Optimal strategy dice game: Dutch TrickTrack

I play a 2 dice game with 2 up to 7 players (could even be more) and I just can not figure out what the optimal strategy is. It is a really fast and easy game but the math is more difficult than you expect. We call it TrickTrack

Rules are as follows:

The first player has to throw the 2 dice together to get a score and can decide to throw a 2nd or a third time to get a better score. Last score stands. Depending on how many times the first player has thrown, the player(s) after him have the same amount of throws to beat his score. If for example the first player throws 3 times and the player after him beats his score with his first throw and decides to stand, then the player(s) after him have only 1 throw each to beat the score of the 2nd player. (with more than 2 players, it plays clockwise) Whoever wins, gets the pot and the advantage to start a new game. If 2 or more players tie when the round is over then they each throw 1 dice and see who throws the highest. Repeat if necessary.

The score is as follows:

With 2 dice you can throw 36 combinations but if the 2 die are different (a three and a two for example) then the highest die goes first, which in this example gives a score of 32. So a two and a six gives a score of 62. A score of 62 will beat 32 because its higher. A double gives a score of 100 times the die, so 2 fives gives 500 which beats 62 and two deuces for 200. The exception is TrickTrack which is the lowest score possible (2 and 1), In this game it beats all other scores and comes right after 6-6 for 600. Making 3-1 the lowest score possible. This basically reduces the combinations to 21 because of all the doubles.

Question:

Now I try to calculate what the optimal strategy is when I play this against 1 or up to 6 players. It basically comes down to a sheet that shows the 12 dice combinations when to stand after 1 or after 2 throws with 1 or more players behind me. So I know if I have to throw a 2nd or third time.

An example of why this is difficult:

If I play against 1 player, I know that with 36 combinations, the average throw is between 5-4 and 6-1. So it is easy to say that if I throw a 5-3, that I have to throw a 2nd time. But the average throw for my 2nd attempt is between 5-4 and 6-1 again, while my opponent gets 2 attempts which will increase his chances to throw a higher combination than average. I might have to give up a little % after my first throw in order to avoid giving my opponent even better odds when he gets 2 attempts.

How can this be solved?

• Thnx Ross, Added – youriw21 Apr 24 at 23:39
• Can the second player accept a tie and go to the tiebreaker, or must he keep throwing to try to win? – Ross Millikan Apr 24 at 23:48
• He can accept a tie, and any players behind him as well – youriw21 Apr 24 at 23:50

Let us first eliminate ties, which simplifies the analysis considerably. Instead of rolling dice, have people draw a random number uniformly on $$[0,1]$$. The highest one wins. We will do a 2 person game.

As usual, the place to start is at the end. If you draw a third number your chance of winning is $$\frac 14$$ because it has to be the highest of yours plus the three of your opponent. On the other hand if you keep a second number $$x$$ your chance of winning is $$x^2$$, so you should keep a second number of $$\frac 12$$ or more.

Now we can figure your winning chance if you draw a second number. If the number is less than $$\frac 12$$ you draw again and have $$\frac 14$$ chance to win, if the number is greater you have $$x^2$$ chance to win. Integrating over $$0$$ to $$1$$ gives $$\frac 18$$ from drawing less than $$\frac 12$$ and winning and $$\frac 13-\frac 18=\frac 5{24}$$ from drawing better than $$\frac 12$$. That totals $$\frac 13$$.

That means on the first draw you should keep any number greater than $$\frac 13.$$ Your chance to win is then $$\frac 13 \cdot \frac 13$$ from drawing less than $$\frac 13$$ and winning plus $$\frac 12-\frac 1{18}=\frac 49$$ from drawing more than $$\frac 12$$ and winning, for a total of $$\frac 59$$. An advantage for the first player.

Going back to the dice game, the only time the second player should not accept a tie is if he has one more roll and the tie is at $$54$$ or less or if he has two more rolls and he can beat your number $$\frac {11}{36}$$ of the time, which is $$63$$ or less.

The strategy for the first player is to accept $$52$$ or better on the first roll and $$61$$ or better on the second.

• Thnx I will try to apply this when playing against more players. But I am not sure you if you included the possibility of a tie. I can not change the rules of the game. And is there a difference between playing against 1 player who has 2 throws left or if I play against 2 players with 1 throw each? Because now 3 can tie. – youriw21 Apr 26 at 0:38
• I don't think the ties are enough to change the first player strategy, but I haven't checked it. Playing with more people the first player should insist on higher numbers, but I don't have a good way to tell how much. It is quite complicated. – Ross Millikan Apr 26 at 1:06