A friend of mine showed me an interesting math puzzle on xkcd.

1. Alice picks two real numbers and puts them in (abstract) envelopes. No information is given as to how she chooses them.
2. Bob picks one of the envelopes by tossing a fair coin.
3. Alice shows Bob the number in the envelope he chose.
4. Bob is forced at gun point to say if the other number in the other envelope is "higher" or "lower" than the number he's already seen.

The question is, does Bob have a strategy which would allow him to guess if the unseen number is higher or lower with better odds than 50-50?


The xkcd post says, "yes" and gives the following strategy.

1. After Bob has seen the number in the envelope, he calculates p(x) = 1/(1+e^-x)), where x is the number he saw.
2. Bob randomly chooses a real number t from 0 to 1, uniformly.
3. If t is lower than the value calculated in 1, Bob says "lower"; if t is higher than the value calculated in 1, Bob says "higher". Thus, Bob says "lower" with probability p(x), and "higher" with probability 1-p(x), since p(x) is bounded between 0 and 1.

The post then computes the probability of guessing correctly:

So for any pair of numbers, if you call the smaller one A and the larger one B, you have a (slightly) better chance of guessing “lower” if you saw B than if you saw A. Your overall chances of guessing correctly — given that there is a 50% chance you’re seeing A and a 50% chance you’re seeing B — are:

N = 0.5 * (1-p(A)) + 0.5 * p(B)
Since we're saying A < B, we can say that N > 0.5, because we chose p(x) cleverly. Since p(x) is always increasing, p(A) < p(B). Thus 1-p(A)+p(B) > 1, so N = 0.5 * (1-p(A)+p(B)) > 0.5 * 1 = 1/2.

The result is very unintuitive, though, and the OP says that he initially thought there wouldn't be a workable strategy. So, my question is, do you buy this, or do you think there's a hole?