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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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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 '13 at 18:24
up vote 0 down vote accepted

This could be solved using the Bellman's equation if you cast this as an infinite horizon dynamic programming problem.

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.

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x is the number of throws? – ulead86 Mar 12 '13 at 20:58
$F$ is the value function and $x$ is the earnings or the amount of money you have. – Bravo Mar 12 '13 at 21:52

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