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Here's a question from my probability textbook:

A casino comes up with a fancy dice game. It allows you to roll a dice as many times as you want unless a $6$ appears. After each roll, if $1$ appears, you will win $\$1$; if $2$ appears, you will win $\$2$; $\ldots$; if $5$ appears, you win $\$5$, but if $6$ appears all the money you have won in the game is lost and the game stops. After each roll, if the dice number is $1$-$5$, you can decide whether to keep the money or keep on rolling. How much are you willing to pay to play the game (if you are risk neutral)?

It's been asked before on MSE multiple times:

When to stop rolling a die in a game where 6 loses everything

Dynamic dice game: optimal price to enter

A dynamic dice game

Here's the answer to the question in my book:

Assuming that we have accumulated $n$ dollars, the decision to have another roll or not depends on the expected profit versus expected loss. If we decide to have an extra roll, our expected payoff will become$${1\over6}(n + 1) + {1\over6}(n + 2) + {1\over6}(n + 3) + {1\over6}(n + 4) + {1\over6}(n + 5) + {1\over6} \times 0 = {5\over6}n + 2.5.$$We have another roll if the expected payoff ${5\over6}n + 2.5 > n$, which means that we should keep rolling if the money is no more than $\$14$. Considering that we will stop rolling when $n \ge 15$, the maximum payoff of the game is $\$19$ (the dice rolls a $5$ after reaching the state $n = 14$). We then have the following: $f(19) = 19$, $f(18) = 18$, $f(17) = 17$, $f(16) = 16$, and $f(15) = 15$. When $n \le 14$, we will keep on rolling, so $E[f(n) \mid n \le 14] = {1\over6} \sum_{i = 1}^5 E[f(n + i)]$. Using this equation, we can calculate the value for $E[f(n)]$ recursively for all $n = 14, 13, \ldots, 0$. After laboriously calculating or writing a program, we get $E[f(0)] = 6.15$, and so we are willing to pay at most $\$6.15$ for this game.

However, I'm wondering if there's a quick way by hand to get a reasonable/"good enough" estimate for $E[f(0)]$ without having to do multiple "average-five-numbers-and-repeat" as the book suggests. Is there?

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You can estimate it as follows: There are two possibilities - Either you roll a six, or you don't. If you don't, the expected gain is $\frac{1+2+3+4+5}{5}=3$ (Note that this is the expected gain GIVEN that you didn't roll a $6$). Now suppose that no matter what number you get from $1$ to $5$ you gain $\$3$. As stated in the solution you provided, you should keep rolling until you have $\ge 15$, so in this case you roll exactly $5$ times. If you are successful with all five rolls, you gain $\$15$, but if you rolled a $6$ in any one of these rolls you gain $\$0$.
So the estimate is $15*\left(\frac{5}{6}\right)^5=6.03$.

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  • $\begingroup$ Thanks @ItsMe. (1) But isn't the expected gain $(1 + 2 + 3 + 4 + 5)/6 = 2.5$, and not $(1 + 2 + 3 + 4 + 5)/6 = 3$? (2) Assuming we get an estimate, how do we know if it's greater or lesser than the actual value of $\$6.15$? $\endgroup$ Jul 21 at 2:18
  • $\begingroup$ I've added some detail to the answer - that is the expected gain given you didn't roll a six (hence there are only 5 options). Now I'm not sure how to determine whether this estimate is more or less without knowing the actual value, but some intuition is that in the estimate the maximum payoff is 15 whereas in the actual problem you could get up to 19, suggesting that the estimate may be lower than the actual value. $\endgroup$
    – ItsMe
    Jul 21 at 4:23
  • $\begingroup$ Here's a way to argue why $6.03$ is an under-estimate, i.e. the modified game is worth less: the intuition is that variance in payoff helps you. There is some threshold, and you keep rolling until you meet it. Say you are close. If you roll a big payoff, you exceed the threshold by a lot (and stop), whereas if you roll a small payoff, you might stay below threshold and simply keep playing. So the variance increases the average amount by which you exceed the threshold when you stop. $\endgroup$
    – antkam
    Jul 23 at 5:38
  • $\begingroup$ Consider an even higher variance game where $1$ pays $15$, and $2,3,4,5$ pays nothing. This game has even higher variance, and is worth $0$ (if you roll $6$ before $1$) or $15$ (otherwise), for an average of $7.5$. $\endgroup$
    – antkam
    Jul 23 at 5:39

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