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.
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?