I feel sleepy, so there might be mistakes here. At least it'll give you guys something to check and criticize if I made a mistake
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Notice how each "value" is possible after successes (0.8) in previous times and a fail (0.2) at the end.
Expectation = sum {value_i x p(value_i) }
5 * 0.2 + 10 * 0.8 * 0.2 + ... + (5*n * 0.8^(n-1)*0.2) + ..., as n-> inf
= sum {n*0.8^(n-1)}, n=1...inf
Now we know that sum{k^n} = k/(1-k}, when n=1->inf
Take the derivative of both sides.
sum{n*k^(n-1)} = -k(-1)(1-k)^(-2) +(1-k)^(-1), n=1->inf
Sub in this formula to our problem:
k=0.8
Therefore, expectation = 0.8*(0.2)^(-2)+(0.2)^(-1) = 25 years
hmm... well.. I guess it "could" be reasonable. I will check back later after my nap
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