The solution for this weeks FiveThirtyEight riddler [https://fivethirtyeight.com/features/should-you-pay-250-to-play-this-casino-game/] is pretty nutty.
The riddle is essentially:
If you drew random numbers from a uniform distribution from [0,1], what is the expected number of draws you would perform until their sum exceeds 1?
So let us define
$$S_{n}$$as the sum of the nth number drawn. Then the probability of the first sum
$$S_{1}$$being less than some value x is:
$$P(S_{1}\leq x) = x$$Then probability of the second sum
$$S_{2}$$being less than x can be calculated by taking
$$P(S_{1}\leq x)$$as a conditional distribution on
$$P(S_{2}\leq x)$$and calculating the marginal prob.:
$$P(S_{2}\leq x) =\int_0^1{P(S_{1}\leq x)}\cdot 1\ dx$$$$ =\int_0^1 x\ dx$$$$ =\frac{x^2}{2}$$Probability of the 3rd is:
$$P(S_{3}\leq x) =\int_0^1{P(S_{2}\leq x)}\cdot 1\ dx$$$$ =\int_0^1 \frac{x^2}{2}\ dx$$$$ =\frac{x^3}{6}$$Extending to n:
$$P(S_{n}\leq x) =\frac{x^n}{n!}$$Since we’re concerned with the sum exceeding $x=1$:
$$P(S_{n}\leq 1) =\frac{1}{n!}$$To find the probability of exactly $n$ draws summing greater than 1, we have to find the probability of $n-1$’s sum exceeding one and subtract that off the probability of $n$’s sum exceeding.
$$P(N\ge n) = \frac{1}{n-1!} - \frac{1}{n!}$$Then we can calculate the expectation on that distribution:
$$E[P(N\ge n)] = \sum_{n=2}^\infty n \cdot (\frac{1}{n-1!} - \frac{1}{n!})$$$$ = \sum_{n=2}^\infty (\frac{n}{n-1!} - \frac{1}{n-1!})$$$$ = \sum_{n=2}^\infty \frac{1}{n-2!}$$$$ = \sum_{n=0}^\infty \frac{1}{n!}$$$$ = e$$The expected number of draws you need to exceed 1 is e!
The convergence to e does make some intuitive sense. You’ll almost assuredly never bust on the first draw, so the minimum # of draws you’ll have to make is at least two. Since we’re drawing from a uniform distribution from [0,1], we have a mean of 1/2. 2x the mean gets us to 1, and 3x the mean moves us to 1.5, so a value in between 2-3 fits the bill.