How many visions does Dr. Strange need?

Question

In Avengers: Infinity War...

Dr. Strange sees 14.5 million futures, in which they only defeat Thanos in one.

Later, when Thanos has the Infinity Gauntlet and all the gems, he kills off half the population of the universe, randomly. He makes quite a deal about how important and fair the randomness of it is.

However, if you removed half the population randomly, there are a great deal more potential outcomes than 14.5 million. If Thanos only removed half the population of Earth (which we can round to 8 billion), how many outcomes could there be that Dr. Strange would have to see?

Answer 1

The number of ways to select half the population is the central binomial coefficient for $4\cdot 10^9$, which is ${8 \cdot 10^9 \choose 4 \cdot 10^9}\approx \frac {4^{4\cdot 10^9}}{\sqrt {\pi 4\cdot 10^9}}$ which is enormous compared to $14.5$ million. The justification for the much smaller number is that Dr. Strange is just looking at "significantly different" futures. Exactly which half is eliminated is probably not "significantly different".

Answer 2

Well, that would be $8000000000\choose 4000000000$, which is about $\frac{8000000000^{8000000000}e}{(4000000000e)^{4000000001}\pi}$, which is a whole lot.

Answer 3

In general, for a given population size $2n$, there are $\binom{2n}{n}$ ways to select half of them. Using Stirling's approximation, we see that for large $n$,

$$ \binom{2n}{n} \approx \frac{2^{2n}}{\sqrt{\pi n}} $$

and for $n = 4 \times 10^9$, this yields a number with about $2.4$ billion digits, give or take.

Answer 4

You would have: $$\binom{8*10^9}{4*10^9}=\frac{(8*10^9)!}{((4*10^9)!)^2}$$ Which can be approximated by $$\sqrt{\frac{2}{8*10^9*\pi}}*2^{8*10^9}$$ Which is effectively $$2^{8*10^9}>>1.45*10^8$$