An infinite line of stepping stones stretches out into an infinitely large lake.

A frog starts on the second stone.

Every second he takes a jump to a neighboring stone. He has a probability, $p$ , of jumping one stone further away from the shore and a probability, $1-p$ , of jumping one stone closer to the shore.

If the expected value for the number of jumps he will take before reaching the first stone is exactly 100, then $p$ can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

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The answer is 299.

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Let $E_n =$ Expected number of jumps to reach the first stone from the $n$ th stone.

From the second stone he has $1-p$ probability of jumping backward (to the first stone) and $p$ probability of jumping forward (to the third stone). So,

$E_2 = 1 + (1-p)*E_1 + p * E_3$

Since, $E_1 = 0$ , this becomes,

$E_2 = 1 + p * E_3$

However, by symmetry, if he gets to the third stone, his expectation value has doubled from when he was on the second stone, since the the expected number of jumps to go from $3$ to $2$ is the same as the expected number of jumps to go from $2$ to $1$ .

So, $E_3 = 2*E_2$ .

Therefore, $E_2 = 1 + p*(2E_2)$

And, $E_2 = 100$

Solving for p, then,

$p = 99/200$

$99+200 = \boxed{299}$