Happy pi day, with "Death's Dice"

Happy pi day folks. With the two projects happening now, there wasn't quite room to do a full video in celebration of the day this year. But I did want to share a fun "surprise pi" occurrence, this one in the form of a simple tweet thread.

---

Death finds you. You plead with him that it's too soon, and he agrees to a concession. Every year, he'll roll a set of dice, and if it turns up snake eyes (both 1's) he'll take your life, otherwise, you get one more year.

But it's not necessarily a normal pair of dice. 

On the first year, both "dice" will only have two sides, numbered 1 and 2. So in that first year, there's a 25% chance of rolling snake eyes and ending things there.

On the second year, he comes with tetrahedral dice, i.e. both are four-sided, numbered 1 through 4, and again only takes your life if he rolls two 1's. The next year, the dice are six-sided, after that, eight-sided, etc., etc.

Each year you have a lower and lower chance of dying, but he'll come back every year with a new set of dice, never stopping.

You might think the question now is something like "what's your expected number of remaining years of life?" But actually, Death gave you a pretty good deal.

The better question to ask here is "what's the probability that you end up immortal?" That is, the probability that Death rolls infinitely many times, with his ever-growing dice, and never once turns up snake eyes.

(Stop reading now if you don't want the answer spoiled)

Empirically, if you go run some simulations (which I encourage you to do!), you may notice that no matter how many years you run this, you're probability of survival never seem to drop below around 0.6366... But what is this value?

As it happens, your chances of immortality are precisely 2/π. Wild!

If you're curious why this is true, it's essentially just a rephrasing of the Wallis product, which I actually did a video on way back when.