Trying to solve a game theory problem with my son. The game goes like this: Players $A$ and $B$ each simultaneously pick a number between 1 and 100. The game is a draw if both players pick the same number. Otherwise, the player who picks the smaller number wins unless that smaller number is one less than the opponent's number, in which case the opponent wins. Find the optimal strategy to win this game. Intuitively I think $(1,3)$ is the best combination ($A$ plays 1, $B$ plays 3) but I'd like to express and validate this mathematically and my game theory knowledge is very rusty. Any pointers on how to find the best strategy that maximizes the chances of winning is much appreciated.
Players $A$ and $B$ each pick a number between 1 and 100. The game is a draw if $A=B$, A loses if B-A=1, A wins if B-A>=2. the best strategy
game theory
Best Answer
Assuming you are trying to maximize expected wins-losses, this game reduces to Rock Paper Scissors with $1$, $2$, and $3$, since $3$ beats $2$ beats $1$ beats $3$, so choosing each number with probability $\frac{1}{3}$ guarantees a win-loss expectation of at least $0$, and strictly greater if the opponent ever deviates by choosing any number greater than $3$. Also, choosing the weights to not be equal allows an opponent to deviate by choosing whatever number beats the number you most frequently pick.*
As the game is symmetric and $0$-sum, this is indeed the best you can do, so the unique equilibrium strategy is for each player to pick $1$, $2$, and $3$ each with probability $\frac{1}{3}$.
*clarification
In the event you deviate by picking two numbers with equal frequency greater than $\frac{1}{3}$, for example picking $1$ and $2$ each with probability $\frac{1}{2}$, opponent can still deviate by picking whichever number beats the other one (in this example $2$.)