Pay \$3k to roll a standard dice and earn \$kn where n is the outcome of the roll. If you can play m times how do you size your bet?

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Assume we have an initial bankroll of $B$ dollars and we can spend $3k$ to play a dice game which has payoff $kn$ where $n$ is the outcome of the roll. I believe that the expected payout of this game is $3.5k$, so the player has an edge of $1/6$, and the variance is $1.2k^2$.

Assume we can play only $m$ rounds of this game. If we bet a the same proportion $p$ of our bankroll on each of $m$ bets we can make a trade-off between the mean and variance of our total expected profit. If $p$ is close to zero we have low variance and low mean profit, and if $p$ is close to one our expected profit and the variance are both high. I know of the Kelly Criterion, which leads to the best expected returns in a scenario with two outcomes.

Is there a best choice for $p$? Is there some sort of generalisation of the Kelly Criterion that applies to this situation?

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Kelly maximises the product of your outcomes.
Bet 3p of your wealth every time, get back (1-2p)(1-p)(1)(1+p)(1+2p)(1+3p) after six.
This is maximised at $$p=0.1756,3p=0.527$$