# Optimal stategy for 2-players d6 game? [closed]

I came across the following problem when preparing for an interview:

Two players each roll a d6, and are not able to see each other's rolls. The player with the higher value wins 1\$(no win in case of draw). After the players roll their dice, each player may either pay$0.25 once to increase their individual roll by 2 or keep their roll(5 becomes 7 and 6 becomes 8). What is the optimal strategy and payout?

The problem is identical to the one stated here: link. However, the optimal strategy I am requested to find should revolve around finding a threshold T so that we pay $0.25 if we roll T or below and we do not pay if we roll above. This is different from the more general strategy proposed in the linked question. The game also shares some similarities with the one proposed here: link. Please help me derive the optimal strategy for the game, thanks! • You should write what you have tried. A good place to get started maybe is to have one player use a fixed strategy, the second player use some naive strategies, and work out the expected winnings for player 1 with each of those strategies. Try and play around with it and post your results so that people can provide more feedback Commented Dec 7, 2022 at 1:43 • Thanks for the suggestion! I in fact tried a few combinations of strategies for both players but I could not find an optimal one. For example, if T=0 for both (0,0), then a player would be incentivized to set T=5 (0,5), so not an equilibrium. From that, the other player would be incentivized to set T=6 (6,5). Then, however, (6,6) becomes better, which leads again to (6,0) and (5,0), assuming I didn't make mistakes. (4,4), (3,3), (2,2) and (1,1) all improve when one player decides to increase its threshold by 1, eventually leading back to the previous loop. Commented Dec 7, 2022 at 11:45 • When a player wins$1, does that come from the other player, or from the house? (I assumed the house in my answer). Commented Dec 8, 2022 at 4:45
• Yes, it does come from the house Commented Dec 9, 2022 at 15:50

This question seems to assume there exists a single "optimum" strategy. There is no dominant strategy in this game, and it seems there is no symmetric pure-strategy Nash equilibrium either. We could find a mixed strategy Nash equilibrium (which always exists, as proved by John Nash), but it is hard to argue that that is the "optimal strategy" in a game like this.

To see that the game has no pure-strategy symmetric equilibria, consider 3 possible cases. ("Pure strategy" means that each player can base the decision to pay on their dice roll, but does not further randomize the decision.)

Case 1: Both players pay whenever they roll a 1.

This can't be an equilibrium, because player A could increase payoff by NOT paying when he rolls 1. Conditional on rolling a 1, this would save A 0.25 for any roll by B, but would gain at most in the $$1 \over 6$$ of the cases, where B rolls a 2 (or possibly not even those if B is paying when B rolls 2).

Case 2: Both players don't pay when they roll 1 but do pay when they roll 2.

This can't be and equilibrium because player A could increase payoff by NOT paying when he rolls 2. Conditional on rolling a 2, this would save A 0.25 for any roll by B, but would gain at most in the $$1 \over 6$$ of the cases, where B rolls a 3.

Case 3: Both players don't pay when they roll 1 OR 2.

This is not an equilibrium because now A could increase payoff by paying when he rolls a 1. It would cost 0.25, but would gain in the $$2 \over 6$$ cases when B rolls 1,2.

Since these cases exhaust all possible symmetric pure strategies, there is no symmetric pure-strategy Nash equilibrium of this game.

A final comment: in some sense what is "optimal" would be for both players to agree to pay zero for any roll. This would maximize joint payout, and would even be a possible Nash equilibrium outcome in the repeated game.

Imo the optimum strategy here would be to pay $$\0.25$$ when we roll a number a number greater than $$3$$ and otherwise not.
That is assuming that both the players are equally intelligent and competent.
More importantly, that one person's gain is other's loss and vice-versa.

Here's how:

The following table provides the expected return for player A in case where none ever pay $$\0.25$$, only one of them pays the amount or both of them pay the amount.

In the above table, for example, when player A rolls a $$3$$ and also pays $$\0.25$$ (thus now effectively transforming the number on the die to $$5$$) then the expected return is calculated as: $$\frac46\times1+\frac16\times(-1)-0.25=0.25$$ (here player B pays nothing for any roll)
In a similar manner, the values have been calculated for other cases as well.

But at this point, our intelligent players must decide whether to pay $$\0.25$$ for $$3$$ or not.
Note that, from the table, we can see that player is getting negative return anyways for rolling $$1$$ or $$2$$.
For $$4,5,6$$ it's wiser to pay $$\0.25$$ to maximize gains or minimize losses.
Now, if I were De Broglie then having full faith on symmetry, would have said that for $$3$$ it's best to not pay as we are paying for $$4$$ but I was a little skeptical and decided to check.

Here are the results:

In the above table, the case $$1$$ is where both the players decide to not pay when rolling a $$3$$ and in case $$2$$ both decide to pay for a roll of $$3$$.

As you can see, the expected return is negative and thus it's rather better not to pay anything for $$3$$.

Hence, we arrive at the optimum strategy of paying nothing for a roll of $$1,2$$ or $$3$$ while paying for $$4,5$$ or $$6$$.

• Can you explain why you think this strategy is "optimum?" It is not a Nash equilibrium--for example, given B's proposed strategy, player A could unilaterally increase his payoff by deciding to pay when he rolls a 1. Commented Dec 7, 2022 at 15:53
• @3rdMoment A wouldn't do that as despite doing that the player A's expected return is negative as shown in the first table above. Commented Dec 8, 2022 at 4:08
• I don't understand. Your first column shows a big increase in the first row (when A rolls a 1 I presume) going from "None+2" to "A+2Only". (I think we interpreted the rules differently too. When a player wins, does the $1 come from the other player, or from the game sponsor?) Commented Dec 8, 2022 at 4:48 • @3rdMoment there should be increase from col 1 to col 2 for 1st row. As in col 2 only A has paid so the likelihood of winning has increased. As for the rules, I assume that when say player B wins$1 then as far as player A is concerned, since it kinda lost so to show that I'd subtract 1 and multiply that the probability of that happening (instead of taking it as 0 as if it wouldn't matter to player A whether B wins or not). This might a little different from the question and so I mentioned it in my assumption at the start. Commented Dec 8, 2022 at 6:59