> it’s definitely possible to quantify how long it would most probably take to succeed
yes.
if success means to roll a 1, and anything else is a failure, you will on average roll the die 6 times before you succeed. E(x) = 1/p for a geometric with p = 1/6. your variance is q/p^2, so 30.
so then, my rational decision as an investor is to quit rolling the die after 17 rolls ( mu + 2 sd = 6 + 2*sqrt(30) = 17 ). I wouldn’t wait until heat death of universe.
yes. if success means to roll a 1, and anything else is a failure, you will on average roll the die 6 times before you succeed. E(x) = 1/p for a geometric with p = 1/6. your variance is q/p^2, so 30.
so then, my rational decision as an investor is to quit rolling the die after 17 rolls ( mu + 2 sd = 6 + 2*sqrt(30) = 17 ). I wouldn’t wait until heat death of universe.