Alex plays a game in which you have to roll a
**
fair
**
50 sided die.

Alex's first throw is to determine his game number $G$ .

If on the second roll he gets his $G$ number he wins, if he gets either 1 less than $G$ or one more than $G$ he loses and if he rolls neither $G, G-1$ nor $G+1$ he rolls again until he gets one of these 3 numbers.

The game ends when Alex either wins or loses.

What is the probability that Alex will win?

If the probability can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

The answer is 67.

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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: $P = \frac{48}{50} \sum_{n=1}^{\infty}(\frac{47}{50})^{n-1}\frac{1}{50} + \frac{2}{50}\sum_{n=1}^{\infty}(\frac{48}{50})^{n-1}\frac{1}{50} = \frac{17}{50}$ .