Sunday, May 23, 2010

Probability 2 - Boy or Girl paradox

I continue to be amazed by the fact that apparently simple probability problems are often so difficult to resolve. It is particularly surprizing in view of the fact that so much science seems to depend on probability these days: the particle physicists' results are small probabilistic anomalies in incredibly expensive experiments, quantum physicists say that nature is inherently probabilistic, the medicine we take is determined by 'random' trials, evolution apparently occurs by 'random' mutation, etc, etc. (I don't necessarily concur with all this, by the way.)

I have been thinking more about this fact since the blog post of Peter Cameron, which derived from another post by Alex Bellos. The particular problem they consider is this: if someone tells you that they have two children at least one of whom is a boy born on Tuesday, what is the probability that the second child is a boy?

Actually, the simpler version of this problem in which there is no mention of the day of birth (possibly introduced by Martin Gardner who died 22nd May), has been discussed and disputed in short and in long, in circles high and low, under the name "The Boy or Girl Paradox".

In short, my view is that the answer to this question is 1/2 (under the simplifying assumptions clearly intended; that is, that the birth rates of boys and girls are equal, etc).
However a closely related problem which I will explain would yield the answer 1/3, the answer preferred by mathematicians.

I think this simple problem benefits from the distinction between states and actions, the states being the quality of being a girl or a boy, and the actions being the declarations made. In the real world problem you need to consider the probability of the action as well as the probability of being in a state.

Let's now consider the problem in detail: the simplifying assumptions mentioned above mean that given a family with two children the following four cases are equally likely: BoyBoy, BoyGirl, GirlBoy, GirlGirl. In the case BoyBoy the likelihood of a declaration that there is a boy is 1, in the case of GirlGirl the likelihood of a declaration that there is a girl is 1, but in the other two cases there is a half chance that the declaration is 'boy', and a half chance that the declaration is 'girl'. Now let's look at the total weight of a declaration 'boy': it is 1+1/2+1/2. The weight which corresponds to the case BoyBoy is 1. Hence the proportion of 'boy' answers which correspond to the state two boys is
1/(1+1/2+1/2)=1/2.

Now a slightly different problem: suppose I interview people with two children, and ask if they have a boy in the family; if they answer yes, what is the probability that the second child is a boy?
Again the simplifying assumptions mentioned above mean that given a family with two children the following four cases are equally likely: BoyBoy, BoyGirl, GirlBoy, GirlGirl. In the case BoyBoy the likelihood that they respond 'yes' is 1, in the case of GirlGirl the likelihood that they respond 'yes' is 0; in both the other two cases the likelihood of 'yes' is 1, not 1/2. Now let's look at the total weight of a response 'yes': it is 1+1+1. The weight which corresponds to the case BoyBoy is 1. Hence the proportion of 'yes' answers which correspond to the state two boys is
1/(1+1+1)=1/3.

If there is also mention of the day of birth there are at least three interpretions:
1) a person declares that they have two children at least one a boy born on Tuesday;
2) when asked if they have a boy, the response is 'yes, and born on Tuesday';
3) when asked if they have a boy born on Tuesday, the response is 'yes'.

The probability of the second child being a boy in the first case is 1/2, in the second
1/3 and in the third 13/27.

In this post I have tried (exaggeratedly) to argue in as simple language as possible, because a simple problem should only require the appropriate distinctions, not a big machinery. However I have earlier proposed (with Sabadini and de Francesco Albasini) a mathematical context, in which many of the perplexities are seen to arise from considering (normalized) probabilities, rather than weights of actions. The precise mathematical point is that normalization does not behave well with respect to sequential operations (which include abstraction). Another simple perplexing example where abstraction does not work well with normalization is Simpson's paradox, the mathematical origin of which is the fact that not always a/b+c/d=(a+c)/(b+d).

An extreme example of Simpson's paradox is the following:
Consider treatments A,B and diseases X,Y.
Treatment A with disease X on one person cures the person (100% cure rate - best possible)
Treatment B with disease X on 99 people cures 98 (worse than 100%).
So A has a better success rate than B with disease X.

Treatment A with disease Y on 99 people cures 1 person (this is better than 0% cure rate)
Treatment B with disease Y on one person fails to cure (0% cure rate - worst possible).
So A is better than B with disease Y.


In both diseases X and Y the treatment A is better than B.

