Answer

At first glance, it seems like the first player (we'll call them player A) will always win because they get the correct answer first.

But then we realize that player B has a turn between player A getting the correct information and getting to win, this guess will be after player B has done a single reduction (presumably of 4 characters to 2) and hence can guess at 50%. This means the game seems to surprisingly still be even with both players at 50%. But it's not.

There's actually a counter-intuitive and better strategy that player A can take to improve their odds of winning to 9/16.

The simple case, using yes-or-no questions to cut characters in half looks like this:

  1. A reduces 4->2
  2. B reduces 4->2
  3. A reduces 2->1
  4. B must guess at 50% or else A will win on their next turn

But instead of cutting the 4 characters in half, A can use a yes-or-no question to ask about 1 character. They have a 25% chance of being correct, and a 75% chance of knowing that it's one of the remaining three.

Let's look at a tree:

A reduces 4-> either 1 (25%) or 3 (75%)

If it's the 1/4 chance that they picked the right answer, then B is forced to guess (or else guarantee losing to A on their next turn). This guess is at 25%.

If it's the 3/4 chance then A has reduced it to 3 possibilities, then B has even odds:

  1. B reduces 4->2
  2. A does not want to pick at this point (it would be 1/3 chance of winning). Instead,
  3. A reduces 3-> either 1 (1/3) or 2 (2/3)
    1. If it's the 1/3 chance, then A now knows the correct answer, and B must guess between 2 at 50% odds
    2. If it's the 2/3 chance, then A knows it's between 2, so whether B guesses or let's A guess it's 50% odds.

But that means it's a 25% chance that B gets only 1/4 pick, or a 75% chance that they get 1/2 odds.

Chance of B losing / A winning: 1/4*3/4 + 3/4*1/2 = 9/16