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I am somewhat stuck on this problem, it should be straightforward but I cannot find a clearly explained solution:

A single dice is rolled as many times as you want. For each throw, you receive n dollars if dice shows $n$, if $n<6$. If dice shows $6$, you lose all the money accumulated and the game stops. What is the expectation and the best strategy for this game?

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  • $\begingroup$ Do you have to stop when you get a 6, or can you continue? If you can continue, you keep on playing until you get by chance an immense run of non-6s. Of course, you have to stop at some point, but you could decide in advance to wait for 100 non-6s and keep playing until you got it. $\endgroup$
    – almagest
    Sep 2 '14 at 10:36
  • $\begingroup$ Good point. I should have added indeed that you stop the game as soon as you get the first 6. Thanks for the comment. $\endgroup$
    – RandomGuy
    Sep 2 '14 at 10:43
  • $\begingroup$ Related: Analysis of the same game, except you lose when you roll a 1, rather than a 6. $\endgroup$
    – MJD
    Sep 2 '14 at 14:49
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The probability that you are still in game after the n-th game is just $(\frac 56)^n$. The average amount of money per non losing game is $(1+2+3+4+5)/5=3$. Hence after $n$ games you are stick with an amount of Money $M(n)$: \begin{equation} M(n)= 3 \cdot n \cdot\left( \frac 56 \right)^n \end{equation}

Now you can maximise this function and get a maximum at $n=\frac{1}{\log(\frac 65)}\approx 5.48$.

So you know that either 5 or 6 is the best number of games. Calculating M(5) and M(6) yields the result that the expectation value of money after 5 or 6 games is acutally the same, namely:

\begin{equation} M(5) = M(6) = \frac{15625}{2592}\approx 6.03 \end{equation}

Edit: (due to other posts) When you however can use the knowledge of your current money, you could do better. You should just stop playing, when in average you lose more money than you win. This happens when the expectation value gets negative. Assume now, that you have already won $N$ coins. You lose per with the probability $\frac 16 \;\; N$ coins. In the same move you win 3 coins with probability $\frac 56$. Hence they are equal, when $\frac N6 = 3\cdot \frac 56$. This yields $N=15$. So you have to stop playing (or could play 1 more time; it doesn't matter) when you reached 15 coins.

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  • $\begingroup$ Isn´t $$\frac{1}{log(\color{blue}{\frac{6}{5}})}\approx 5.48$$ ? $\endgroup$
    – callculus
    Sep 2 '14 at 11:45
  • $\begingroup$ Sure thing. Thx $\endgroup$
    – Peter
    Sep 2 '14 at 12:17
  • $\begingroup$ Thank you for your clear answer. However, it puzzles me that it seems a bit in contradiction with the other answers given below that suggest that the optimal strategy is for $n=15$ rather than $n=5$ or 6. $\endgroup$
    – RandomGuy
    Sep 2 '14 at 14:02
  • $\begingroup$ No, it is not a contradiction as I assumed to fix a strategy in the beginning, i.e. don't react on your results of the current throws. However, as I can read from your problem, it is allowed to use the knowledge about the saved money from this series. Then you should play until you reached 15 coins. When you have more the 15 coins, you loose more coins in average than you win. -> I inserted this in my answers. $\endgroup$
    – Peter
    Sep 2 '14 at 15:12
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Denote by $g(n)$ the expected end amount under optimal play when you have already won $n$ points. Then $n\mapsto g(n)$ satisfies the downward recursion $$g(n)=\max\biggl\{n, \ {g(n+1)+g(n+2)+g(n+3)+g(n+4)+g(n+5)\over6}\biggr\}\ .\tag{1}$$

When $n\geq100$, say, then you should obviously stop playing further. This means that there is some minimal $n_0$ with $g(n)=n$ for all $n\geq n_0$, whereas for $n<n_0$ it pays to play on.

Doing the recursion $(1)$ starting with $n_0=100$ gives the end value $$g(0)={2893395172951\over470184984576}\doteq 6.15374\ .\tag{2}$$ Further experiments then show that in fact $n_0=15$, as suggested in other answers, and the end value is then the value given in $(2)$.

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So if your strategy is to play $n$ times, your chance of not getting a 6 is $\lambda^n$ where $\lambda=\frac{5}{6}$. You expect to get 3 each throw (assuming it is not a 6), so you get $3n$ with prob $\lambda^n$ and 0 otherwise. So you want to take $n=5$ or 6 with an expected gain of just over 6.

But you are free to change your strategy as you go along. So suppose you have accumulated $N$. The question is whether it is worth risking another throw. If you throw you lose $N$ with prob 1/6 and gain 3 with prob 5/6, so it is worth continuing if $N<15$. I leave it to you to figure out the expected gain from following that strategy.

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  • $\begingroup$ I wouldn't be surprised if the problem to find the expected gain is much more difficult than the problem to find the optimal strategy. $\endgroup$
    – drhab
    Sep 2 '14 at 11:43
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Partial answer (only concerning the strategy)

If you have gained $m$ points then $\left(m+3\right)\frac{5}{6}$ is your expectation by taking another gamble.

So it exceeds $m$ iff $m<15$.

Now the question: if you indeed follow the strategy of going on if $m\geq15$ has not been reached yet then what will be your expected gain?

Uptil now I have no answer to that.

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Im not sure but the strategy maybe based on the binomial probability for one success of 6 over n attemps. And the expected value must be the weighted mean from 0 ("success" of 6) and the mean of the other group multiplied n (or n-1) times (what is 3, so n*3).

The poisson distribution is good for infinite attempts but I think the problem is related to a undefined finite (and low) number of attempts.

The probability, as someone said before, is

$$P(k)=\left(\frac{5}{6}\right)^k$$

I dont think that maximizing the function make sense because the amount of money that you gain or lose isnt meaningful for the value of the money, so I prefer to play to a desired level of risk instead of a maximum amount of money.

For me is ok play for a $P(k)\geq 75\%$ because after this quarter you leave the meaning of something moderately deterministic to something more and more random (the pure randomness is over 50%, and pure deterministic states are 0% and 100%). This is a conservative strategy that want the maximum "safe" money.

So

$$P(k)=\left(\frac{5}{6}\right)^k\geq0.75 \rightarrow k=\lfloor\frac{log\ 0.75}{log\frac{5}{6}}\rfloor=1\ where\ E(1)=3$$

A different approach is to do it for a maximum risk of 50%: play over pure luck. In this case:

$$P(k)=\left(\frac{5}{6}\right)^k\geq0.5 \rightarrow k=\lfloor\frac{log\ 0.5}{log\frac{5}{6}}\rfloor=3\ where\ E(3)=9$$

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