Thursday, September 25, 2008

The Sibling Sex Test and other Mathematical Stories

I'll cut to the chase just this one time- and only because I'm so excited to write about this!

A friend asked me this puzzle today:
"I have two children.. one is a boy. what is the probability of the other being a boy as well?"

My immediate reaction was: "They are independent! So 1/2!"

This seems a bit counterintuitive. Imagine you tossed a coin 20 times, and got H 19 times. If someone asked you, "So, are you going to get 20 heads?", you'd shake yours and say "Nah, 19 is enough of a fluke, 20 would be awe-fking-some but impossible." Thus we are implying that the probability of the 20th toss is dependent on the results of the earlier 19 tosses. But that's human psychology. In reality they are just independent.

Or so I was mulling until the friend came back saying, "No! you are wrong, its 1/3." And my attention was dragged towards a Wikipedia-take on the issue.

Now, ladies and gentlemen, I must diverge from the topic and address this crazy rise of Wikipedia-reliance. Just to be clear, I love Wikipedia-surfing. The sheer joy of hopping from one all-you-can-find-on-topic-X page to another is only paralleled by the glee one experiences when one proudly rattles off recently-Wikied information to a guileless hottie at the pub. Get me no wrong, Wikipedia has been one of the most amazing new ventures in the last decade- by warehousing, presenting and classifying information, it has enabled us and opened our minds.

But we often forget that Wiki's open source strength is itself its biggest drawback. It really doesn't matter when it comes to finding out who appeared on the Playboy cover in April 1984, but when you are talking Maths and famous conundrums, the wwwikipedia reliance could be a bit stuttering. As has been proved in this case.

And this is what it said:
"For a single birth, there are two possibilities (a boy or a girl) with equal probability. Therefore, for two births, there are four possibilities: 1) two boys, 2) two girls, 3) first a boy, then a girl, and 4) first a girl, then a boy; all of them have equal probability. We are given that one of the children is a boy. Thus, only one of the four possibilities -- two daughters -- is eliminated. Three possibilities with equal probabilities (1/3) remain. Out of those three, only one -- two sons -- is what we are looking for. Hence, the answer is 1/3."

That did stump me for a while, until a Chai latte later, I discovered the fallacy in the above. Can you?

* WARNING, DISCLAIMERS AHEAD *



Yes, you got it! (Or you are too bored)

Assume there are 2 variants of the problem.

Variant 1: (the puzzle I asked above)
The puzzle says "THE FIRST ONE of them is a boy" (i.e permutations)

Hence your sample set in this case isnt BB BG GB. Thats wrong. You know that the 1st one is a boy. the second one cud be a B or a G. hence sample set is BB and BG simply. GB is ruled out

Variant 2:
Now suppose you were told that "ANY ONE of the 2 children is a boy. what is the probability that the other is a boy?"

Now this is combinations. your sample set now is BB and BG (BG = GB in this case) and the answer again is 1/2.

What I'm trying to get at is that the incorrect explanation (which leads to answer 1/3) selectively uses permutations and combinations hence resulting in an irrational answer. If you are assuming that the order matters then the case GB cant exist. if you are assuming that the order doesn't matter, then BG = GB and GB is thus a simple repetition of BG. In either case the answer is 1/2. And this is consistent with my 1st logical answer i.e 1/2.

I was surprised to find a lot of debate on this on the net- and quite a bit of it wasteful. For example, there has been talk of frames of reference, of ambiguous meanings of "one of them is a boy" etc etc. And people have gone into discussions of conditional probability v/s absolute probability and have defended 1/3 on the basis of that. But that's absolute trash. The two can often be different, yes, but only when events are dependent on each other. In the case of independent events, they have to be equal! Just because you get an erroneous answer of 1/3 you can't defend it by saying that probability depends on sample sets and hence depending on which sample set you choose, you have a different probability. That's wrong. Whatever sample set you choose, your answers must be consistent.

I was given yet another argument by the friend's random friend: BG and GB aren't the same because the boy being elder to the girl is different from the girl being elder to the boy. This is another argument citing permutations.

Again on first glance, you can make out that this is specious, because ages of the 2 siblings shouldn't make any difference to their genders. You haven't been told which sibling is the boy, the elder one or the younger one. Thus age cannot be a basis of permutations.

More formally, the error in this argument is this: If you really wanted to complicate matters in your head and include age, you should consider it completely. Hence if you considered the various combos as Elder/Younger, the total combos would be BB,BG,GB,GG.

Of this one is a boy, but as we said, it could be the elder one or the younger one. Hence there are 2 possibilities, and the final probablity is given by:

P(2nd kid is boy/1st kid is boy) = P(1st kid is elder)*P(2nd kid is boy/1st kid is boy and hes elder) + P(1st kid is younger)*P(2nd kid is boy/1st kid is boy and hes younger)

There's an equal chance of the kid being elder or younger, hence both probablities = 1/2

Hence P= 1/2*(2nd kid is boy/1st kid is boy and hes elder) + 1/2*P(2nd kid is boy/1st kid is boy and hes younger)

Possibility 1: the boy is elder
In this case BB and BG remain and the answer is 1/2

Possibility 2: the boy is younger
In this case BB and GB remain and the answer is 1/2

and the final answer is 1/2 :)

Of course the clever mind will realize that all I've done is prove that the gender probablities are independent of the age.

Probability theory can get really overwhelming at times- conditional probabilities, what sample sets to choose, independent or dependent events- all cause a lot of confusion. But at the end of the day, there is one thumb rule to mastering it: it should all tie up to common sense. Or else the chance that the purpose of mathematics is lost is unity :)

5 comments:

prolificd said...

that's like drawing 2 balls out of the urn at the same time. isn't it? rather than one at a time. with an equal number of both and with replacement.

Atticus said...

hey prolificd, thanks for posting

i dont know which part you are referring to, but this problem can be best understood and grasped by realizing its difference from the usual urn problems.

In the urn problems, the act of taking 1 after the other introduces conditional probability, since the outcome of the 1st draw affects the probablity of the 2nd. this is BECAUSE THE URNS HAS A FINITE NUMBER OF RED AND BLACK BALLS.

In this case however, thats not true. Irrespective of whether you considered them as coincident events, or one happening after the other, the urn has infinite number of boys or girls so as to speak (we arent drawing boys and girls from a finite set of population), and hence irrespective of the 1st kid being a boy or the girl, the probablity of the 2nd kid remains 1/2.

Chrysalis said...

I feel so happy that someone out there agrees that wikipedia though very factual should be used in right proportions while backing up the argument.

p.s you probably arent used to me using my msn id chrysalis before, so just to let you know this is attiya here!

Atticus said...

Oi Attiya! yeah it seems Britain nearly was legalizing the use of wikipedia in classrooms until someone protested...

Anonymous said...

Your blog keeps getting better and better! Your older articles are not as good as newer ones you have a lot more creativity and originality now keep it up!