Find expected value of last roll

Question

Suppose that you play the following game: You roll a fair die at most $N$ times and get an amount of dollars equal to the last number rolled. You can decide to stop the game at any time. What is the (approximate) value of this game for $N=60$?

Here is my approach:

If you can only roll the die once then the expected value is $3.5\$$. Therefore, if you can only roll the coin twice it makes sense to roll again only if your first roll is $1,2$ or $3$. Thus, the expected value is $\frac{1}{2}(3.5)+\frac{4+5+6}{6}=4.25$.

If you have two rolls left, then the expected value of those is $4.25$. Hence, you roll again only if you get 1,2,3 or 4. The expected value is thus $\frac{2}{3}(4.25)+\frac{5+6}{6} \approx 4.67$.

Similarly, if you have three rolls left, the expected value is $\frac{2}{3}(4.67)+\frac{5+6}{6} \approx 4.94$.

With four rolls left, the expected value is $\frac{2}{3}(4.94)+\frac{5+6}{6} \approx 5.13$.

Thus for the first $45$ rolls you stop only if you roll a $6$. It follows that the expected value should be $6(1-(5/6)^{45})+(5/6)^{45}\cdot 5.13 =5.999762..$

Is this correct? Somehow this seems too high. Thanks!

Answer 1

Just to add a bit more insight: The value is high since the probability of not hitting $6$ in $60$ trial is quite small - more precisely it is $(\frac{5}{6})^{60} \approx 0.0000177$

This means that the probability of hitting $6$ is really close to $1$ and hence, you can stop at $6$ with a really high probability

Answer 2

Well, it's in fact too low :-), because you have $55$ more rolls, not $45$. Other than that, it's a good and correct solution.

The exact value is

$$ 6-\frac{13045120539345589349977672100067138671875}{339400541532564288120296890790101415733755904}\approx6-3.8\cdot10^{-5}\;. $$