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This question is similar to This Dice Blackjack game - is there Nash equilibrium?, but the game rules are completely different.

This Dice Blackjack is game for two. It is also played with fair six-sided dice. In this game is no bettor or dealer, both players have the same role.

Rules: Any player can roll dice as many times he/she desires while the sum of rolls is lower than 21. Players does not see the sum of rolls of the other player till both finish (they can only guess what other player has). When both players finish, they compare their sums of rolls.

Winner: Winner is the one with lower penalization $ = \begin{cases} 21 - \sum,& \text{if } \sum \leq 21\\ 100, & \text{otherwise} \end{cases}$

where $\sum$ is sum of player rolls. If tie, nobody wins. Goal is to achieve as many wins as possible in huge number of games.

Example games:

  1. game - player A stops at 18, player B stops at 16 (4, 4, 4, 4). Player A wins. (player B was too afraid to risk)

  2. game - player A stops at 19, player B reach 22 (5, 6, 1, 6, 5). Player A wins. (player B start risking but has no luck)

  3. game - player A get to 23, player B stops at 17 (4, 2, 6, 5). Player B wins. (player B become again careful and this time it payed off)

Primary question: Is here the Nash equilibrium? That means existence of optimal strategy, that you should play no matter what the opponent plays? Or, can be proficient to guess the opponent strategy and adjust your strategy according to what he/she is probably playing?

I think it is similar to rock/paper/scissor. As I read somewhere there is "optimal" strategy, but if your opponent strategy is biased, you can get better outcome with adjusted strategy.

If anything is not clear, please comment, I will do my best to improve the question.

Secondary question: What should be name of this game? Blind dice Blackjack? Is there any name convention for game modifications like this one? Please comment if you have idea or mention the idea in your answer for primary question.

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    $\begingroup$ Every game with a finite amount of possible strategies has a Nash equilibrium. This includes rock-paper-scissors, in which the equilibrium is for both opponents to play completely at random. Of course, if the strategy of the other player is different, say play always paper, then your optimal strategy changes, but that requires knowing what the other strategy is. $\endgroup$ – ffao Jul 2 '17 at 18:30
  • $\begingroup$ Each player doesn't know the other player's sum, but does s/he know the number of rolls the other player has made so far? That could alter the strategy. $\endgroup$ – Penguino Jul 3 '17 at 0:33
  • $\begingroup$ @Penguino No, they do not know the number of rolls. $\endgroup$ – matousc Jul 3 '17 at 6:53
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They would rapidly each settle on

18

I ran a simulation and that number wins against any other number.
In some cases it is not the optimal for example against 15, 16 and 19 you should play

17

You have a play that is not exploitable so you can just plain solve it.

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