2

I have an $100$-sided die. Every round, I roll the die, and can either choose to receive the amount shown by the die, or pay $1 to reroll the die and continue.

If I want to find the expected value of this game, say $E$, we can condition on rolling less than $E-1$ or more to derive the equality $$E = \frac{\lfloor{E-1}\rfloor}{100}(E-1) + \frac{1}{100} \sum_{k=\lfloor E-1\rfloor+1}^{100} k$$
which gives approximately $E \approx 87.4$ I believe. This corresponds to a strategy of rerolling every time I roll below 87, and keeping the higher values. How does the expected value and strategy of this game change with the following modifications?

a) This time, it costs nothing to reroll, but you can have at most 100 rerolls.

b) You still have at most 100 rerolls, but it costs $1 to reroll.

Intuitively, it seems that we should approach (a) using a 'backtracking' approach. If I have just one round, my expected value is $50.5$. If I have two rounds, let $X$ be the expected value, and suppose we reroll the first roll only if we roll $a$ or lower. We have $$ \mathbb E(X) = \mathbb E(X|X>a)\mathbb P(X>a) + \mathbb 50.5\cdot \mathbb P(X\leq a) = \frac{100+a+1}{2} \cdot \frac{100-a}{100} + 50.5\cdot \frac{a}{100} $$ The right hand side is a quadratic in $a$, maximised at $a=50$, giving an expected value of $63$.

Then if $Y$ is the expected value of the third round following a strategy of rerolling the first die only if we roll $b$ or lower, we now have $$ \mathbb E(Y) = \mathbb E(Y|Y>b)\mathbb P(Y>b) + \mathbb 63\cdot \mathbb P(Y\leq b) $$ which gives $\max \mathbb E(Y) = 70.03$ for $b\in \{62,63\}$.

Clearly, I can continue and iterate up to 100, but this seems hard without a computer. Is my approach correct, and is there an easy way either extend this procedure by hand, or to approximate the value given for higher $n$?

Furthermore, how does the situation in (b) differ? It seems to me that not much changes: I simply have some $-1$s in my expectation equations, which present some computational difficulties, but the same questions still apply.

RDL
  • 920
  • Note: I don't believe there is any sensible way to do this without mechanical assistance. – lulu Nov 23 '23 at 10:39
  • @lulu this is not a duplicate. OP is asking about two variations of the linked question, not the linked question itself which they already know how to do. – Especially Lime Nov 23 '23 at 10:46
  • 1
    @lulu While I was aware that the initial question has been posted on this site, I was not aware of any consideration of my variants, and don't see why it's a duplicate - the initial problem can be solved by establishing a direct equation on $E$, while I don't see an analogue for the two variants beyond the round-by-round consideration I outlined above. – RDL Nov 23 '23 at 11:28
  • Ok, I released the duplicate. But I don't see what more there is to say about the question...the duplicate goes through the computation for part $b$ in detail. These are both examples of so-called American Option pricing, the topic is well known and well studied, backwards induction is optimal. – lulu Nov 23 '23 at 11:35
  • Ah, since I removed the duplicate, I should reference it here. Or here No doubt there are others, this is a fairly frequent question. – lulu Nov 23 '23 at 11:37
  • @lulu Forgive my blindness: when you say 'the duplicate goes through the computation for part b in detail', which 'duplicate' are you referring to? – RDL Nov 23 '23 at 12:56
  • sorry, the duplicate appears to only handle the infinite recursion case which, oddly, is easier since exercising the option gets you back to the start (with $$1$ deducted, of course). And having $100$ options is effectively infinite so this is a good approximation. To do it exactly, use the same recursive step they do, but keep track of the state and proceed backwards exactly as you did. – lulu Nov 23 '23 at 13:04
  • To your question about use of automation, even the simpler problem (rolling a $6$ sided die with an option to re-roll) pretty much requires a machine if you have more than a few options. Sure, you can look at the growth of the function invovled and prove various approximations but it's usually not worth it, as minor modifications to the problem tend to invalidate the analytic methods. – lulu Nov 23 '23 at 13:06

0 Answers0