# Optimal strategy of a dice game

$$A$$ and $$B$$ are playing a dice game. They both roll a standard 6 sided dice once, but they cannot see the outcome. Then, there is a box containing money equal to the sum of the outcome of the two dices. Then $$A$$ and $$B$$ need to bid to buy the box.

For example, if $$A$$ rolls 3 and $$B$$ rolls 4. Then the box will contain 7 dollars. However, for now, without knowing any outcome of their rolls, $$A$$ needs to give a price to buy the box. Then $$B$$ can give another price. The one with the highest price will get the box and dollars in the box. Do you want to be the first to bid or the second to bid? What is the optimal strategy?

If now $$A$$ and $$B$$ can see the outcome of their rolls respectively, but cannot see the other player's outcome. Do you want to be the first to bid or the second to bid? What is the optimal strategy?

I suppose for the first scenario, being the first to bid is advantageous since you can bid 7, which is the expected value of the sum of two dices and there is no reason for the second player to bid a price higher than this.

For the second scenario, being the second to bid seems advantageous since you can obtain information from the bid of the first player.

• Actually in the first scenario there is no advantage to bidding first. The first to bid will bid 7 and the expected value for buying a box for 7 is 0, so no advantage gained. Oct 24, 2019 at 8:31

• Firstly, if Alice bids $$d$$ on the box, then Bob will either let Alice buy the box or he will outbid Alice buy a negligible ammount (say one cent). For simplicity, I assume that he gets to buy the box for $$d$$ dollars.
Now, let $$A$$ and $$B$$ be the values of the dice rolls of Alice and Bob and $$d ∈ ℝ$$ the amount that Alice bids. Bob does not know $$A$$, but knowing her strategy, he can derive $$E[A\mid d]$$, the expectation of $$A$$, given $$d$$. The choice that presents itself to Bob is now very simple: He can choose to buy the box for $$d$$ dollars, which is a gamble with expected payoff $$p = E[A \mid d] + B - d$$ that he can calculate. If $$p > 0$$, he will take the gamble, if $$p < 0$$ he will leave it and essentially force this gamble on Alice.
Now what about Bob's payoff, assuming that Alice plays some optimal strategy for herself? To ensure that her expected payoff is not negative, Alice's strategy must yield $$p ≥ 0$$ with probability $$1$$. Since $$B$$ is independent of $$d$$ and $$E[A \mid d]$$, this requires $$E[A \mid d] - d ≥ -1$$ with probability $$1$$. Thus $$E[p] ≥ E[B] - 1 = 2.5$$. An optimal strategy of Alice where this holds with equality is the one where she always bids $$d = A + 1$$. In that case Bob always buys the box and gets a payoff of $$B-1$$.