# Dice game modelling: Lose everything on “3”, double everything on “1” or “6”

I was recently playing a quite easy dice game:

You trow a fair dice: if you get a "3" the next player continues, if you get something else it is up to you to continue. If you continue and you throw a "1" or a "6" you get twice the result from the first trow. If you get a "3" you lose everything and the game continues (you always can double your result with a "1" or "6" and lose everything with a "3").

I was wondering if you could tell, at what point it is better to quit than to continue (I'm pretty sure you can, but I don't know how).

Any thoughts are welcome.

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Gambling fits loose people, but they usually end up losing more than their money. –  Asaf Karagila Mar 12 at 18:18
@AsafKaragila Speak for yourself! I've never lost while playing craps! (Fortunately, I don't live near a casino ;) ) –  Arkamis Mar 12 at 18:19
@Arkamis: Me neither. But vacuously. –  Asaf Karagila Mar 12 at 18:23
If money doesn't have diminishing marginal utility for you, then it is never better to quit than to continue. The question of when to quit depends on what your utility function is. –  Tanner Swett Mar 12 at 18:24
@AsafKaragila You missed out on all the fun, then! –  Arkamis Mar 12 at 18:24
The value function $F(x)$ when your current earnings are $x$ will be of the form: \begin{align} F(x)&=\max\left\{x,\frac{3}{6}x+\frac{2}{6}2x+\frac{1}{6}0\right\}\\ F(x)&=\max\left\{x,\frac{x}{2}+\frac{2x}{3}\right\} \end{align}
Since you are always better off in an expected sense, (i.e. since $\frac{x}{2}+\frac{2x}{3}>x\ \forall x$, it is never optimal to stop playing. In other words, the loss of the money occurs with a probability 1/6, which is never high enough to forbid one from playing.
$F$ is the value function and $x$ is the earnings or the amount of money you have. –  Bravo Mar 12 at 21:52