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, , of jumping one stone further away from the shore and a probability, , 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 can be expressed as , where and are coprime positive integers, find .
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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 ∗ ( 2 E 2 )
And, E 2 = 1 0 0
Solving for p, then,
p = 9 9 / 2 0 0
9 9 + 2 0 0 = 2 9 9