Just One Jump!
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 50% chance of jumping one stone closer to the shore and a 50% chance of jumping one stone further away from the shore.
What is the expected value for the number of jumps he will take before reaching the first stone (the one closest to the shore)?
Other Expected Value Quizzes
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Relevant wiki: Expected Value - Problem Solving
Let E n = Expected number of jumps to reach the first stone from the n th stone.
From the second stone he has a 0 . 5 probability of jumping backward (to the first stone) and a 0 . 5 probability of jumping forward (to the third stone). So,
E 2 = 1 + 0 . 5 ∗ E 1 + 0 . 5 ∗ E 3
Since, E 1 = 0 , this becomes,
E 2 = 1 + 0 . 5 ∗ 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 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 + 0 . 5 ∗ ( 2 E 2 )
Or, E 2 = infinity