If At First You Fail, Try And Try Again
Suppose you are running a trial that independently has a 1 in chance of succeeding. You know that if you attempt it times, then the expected number of success is 1.
As tends to infinity, what is the probability (to 3 decimal places) that you have at least 1 success during those trials?
As an explicit example, for , each trial independently has a 0.5 chance of succeeding, and you have a 0.75 chance of succeeding at least once during the 2 trials.
The answer is 0.6321.
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1 solution
The probability that you succeed at least once during the n trials, S n , is equal to, by complementary counting,
= 1 − P ( you fail during each trial ) = 1 − ( 1 − n 1 ) n .
As n tends to infinity, we get
lim n → ∞ S n = 1 − lim n → ∞ ( 1 − n 1 ) n = 1 − e 1 = 0 . 6 3 2 1 .