This is Problem 1.25 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.
You are handed two envelopes, and you know that each contains a positive integer dollar amount and that the two amounts are different. The values of these two amounts are modeled as constants that are unknown. Without knowing what the amounts are, you select at random one of the two envelopes, and after looking at the amount inside, you may switch envelopes if you wish. A friend claims that the following strategy will increase above $1/2$ your probability of ending up with the envelope with the larger amount:
Toss a coin repeatedly. Let $X$ be equal to $1/2$ plus the number of tosses required to obtain heads for the first time, and switch if the amount in the envelope you selected is less than the value of $X$ . Is your friend correct?
The answer given in the solution manual claims that this indeed helps, and that the probability of getting the better envelope is given by $$p = \frac{1}{2} + \frac{1}{2} P(B)$$
where $B$ is the event that $a<X<b$, with $a,b$ being the smaller and larger amount of dollars, respectively.
I do not buy this solution for the following reason: tossing a coin has nothing to do with the contents of the envelopes. You do not gain any information by doing it. You could just as well count the amount of leaves on a nearby tree instead and use that for $X$.
Similarly, opening the first envelope also gives you no useful information about the ordering relation between $a$ and $b$, so surely that's another red herring. Even if you forget the coin tossing, the probability of "winning" is still $1/2$, swap or no swap.
I suppose maybe the catch is in interpreting the following sentence: "The values of these two amounts are modeled as constants that are unknown". I take it to mean that they're just two randomly and independently generated numbers.
Am I out of my mind? Surely the solution manual is wrong.
The manual is right. The coin flips just generate the number $X$. If both envelopes contain numbers smaller than $X$, you have a random choice among the envelopes because you will take the second one. If both envelopes contain numbers greater than $X$, you have a random choice because you will take the first one. If one is greater than $X$ and the other is less, you will certainly pick the correct one, so if there is any probability this situation obtains you have a strictly greater than $\frac 12$ chance of picking the correct envelope. You can generate $X$ any way you want with the same effect as long as it has positive probability to be in each interval $(k,k+1)$. The argument tacitly assumes that it is possible one is below $X$ and one is above $X$. The reason to do the coin flipping is to get a distribution where all naturals plus $\frac 12$ have some chance to be chosen. This guarantees that there is some chance $X$ is between the two numbers.