100 jumps

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 p , of jumping one stone further away from the shore and a probability, 1 p 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 p can be expressed as a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


Other Expected Value quizzes.

Image credit: smwmarketingny.com


The answer is 299.

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1 solution

Geoff Pilling
Jun 10, 2016

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

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

E 2 = 1 + ( 1 p ) E 1 + p E 3 E_2 = 1 + (1-p)*E_1 + p * E_3

Since, E 1 = 0 E_1 = 0 , this becomes,

E 2 = 1 + p E 3 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 3 to 2 2 is the same as the expected number of jumps to go from 2 2 to 1 1 .

So, E 3 = 2 E 2 E_3 = 2*E_2 .

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

And, E 2 = 100 E_2 = 100

Solving for p, then,

p = 99 / 200 p = 99/200

99 + 200 = 299 99+200 = \boxed{299}

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