strange about this reasoning? It's the first obvious answer given here.
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