A 50 sided die

This is a very nice problem!

So, I guess the first thing we need to remark is that if Alex gets a 1 or 50 the first time he rolls, the "rules" are a little bit different. He can only lose if he gets a 2 (if he rolled a 1 the first time), or a 49 (if he rolled 50). If Alex had gotten a 2, 3, ..., or 49 on the first roll, he would have lost on 2 different outcomes (G-1 and G+1).

Once we get that out of the way, let's calculate the probability that Alex wins after n rolls: $P(n) = \begin{cases} (\frac{47}{50})^{n-1}\frac{1}{50} &\mbox{if } G \not\in \{1, 50\} \\ (\frac{48}{50})^{n-1}\frac{1}{50} & \mbox{if } G \in \{1, 50\} \end{cases}$ .

Now, note that the probability that Alex rolls a number different than 1 and 50 on the first roll is 96%, and, obviously, the probability that G becomes 1 or 50 is 4%.

Furthermore, note that the probability of Alex winning is: $P = \begin{cases} \sum_{n=1}^{\infty}(\frac{47}{50})^{n-1}\frac{1}{50} &\mbox{if } G \not\in \{1, 50\} \\ \sum_{n=1}^{\infty}(\frac{48}{50})^{n-1}\frac{1}{50} & \mbox{if } G \in \{1, 50\} \end{cases}$ .

Combining the previous two paragraphs, we get the final solution, using the formula for the sum of a geometric series: $P = \frac{48}{50} \sum_{n=1}^{\infty}(\frac{47}{50})^{n-1}\frac{1}{50} + \frac{2}{50}\sum_{n=1}^{\infty}(\frac{48}{50})^{n-1}\frac{1}{50} = \frac{17}{50}$ .

The prob of a win is $\frac{1}{3}$ except for when the first roil is a 1 or a 50 - then it is one half. $\frac{1}{2}$

The total prob of a win is therefore $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{48}{3}$ all divided by 50 which is $\frac{17}{50}$ for a sum of 67

I also calculated with the two cases ( $\frac {1} {2}$ probability of winning, when G is either 1 or 50, and $\frac {1} {3}$ otherwise, like Terry Smith did), but my formula looks a bit simpler:

$\frac {48}{50} \times \frac {1}{3} + \frac {2}{50} \times \frac {1}{2} = \frac {17}{50}$ . Therefore, the solution is 17 + 50 = $\boxed {67}$ .

This way, we also get the same general formula for the n-sided die as Paul Fournier:

$\frac {n-2}{n} \times \frac {1}{3} + \frac {2}{n} \times \frac {1}{2} = \frac {(n-2)+3}{3n} = \frac {n+1}{3n}$