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So, I was playing this video game where you could use a jewel to add value to an item. Item values goes from +0 to +6. Each time you use the jewel you have a 50% chance of success, now, the Expected value of jewels I should use to get it to +6 would be easily calculated, it's 12. (6/0.5).

But here's the thing, each time you miss, it subtracts one, so if your item is +2 and you fail, it goes back to being +1. The only exception is when it is +0, as it cannot go any lower.

So my question is, how do i calculate this number analytically? I ran it through python and the result is 42. There's a second case where the chance of success is now 75%, in which my script gives me an expected value of 11.

Thanks for the help!

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For $k=0,1,\dots,6$ let $e_k$ be the expected number of rolls to get to $6$. We have $$\begin{align} e_6&=0\\ e_5&=1+\frac12e_4+\frac12e_6\\ e_4&=1+\frac12e_3+\frac12e_5\\ e_3&=1+\frac12e_2+\frac12e_4\\ e_2&=1+\frac12e_1+\frac12e_3\\ e_1&=1+\frac12e_0+\frac12e_2\\ e_0&=1+\frac12e_0+\frac12e_1 \end{align}$$

and we want the value of $e_0$.

The equation for $e_5$ for example, comes from the fact that we always have to make $1$ roll and after that the value will be either $4$ or $6$, each with probability $\frac12$.

To solve, this system, use the last equation to express $e_1$ in terms of $e_0$, then the previous one to express $e_2$ in terms of $e_0$ and so on. Eventually, we get $e_6$ in terms of $e_0$, and $e_6$ is known, so we can calculate $e_0$.


I get, successively, $$\begin{align} e_1&=e_0-2\\ e_2&=e_0-6\\ e_3&=e_0-12\\ e_4&=e_0-20\\ e_5&=e_0-30\\ e_6&=e_0-42\\ \end{align}$$ so that the expected number of rolls is $42$.

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So, messing up with the numbers, and saying I have to get the item to just +1, or +2, and so on, i get the series, 2 6 12 20 30 42 and so on, which represents exactly n^2 + n So i guess that's the analytic result, still not sure how to get there.

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  • $\begingroup$ Now I see that you have the right sequence. If you mean that you want to prove that the expected number of rolls to get to $n$ is $n^2+n$, you can do it by strong induction. $\endgroup$
    – saulspatz
    Apr 12 '20 at 16:45

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