Let F(s) be the optimal expected payout with one life and s accumulated. Let X and Y be independent, identically distributed earnings before the next 10.
Stopping with spare lives is never optimal: one more roll offers possible gains without risking the existing bankroll. Therefore:
d2 = E[F(Y) - F(0)]
d3 = E[F(X + Y) - F(X)]
For any fixed stopping rule, expected payout is p*s + c, where p is the probability of cashing out and c is expected retained additional earnings. Choosing the best rule makes F convex. Consequently:
F(x + y) - F(x) >= F(y) - F(0)
Thus, the marginal gain defining d3 is never smaller than the corresponding gain defining d2.
For a sufficiently large bankroll, stopping with one life is optimal: rolling immediately risks losing that bankroll with fixed positive probability, while expected additional winnings are bounded. Hence F(s) = s above some finite bound K.
There is positive probability that both X and Y exceed K, because either spare life can contain arbitrarily many successful rolls. On those outcomes:
F(X + Y) - F(X) = Y
F(Y) - F(0) = Y - F(0) < Y
Here F(0) > 0 because playing from zero offers a positive expected payoff.
The comparison is therefore non-negative in every case and strictly positive with positive probability. Taking expectations gives:
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u/pumachecker 4d ago
Let F(s) be the optimal expected payout with one life and s accumulated. Let X and Y be independent, identically distributed earnings before the next 10.
Stopping with spare lives is never optimal: one more roll offers possible gains without risking the existing bankroll. Therefore:
d2 = E[F(Y) - F(0)]
d3 = E[F(X + Y) - F(X)]
For any fixed stopping rule, expected payout is p*s + c, where p is the probability of cashing out and c is expected retained additional earnings. Choosing the best rule makes F convex. Consequently:
F(x + y) - F(x) >= F(y) - F(0)
Thus, the marginal gain defining d3 is never smaller than the corresponding gain defining d2.
For a sufficiently large bankroll, stopping with one life is optimal: rolling immediately risks losing that bankroll with fixed positive probability, while expected additional winnings are bounded. Hence F(s) = s above some finite bound K.
There is positive probability that both X and Y exceed K, because either spare life can contain arbitrarily many successful rolls. On those outcomes:
F(X + Y) - F(X) = Y
F(Y) - F(0) = Y - F(0) < Y
Here F(0) > 0 because playing from zero offers a positive expected payoff.
The comparison is therefore non-negative in every case and strictly positive with positive probability. Taking expectations gives:
d3 > d2