However treatment A saves 2 people in 100, B saves 98 in a hundred.
B is worse than A???

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5 Comments:

Blogger dave said...

I was interested to read in The Lady Tasting Tea that Kolmogorov was himself unhappy with the idea of sample spaces, and was trying to reformulate (again) probability.

Here's another weird probability issue: "probability dilation."

5:57 AM  
Blogger Sean Carmody said...

Bob, it's been a while since I've thought about these probability paradoxes (probably the last time would have been one of a number of extended Monty Hall discussions I've had over the years). Thinking about it some more, I'm leaning more and more to the notion that the prob=1/2 rather than 1/3 answer actually makes the most sense.

3:45 PM  
Blogger Sean Carmody said...

Bob, in the spirit of what you have written here, I just posted the following comment on Peter Cameron's blog (it is awaiting moderation):

Peter, in one of your comments you write:
"If I knew that a different algorithm was being used, that would (by an application of Bayes’ theorem) give me a different measure, and the calculation would give a different answer. But if you want a different answer to this question, you had better make this algorithm public, so that I can use this knowledge to adjust the probabilities."

The implication here is that some kind of minimal interpretation of the information contained in someone saying “I have two children, and (at least) one is a boy” is precisely the information in the formal mathematical event represented by the subset {BB, BG, GB} of the sample space {GG, BB, BG, GB}. I'm not sure that's necessarily the best interpretation.

To explain why, I'll stick with this simpler problem rather than the Tuesday version, but the same argument applies there. I will have to assume that we are working with a bigger state space, but that doesn't mean I have to start worrying about whether the person is lying or the probability that they would have said something in Klingon instead. I won't say anything yet about this bigger space other than the fact that it includes events BB, BG, etc, each of which has probability 1/4.

I'll denote by M the "mathematically formal" event represented by {BB, BG, GB}--strictly speaking this means the union of the events BB, BG, etc in the bigger space--and H the event that a "human" tells you “I have two children, and (at least) one is a boy”. The probabilities we are interested in are P(BB | M) and P(BB | H).

In both cases, we can appeal to Bayes theorem and so have

P(BB | M) = P(BB) P(M | BB) / P(M)

and

P(BB | H) = P(BB) P(H | BB) / P(H)

Now P(M) = 1/3 and P(M | BB) = 1 and so P(BB | M) = 1/3, which is the classical answer. No surprise there. What about the case for P(BB | H)? To have it line up with the M case, we have to accept that P(H) = 1/3 is a minimal, neutral probability to assign to the event that someone tells you “I have two children, and (at least) one is a boy”. Even without worrying about complex or far-fetched algorithms, that doesn't seem quite right.

I would argue that a more reasonable probability would be 1/2. One algorithm that would give this outcome is that a person with BB would say “I have two children, and (at least) one is a boy”, a person with GG would say “I have two children, and (at least) one is a girl” and a person with BG or GB would say “I have two children, and (at least) one is a boy” or “I have two children, and (at least) one is a boy” at random, each with probability 1/2. In this case P(H) = 1/2, which is evident if we calcalate P(H) = P(H | BB)P(BB) + P(H | GG)P(GG)+...
Also, since P(H | BB) = 1, we get P(BB | H).

An important point to emphasise here is that I don't really have to assume I know precisely what algorithm is being used. Rather, I need to be able to come up with the probabilities P(H) and P(H | BB). To me, values of 1/2 and 1 seem like very reasonable minimal interpretations of this information.

What does all this mean for the Tuesday problem? Here we can again distinguish the mathematical event M and the human utterance event H for the statement “I have two children, and at least one of them is a boy born on Tuesday". P(BB | M) = 13/27 for the reasons discussed in the post, but in the case of the human utterance, I think you can quite reasonably assign P(H | BB) = 1/7 (all days are equally likely to have been spoken) and P(H) = 1/14 (the day and the gender are independent), which means that P(BB | H) = 1/2!

4:25 PM  
Blogger Sean Carmody said...

Mis-typed my numbers for P(M):

Now P(M) = 3/4 and P(M | BB) = 1 and so P(BB | M) = 1/3

4:43 PM  
Anonymous Sean Carmody said...

My blog post on the puzzle is now up: http://www.stubbornmule.net/2010/06/eliminating-the-irrelevant/

10:38 PM  

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