Infinite Coin Flips

Let's assume that John does $n$ coin flips, each of which have a $\frac{1}{n}$ probability of landing Death. Then the probability of Charlie surviving one flip is $1-\frac{1}{n}$ . So the probability of him surviving $n$ flips is $(1-\frac{1}{n})^n$ , and we need to find $\lim_{n \to \infty}(1-\frac{1}{n})^n$

Notice this is very similar to the limit for $e$ : $e = \lim_{n \to \infty}(1+\frac{1}{n})^n$ . This can actually be generalized to $e^x = \lim_{n \to \infty}(1+\frac{x}{n})^n$ .

In our case, we have $e^{-1} = \lim_{n \to \infty}(1+\frac{-1}{n})^n = \lim_{n \to \infty}(1-\frac{1}{n})^n$ . Therefore, our answer is $\frac{1}{e}$ , or ~0.368.