# Bet on the sum of two dices

There are two players and each one has a dice with six sides from 1 to 6. The probability of each side is equal. Now two players roll their dice, and they only know the number of their own dice. They will give a price of the sum of these two dices in turn, until one of them doesn't provide a higher price. The winner will get the moeny equals to the sum of these two dices and pay the price he/she provided. What is the optimal strategy of playing this game?

• can any number be bid or only integers? – Jagol95 Dec 15 '18 at 12:45
• Does the loser in the auction get $0$ or the negative of what the winner gets? – Henry Dec 15 '18 at 16:40
• I think any number can be the price. @Jagol95 – J. Z Dec 21 '18 at 6:34
• The loser's payoff equals to zero as he/she won't pay any money and won't get anything. The winner's payoff equals to (sum of two dices-the bid price)@Henry – J. Z Dec 21 '18 at 6:36

## 1 Answer

I don't know much game theory, but here is a way that it could maybe play out.

First player bids his number plus a uniformly random number from (-3,3) ( 6 is the highest 1 the lowest). Second player bids the first persons bid plus a uniformly random number between 0 and his number. Now the first player infinitesimaly overbids x*33% of the time if not he passes. x is (the expected value of the opponents roll + my roll)- the previous bid. The opponent now uses the same strategy and so on.

I have no idea how one could go about solving this game analyticaly. I just tried to intuitively think up this solution based on the assumptions that the game should terminate and each players move has to involve a random choice