Last month, our puzzle was inspired by failed crypto tycoon Sam Bankman-Fried, who once said he would take a bet with a 51% chance to duplicate the Earth, and a 49% chance to wipe it out completely. The problem we posed added an extra twist:

Suppose Earth had probability $x$ to double, and probability $(1-x)$ to get wiped out. Further, once a year, the same bet would be taken for any resulting duplicated Earths.

What would $x$ have to be for there to be a 50% chance of everlasting life in the universe?

Solution

First, let’s get a feel for the process.

On the annual bet day, Earth faces a binary fate—either it doubles or vanishes. In the event of vanishing, humanity faces permanent loss. Conversely, if Earth doubles, both instances continue the bet, creating a potential lineage that persists indefinitely.

Modeling the problem

Though there are many different possibilities, with lineages surviving for varying durations, there are essentially only two outcomes leading to extinction: immediate wipeout or doubling with subsequent extinction for both Earths:

$$ \begin{align*}P(\text{extinction}) = &\ P(\text{immediate wipeout}) \\ &+ P(\text{doubles, but both lineages go extinct}).\end{align*} $$

The probability of immediate wipeout is the probability the bet is lost, $(1-x)$. By complement, the probability of Earth doubling is $x$.

If Earth doubles, each child Earth faces the same situation as the parent which means the original Earth has the same extinction probability as either of its children.

So, the probability that the first Earth doubles and both lineages go extinct is

$$ x P(\text{extinction})^2. $$

If we substitute both of these results into the original equation, we get

$$ P(\text{extinction}) = (1-x) + x P(\text{extinction})^2. $$