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You are given a die with 100 sides. One side has 1 dot, one has 2 dots and so on up until 100. You are given a chance to roll the die and, however many dots come up, you can choose to (a): take that many dollars, or (b): pay $1 and roll again. You can continue to reroll as many times as you see fit, but you only keep the money of the one roll that you choose to end with. What is the optimal strategy and expected value?

Here we may consider N sided dice with n continue cost. Denote X is expected value under the strategy. There are two strategies:

  1. Given an fixed number $k,$ when we toss the number $<k:1,\cdots,k-1,$ we continue to toss with cost $n;$ when we toss the number $\geq k:k,\cdots,N,$ we stop and accept the number. Then the expected value can be computed as following: $$X = \dfrac{1}{N}(k+2+...+N) + \dfrac{k-1}{N}(X-n).$$

Then we find the optimal $k$ to maximize $X.$

  1. Another solution is replacing fixed $k$ in solution 1 by $X$ itself: $$X = \dfrac{1}{N}(X+2+...+N) + \dfrac{X-1}{N}(X-n),$$ and solve the value of $X.$ I am not sure if it makes sense.

I guess solution 1 is correct (since I cannot explain the meaning of solution 2) and I numerically tested the cases where the solution 1 is always larger than solution 2. What's wrong with the solution 2?

user6703592
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  • Is $n$ meant to be $m$? Should say, I don't understand what strategy you are describing in either case. – lulu Jan 12 '22 at 17:38
  • Perhaps it would clarify things if you worked an example with small numbers explicitly, and illustrate what your strategy would direct you to do in each case. – lulu Jan 12 '22 at 17:44
  • @lulu pls see my update – user6703592 Jan 12 '22 at 18:12
  • I don't see where you added any examples. In any case, why would you assume that using the expectation as a threshold was a good idea? And, if $X$ is not an integer, is that really the expression you intend? – lulu Jan 12 '22 at 18:15
  • Examples are always a good idea. Simplest non-trivial example is a two sided die with cost $1$. In that case you should clearly take whatever you get on the first roll, right? That gives you an expectation of $1.5$ which you can check against your strategies. – lulu Jan 12 '22 at 18:21
  • @Masacroso I don't think the continue value is simple 99/2. Suppose the times of rolling is fixed say 3 times, you should use the backward pricing to calculated the continue value. Here I use the strategy of optimal threshold since the times can be infinite. And such method is also used for perpetual American Option pricing in financial engineering. – user6703592 Jan 13 '22 at 08:13
  • @Masacroso but the expected value based on the optimal thresholds is about 87 under thresholds k=87 for N=100, cost = 1. – user6703592 Jan 13 '22 at 10:15
  • Seems a duplicate of 100-sided die probability. – CiaPan Jan 14 '22 at 09:49

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