Re: Probability question
(1- 1/n)^n is about 1/e, where e is Euler's number 2.7182818... More precisely, it is about 1/e(1-1/(2n)). You can get this from the binomial theorem and the power series e^x = 1+x+x^2/2+x^3/6+x^4/24...
For both situations, the probability of losing all games is about 1/e, so the chance of winning at least once is about 1-1/e ~ 63.21%. The exact values are 63.40% for 100 trials with 1% success rate and 64.15% for 20 trials with 5% success rate.
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