How to win this dice probability game? The game is a variation of Pig. Here is how the game works:
There are about 20 players. Each round, a single six sided die is rolled. All players add that rolled number to their "bank." However, if a 2 is rolled starting at the 3rd round, all banks are set back to 0. Before a die is rolled, a player can choose to sit down and add their bank to their score, keeping it permanently. The game is played until scores are reset a total of 6 times, at which point the player with the highest score wins.
What is the best time to sit down each round to have the highest chance of getting the greatest score?
 A: This is how I look at it, but I'll admit I may have misunderstood the game!
Assuming you have a current banked score of B, the expected return for any given round is:
$$E(return)=\frac{1}{6}(-B)+\frac{1}{6}(1+3+4+5+6)$$
$$E(return)=\frac{1}{6}(19-B)$$
So once you have a bank of 19 points, it is better to get out than take the chance.
I believe this will maximize your average score in the long run.  However, when it comes to games, sometimes things are more complicated than simple optimization.  For a 2 player game, I think I would follow the advice of my analysis above.  However, for a 20 player game, it is clear that you will need some luck, since 2nd place is the first loser, you want to give yourself a chance at a very high score, not just try to avoid a very low one.  Intuitively, I expect that this means you need to push your luck passed the 19 score mark, but I'll have to think harder about how to quantify that for a game of n people.
Running a simulation, I find that the mean is, indeed, optimized near a threshold of 19.  However, as I also suspected, the lucky game (mean + 2$\sigma$) is actually optimized out around 29 or 30.  So if you need to beat 19 other players wait till then.

A: The following results are from my simulation in R.
Assuming there are 6 rounds in total, the first two throws are always performed in each round, and after each round (the first 2) everybody resets.
I will simulate the game for only one person, since we assume the players play independently. I will test which strategy performs better, always banking on the first throw, second, third, and so on.
result=replicate(1e4,{
  bank=rep(0,20)
  for(i in 1:6){
    round_draws=sample(1:6,20,replace=T)
    first_two=which(round_draws[3:20]==2)[1]+2
    if(!is.na(first_two)){
      round_draws[first_two:20]=NA
    }
    new_bank=cumsum(round_draws)
    new_bank[is.na(new_bank)]=0
    bank=bank+new_bank
  }
  return(bank)
})

And the results look like this.

The best result (in the long run, on average) is banking every 6 throws.
