Optimal strategy for dice game

Slightly modified repost of an older question that I don't think was answered fully.

The Game: You are competing in a game with 2 other players with a 21-faced dice (faces labelled 1-21). All three of you gets to choose a number in secret and then the dice is rolled. Whoever chose the number closest to the number on the rolled face wins. In the event of any ties no player wins. You only play once. What is your strategy?

What I've tried: I'm quite the beginner at game theory so please excuse (and correct) any mistakes in the following.

I understand that if all players are rational, there should be no strategy that gives any one player an advantage due to symmetry. So I'm trying to find a nash equilibrium for the game where no one has an advantage.

An initial thought I had: Players could pick 11 as it's the middle number and close to as many numbers as possible. But if all players do that then one could use a strategy where they choose 10 or 12 (or a mixed one where they alternate between the two) to increase their expected payout (they win on 10/21 of the numbers). However, another player could then choose 9 or 13 to have an advantage, then another player could pick 8 and 14 etc. trying to be "outside" the other players.

This would continue with players trying to increase their expected payout, until the value of being "outside" the numbers the others pick is not worth the fewer numbers you are close to. e.g. if another player picks 3 as their number, you wouldn't want to pick 2 just to be "outside" them. It's better to pick 4 to be closer to more numbers. But I feel this is a bit too complicated to develop a mixed strategy from.

If any of the players knew that other players were using a pure strategy and only choosing 1 number, they could easily change their number to get a higher payout which would repeat so there shouldn't be a pure strategy nash equilibrium. Hence, if there exists a nash equilibrium the players should be using mixed strategies.

Intuitively, one solution could be if players picked 4, 11, 18 at a 1/3 frequency each. This divides the numbers into 3 even sections (1-7, 8-14, 15-21). I do feel that this is a nash equilibrium but how can I go about proving that? And even if it's a nash equilibrium, how do I go about proving that it's unique and that there are no other strategies I could use?

You are correct, that you guessed that the pure strategy won't work and that you have to pick up other numbers too. However, your strategy still has the same flaw: it allows for ties to happen to often, thus, the expectation of winning is very low.

The computations show that if all 3 players use your strategy $$S_0=\frac{1}3i_4+\frac13i_{11}+\frac13i_{18}$$, then the probability of winning is $$p_0=\frac{104}{567}=18.3\%$$.

If one of the players switches to strategy $$S_1=\frac15 i_4+\frac15i_{10}+\frac15i_{11}+\frac15i_{12}+\frac15i_{18}$$, then he starts to win in $$p_1=254/945=26.9\%$$ of cases.

I don't believe there is a good manual method to find the equilibrium. But with standard methods on a computer one can find, the following distribution to be optimal:

It has 2 peculiar features. First, the tails $$|i-11|>6$$ are exactly 0. Secondly, there is a small dip in the centre ($$p_{10}=p_{12}=0.0866973$$, $$p_{11}=0.0863136$$).

The probability of winning for equilibrium strategy is $$p_{e} = 28.2\%$$

• Assuming that each player uses your optimal distribution, what is the probability of winning? Jan 22, 2022 at 15:26
• It's about 28.2% Jan 22, 2022 at 17:15
• Interesting. With each player using uniform distribution, the probability of winning is about 29.67% Jan 22, 2022 at 17:44
• If 2 players keep playing uniformly, and 3rd player switches to $1i_{11}$ they start winning in 37.1% cases. If 2 players play optimally and 3rd player switches to $1i_{11}$, their probability of winning doesn't change. Jan 22, 2022 at 17:50
• It's also not surprising that with uniform strategy ties happen less often. Jan 22, 2022 at 17:53