Mathematics Asked by student on December 5, 2020

**Problem**

A player plays a game with the following rules. He throws a die until $1$ appears or he wishes to finish. The aim of the game is to maximize the last thrown number. What would be the best strategy and what is the expected number of the last throw.

**Ideas**

I reckon that after the first trow we need to look at the number. If it’s smaller than $4$, it makes sense to throw the dice again as the probability of getting four and higher is bigger than the probability of getting two and one.

Let $S_n$ denote the strategy of stopping if you throw $n$ or more, and let $E_n$ be its expected value. Then whenever you throw, you have:

- a $frac16$ chance of throwing a $1$;
- a $frac{7-n}{6}$ chance of throwing $n$ or more;
- an $frac{n-2}{6}$ chance of re-throwing.

And if you throw $n$ or more, your expected value is $frac{n+6}{2}$.

So if $E_n$ is the expected value of strategy $S_n$, we have $$E_n=frac16+frac{7-n}{6}frac{n+6}{2}+frac{n-2}{6}E_n$$

Now you can easily evaluate $E_n$ for $n=2,3,4,5,6$.

Correct answer by TonyK on December 5, 2020

If you roll a 4, you have a decision to make.

If you keep rolling until you roll a 1,5, or 6

then you should get each of those events 1/3 of time.

If you stop, you have accepted the 4.

So the decision whether to keep rolling comes down to whether after 3 games you prefer

4,4,4 or 1,5,6

I don't see any way of justifying stopping after you roll a 3.

Answered by user2661923 on December 5, 2020

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