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:

- A reduces 4->2
- B reduces 4->2
- A reduces 2->1
- 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:

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

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