Here is the analysis using Bayes’ Theorem of the probability puzzle I posted below. In this post I’m changing the puzzle so that there’s an odd number of tosses, which just rules out the need for a special case when there’s an exact 50:50 split. Here’s a shortened (and maybe clearer) statement of the puzzle.
I go into another room and do the following, all of which you fully know that I am doing. I flip a coin 101 times. I keep track of the number and outcome of each toss. I randomly select 51 numbers corresponding to tosses that landed with the majority outcome, generating a set S. When I return I list the numbers in S and tell you the outcome of the corresponding tosses. This process returns an output of the following form <S, ____>, where S is a set of 51 numbers and ____ is either heads or tails, depending on the outcome of the tosses corresponding to numbers in S. For example, I might return and tell you that all of the odd tosses landed tails: i.e., <{odds<102}, tails>. The question is this: Suppose I tell you that all of the tosses corresponding to numbers in some set S landed tails: i.e., assume the outcome <S, tails>. Should your probability that all of the tosses corresponding to numbers not in S were heads be higher, lower, or the same as your probability that all of the tosses corresponding to numbers not in S were tails?
I say it should be higher. Consider the following propositions.
p = “The outcome <S, tails> was produced by the process described above.”
q = “The distribution was such that all of the tosses corresponding to numbers in S were tails and all of the rest landed landed heads.”
r = “Every toss landed tails.”
My claim is that P(q|p) > P(r|p). As I explained in the previous post this looks like it might be an instance of the gambler’s fallacy, but using Bayes’ Theorem we can prove that it is not.
By Bayes’ Theorem:
P(q|p) = P(p|q) * P(q) / P(p)
P(r|p) = P(p|r) * P(r) / P(p)
Since the unconditional probability for any particular sequence of outcomes is the same as for any other P(q) = P(r), so to prove P(q|p) > P(r|p) it suffices to show that P(p|q) > P(p|r).
To begin, P(p|q) = 1. Why? If q is true then S is the only set that could be produced by randomly selecting 51 numbers corresponding to tosses that landed tails.
Next, P(p|r) = 51!*50!/101!. Why? If r is true then the chance that S is the set generated by randomly picking from numbers corresponding to tosses that landed tails is calculated thusly: On the first draw the chance of drawing a number in S is 51/101, on the second 50/100, on the third 49/99, etc. . . take the product.
Hence, P(p|q) > P(p|r); so P(p|q) * P(q) / P(p) > P(p|r) * P(r) / P(p); and P(q|p) > P(r|p).
The difficult part of this puzzle is stating p, q, and r in the correct way. In particular, if you update only on the information that all of the tosses corresponding to numbers in S were tails you will be tempted to think that the correct answer commits the gambler’s fallacy by supposing that the outcomes of tosses corresponding to numbers not in S are dependent on the outcomes of tosses corresponding to numbers in S. However, that is not all of the information that you have. You also know how the outcome <S, tails> was generated and the requirement of total evidence demands that you update on a proposition that expresses this information.
BTW, I came up with this puzzle on a car trip back from a conference, but it has a ring of familiarity. This leads me to suspect that I had seen it somewhere before and have forgotten where. It’s certainly just the kind of example one would use to illustrate the usefullness of Bayes’ Theorem and the requirement of total evidence. So, if anyone knows where I might have read this puzzle or something like it before, I’m interested to find out. Also, if someone can confirm that P(p|r) = 51!*50!/101! is correct, I’d be much obliged. My confidence with combinatorial/probability expressions is a little low, not having ever taken a course in probability or statistics. In any case, it’s clear that P(p|r) < 1.
I think that the commenters that pointed out that the lecturer is inconsistent in double-counting reflections make roughly the same point but in a less logicky way.
jrshipley
The good news is that the new problem statement is easier to follow (and agrees with what I thought you meant).
The bad news comes in two parts. The first is that the analysis is too complex. The second is that the answer is not correct.
In general, the probability of any number of heads greater than 50 is 0, and the probability of any smaller number of heads, h, is C(h,101)/(2^100). Notice that the sum of the probabilities is 1.
In particular, the probability of 0 heads is 1/(2^100) and the probability of 50 heads is C(50,101)/(2^100).
Why? Because the selection of S removes half of the distribution. We choose binomially. And then we announce which side of the distribution the outcome lies in.
The denominator has just been cut in half.
IOW, we could ask, what is the probability of all 0 heads given that the majority were tails? What is the probability of 50 heads, given the majority were tails.
(the selection of S is not a random selection from all 101 coins, but only from those in the majority pool)
Jonathan