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?

  • 2
    $\begingroup$ Just to be sure I understand correctly, players who sat down all stand up again once the scores reset? $\endgroup$
    – Glen_b
    Sep 24, 2019 at 13:58
  • 1
    $\begingroup$ This can be solved using combination of Dynamic Programming and Game Theory. You need Game Theory under the assumption that you can see all the other players, and know their current scores plus value in the bank. Policies that maximise your expected final permanent score will be subtly different from policies that beat all other current players - e.g. you will be incentivised to sit down earlier once in the lead, because other players are forced to take more risk. $\endgroup$ Sep 25, 2019 at 10:17
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    $\begingroup$ Actually with 20 players this could be beyond DP, and require an approximate answer. Also policies that simply maximise your expected score are likely to be reasonable first order estimates. $\endgroup$ Sep 25, 2019 at 10:19

2 Answers 2


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.

enter image description here


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.

  for(i in 1:6){

And the results look like this.

enter image description here

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


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