Expected value of suming up to 6 tossing a coin, where each tail adds one and each heads subtracts one(can't go less than zero points) 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!
 A: 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 $42$.  
A: 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. 
