Dice game: bidding for sum

Question

Consider the following two-player game.

• Both players roll a fair die. They can see their own roll, but not their opponent's roll.

• Then, both players simultaneously choose a (possibly fractional) amount to bid.

• Whoever bids higher wins the sum of the two die rolls minus their bid, while the lower bidder gains nothing. When they bid the same amount, both players get nothing.

What is the best strategy for this game? What are the Nash equilibria?

Let n be the number you roll, then the random variable for the sum is uniform between $[n+1, n+6]$. The expected value for the opponent’s roll is 3.5 and so the expected sum would be $n + 3.5$, I will bid as close to the expected sum as possible (so the first whole number that is below the expected sum $n + 3.5$), which would give me an expected payoff of $0.5$. Is this solution correct? I don't know if adding the expected value part is the best strategy, but I assume I want to lower the chance of the opponent getting around the expected value.

Edit: The question was not clear about whether the bid is announced to the other player, but since if it is announced we have to consider the sequence (whether I am first or second), I assume the bids are secret. Though I want to know what the strategies would be for the two different cases (secret and non-secret), if I can choose to bid first/second? You can bid any amount and the opponent is a rational player.

To give an example of the run, you got 3 and your‍‍‌‌‍‌‌‌‍‌‌‌‍‍‍‍‌‍‍‍ opponent got 6, and if you bid 7 and the opponent bids 8, whoever wins the bid will get 9 points minus the points they bid, so the opponent wins and get 1 point in this case. If there is a tie in the bids, then there is no prize. If you bid a 10 instead of a 7, then you would win the bid, and get -1 point.

Answer 1

We are looking for a Nash equilibrium, which I think means that both players are assumed to have the same strategy and we look for the strategy which maximises the player's expected profit. By symmetry, this will also maximise the opponent's expected profit. This is fine: both players can have a positive expected profit as this is not a zero sum game.

To aid our understanding, the following table assumes that the strategy is $bid=n+1$. In other words the player (and the opponent) bids 1 more than their roll. This is not a serious candidate for an optimal strategy, it is just for illustration.

We can now play with the strategy (yellow box) to try to increase the expected profit from 0.56. We make the following observations, for any strategy:

• Of the 36 possible outcomes, 6 of them will be a tie, resulting in zero profit.
• Of the 30 remaining outcomes, half will result in a win for the player and half in a win for the opponent.
• It is inefficient for two of the bids to be the same. If they were the same (say we bid the same stake when we roll a 4 or a 5) then this increases the number of ties from 6 to 8, reducing the prize pot. Thus the optimal strategy must have 6 different bids.
• It makes sense for the bids to get bigger as the roll gets bigger.
• Reducing the bid for any roll by a small amount (not so much as to match another bid) increases the expected profit because it doesn't change the winner but does reduce the bid.
• We therefore arrive at a somewhat degenerate strategy, where the bids increase with the roll, but only very slightly:

Solution: $bid = n \times 0.01$

We can reduce the $0.01$ increment to approach a sort of supremum optimal solution. In the limit, the expected profit is $35/12 \approx 2.92$.

This isn't necesarily the strategy that the questionner sought, but I think it is consistent with the rules as presented.

It may seem surprising that one of the players does not increase their bid in order to win a higher number of cases. But of course the strategies are assumed to be symmetrical, so it is not possible to gain an advantage in this way. To maximize their return, the players simply reduce their bids and share out the proceeds.

Answer 2

A "strategy" in the simultaneous move game consists of a function $f(n)$ that specifies what a player will bid for each possible dice roll $n$. (So just a list of six numbers.)

A Nash equilibrium is a pair of strategies such that neither player could unilaterally improve his payoffs by changing to a different strategy (holding the other player's strategy fixed). In other words, each player's chosen strategy is a "best response" to the other player's chosen strategy.

One possible Nash equilibrium (not the only one) is for both players to always bid "100" regardless of their dice roll. Both players will get zero payoff, but neither can improve his payoff by unilaterally changing strategy.

You will notice that, although this is a Nash equilibrium, and both players are playing a "best response," it would be strange to claim that bidding 100 is the "best strategy."