# Guess Who?

(Hasbro’s) Guess Who? is a “search race” game between two players. Each player selects a character from a pool of 24 cards. The objective is to be the first to find the opponent's character by asking in turn YES/NO questions about the (physical) attributes of the character.

We would like to know which gameplay strategy is the best for this game? Let's consider with the most intuitive one: asking question that eliminates half of the candidates. Since we always eliminate half of the candidates then of course the expected number of eliminated possibilities is 1/2. Is this the best we can do, what about other strategies, for example what if we ask a question that splits the candidate list into \( \frac{N}{3}\) and \( \frac{N}{3}\) where \(N\) is the number of total candidates. The expected number of eliminated candidates is $$ \frac{\frac{N}{3} \frac{2N}{3} + \frac{2N}{3} \frac{N}{3}}{N} $$ which can be simplified to \( \frac{4N}{9} \), and that number is less than \( \frac{N}{2} \). The general formula for the expected number of eliminated candidates if we ask splitting the list into \( \frac{N}{x}\) and \( 1 - \frac{N}{x}\) is $$ \frac{ \frac{N}{x} \left( 1 - \frac{N}{x} \right) + \left( 1 - \frac{N}{x} \right) \frac{N}{x} } {N} $$ when we simplify the terms we end up with: \( 2N \left( \frac{1}{x} - \frac{1}{x^2} \right) \). Then which number \(x\) number maximizes the expected value? When we take the derivative with respect to \(x\) we get \( 2N \left( \frac{-1}{x^2} - \frac{-2}{x^3} \right) = 2N \left( \frac{2-x}{x^3} \right) \). Then we have \( x^{\star} = 2 \) which verifies that our intuition at the beggining was right.