Okay, still tossing Toss Up around in my head... Let me back off for a minute to the simpler question of "solitaire Toss Up" (which is what Thane looked like he was playing "against" his kids anyway!). In this version, you're just trying to maximize your score on each turn; we ignore the race-to-100 aspect entirely. If you're playing with the original bankruptcy rule, where you only risk the points you've accumulated on the current turn, then in solitaire your cumulative score is irrelevant: we're just asking for which numbers of points n is the expected value of rolling again positive. This should be easy to answer, right? If you have n points so far, the value of stopping is n, while the expected value of rolling is E_roll(n) = p(bust)*0 + sum_i>0 p(roll i green) * E(n+i) Here E(n) is max(n, E_roll(n)), and the number of dice you roll is 10 - (n mod 10). That's because of the very important rule that if you roll until all ten of the dice are green, then you get to start over with rolling all of them, and the number of points you're risking is now over 10 -- and is potentially unbounded. This makes it tricky to solve the above equations for E(n): you'd like to do some downwards induction, starting from some values of n that are so large that you'd *never* roll the dice, so you know E(n)=n. But how large is large enough? Intuitively it ought to be something over 5000! (If you're allowed to start a sentence with "intuitively" and end with an exclamation mark.) Becuase with 5000 points, you have a 1/1024 chance of going bankrupt versus an expected outcome of rolling 5 green dice out of ten, with the deal sweetened by the fact that you also have a 1/1024 chance of making it to 5010 points and getting to decide again. With a computer, we can afford to be empirical about it: cap the upward induction of E_roll(n) at some specified value N, declaring arbitrarily that we never reroll if n>N, and then see how the expected value function changes with N. If we pick N=10,000 things have indeed stabilized, and we can compute the expected value of the game: E(0) is approximately 13.2393. So the optimal strategy beats Thane's 4-5 points per turn by quite a bit! And what is the optimal strategy? It turns out that you always roll if your number of points so far is less than 27! If you are lucky enough to get that high, here are the smallest numbers of points n for which you do not roll despite having 1,2,3,...,10 dice available: 29, 28, 27, 56, 105, 224, 483, 1072, 2371, 5220. That's all you need to know to score 13.23 points per turn on average -- which should probably be enough to beat sub-optimal players, even without solving the two-player version of the game. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.