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The game "Luck of the Dice" is described at the end of page 1 and the beginning of page 2 of this problem set. This question uses a slightly modified version of the game, called "Luck of the Dice 2", with the only change being that rule 2 is replaced with the following rule:

(2*) On each of his turns, player i rolls his die and moves his piece to the right by the number of squares that he rolled, say x. If his move ends on a square marked with an arrow, he moves his piece x squares forward. If that move ends on an arrow, he moves another x squares, repeating until his piece comes to rest on a square without an arrow.

Assume exactly one board square, say square number n, is marked with an arrow.

Question: Find, with proof for all $1\leq s_1\leq 15$, all choices of n that maximize the expected distance in squares the first player will travel in his first 2 turns.

To clarify, if the first player reaches the end in the first or second move, he is considered to have travelled a distance of $13$. That may complicate things slightly, but it'll make the expected value calculations more accurate.

For $s_1 = 1,$ the answer is either $n=1$ or $n=2$. For $s_1=2$ or $3$, we can just compute the expected value directly from the definition for all $n$ reasonably small (the maximum distance one can travel for $s_1 = 2$ is $6$ if there's an arrow on square 2 and $9$ if $s_1 = 3$ and there's an arrow on square 3). There's most likely a more efficient way.

I'm not sure if we can induct on $s_1$ somehow. If $s_1 \ge 12$, then as $s_1$ increases, the probability of the first player reaching square $12$ on his first move clearly increases.

For $s_1 = 4$, I think the answer to the original problem is still 4, but I can't seem to prove it in an efficient way. One can literally just use brute force to compute the expected values for $n=1,2,\cdots, 8$ (we only need to do it up to $n=8$ because for larger n, adding the arrow will result in the same expected value as if there was no arrow at all). For instance, I get expected values of $89/16,95/16,102/16$ for $n=2,3,4$.

user33096
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  • I don't think induction is the way to go; the endpoint is at $w=13$, so for sufficiently large values of $s_1$ the expected value is maximized around $\tfrac{w}{2}$, so at $6$ and/or $7$. This happens because for a first throw smaller $\tfrac{w}{2}$, as $n$ increases, the expected value increases strictly. On the other hand, for a first throw larger than $\tfrac{w}{2}$, you are exceedingly likely to reach the end anyway, so increasing $n$ further adds very little to the expected value for high throws, while taking away a lot from the expected value from lower throws. – Servaes Dec 22 '22 at 13:43
  • It seems likely that $n=6$ and/or $n=7$ for all $s_i\geq6$. The cases $s_1\leq5$ can then be checked by brute force; it seems likely that $n=s_1$ in these cases. – Servaes Dec 22 '22 at 13:44

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