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There are two players, and each one has a die with six sides from $1$ to $6$. The probability of each side landing is equal. Now, the two players roll their dice, and they only know the number of their own die. They will propose prices in turn, until one of them doesn't provide a higher price. The winner will get the money equal to the sum of these two dice minus the price they provided.

What is the optimal strategy for playing this game?

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  • $\begingroup$ can any number be bid or only integers? $\endgroup$
    – Jagol95
    Dec 15, 2018 at 12:45
  • $\begingroup$ Does the loser in the auction get $0$ or the negative of what the winner gets? $\endgroup$
    – Henry
    Dec 15, 2018 at 16:40
  • $\begingroup$ I think any number can be the price. @Jagol95 $\endgroup$
    – J. Z
    Dec 21, 2018 at 6:34
  • $\begingroup$ 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 $\endgroup$
    – J. Z
    Dec 21, 2018 at 6:36
  • $\begingroup$ Suppose the winner proposed $5$ and the dice are $3,1$. Does the "winner" lose $1$? $\endgroup$
    – Ross Millikan
    Jan 15 at 3:48

1 Answer 1

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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

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