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Thanos snaps his fingers and half the Earth's population disappear. He snaps again and half of the remaining half disappear. Now, what is the probability of any given person (for example, myself) disappearing after 2 snaps?

My simplistic reasoning was that, since after 2 snaps 75% of the population are gone, it means that any one person (e.g. me) has a 75% chance of having been eliminated as well.

My friend's asking for a formula, i.e. mathematical proof of this calculation and the best I can come up with is:

  1. event (snap): 50% chance of being obliterated
  2. event (snap): 50% chance of having survived the first snap and another 50% chance of perishing so 50% x 50% = 25% Sum of two events = 50% + 25% = 75%

Could someone please verify if my logic is sound and what's the correct math behind it.

Thanks so much!

Enic
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2 Answers2

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Let $E_1$ denote the event that the person was not eliminated by the first snap.

Let $E_2$ denote the event event that the person was not eliminated by the second snap.

Then, a person survives if and only if events $E_1$ and $E_2$ both occur.

$p(E_1,E_2) = p(E_1) \times p(E_2|E_1).$
You are given that :

  • $p(E_1) = (1/2).$
  • $p(E_2|E_1) = (1/2).$

Note that this approach bypasses any need to consider whether any events are independent of each other.

The confusing part of this problem is that you are actually given $p(E_2|E_1)$, rather than being given $p(E_2)$.

user2661923
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  • This answer is much clearer than the other one. $A$ and $B$ in the other answer are not independent, though of course $P(B)=P(A)P(B|A);$ this is akin to how we are able to multiply edge probabilities in a probability tree whose trials are not independent. – ryang Mar 25 '22 at 03:55
  • Neatly done, the conditional probability observation was very important because one can't disappear following "both" snaps, so clearly the events $E_1,E_2$ aren't independent as stated. – Sarvesh Ravichandran Iyer Mar 25 '22 at 08:35
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Given two independent events $A$ and $B$ the joint probability is given by multiplication: $P(A \wedge B) = P(A)P(B)$.

Now, the probability of surviving a snap is $0.5$ independent of when the snap happens. So if $A$ is the event of surviving the first snap and $B$ of surviving the second snap, the probability of surviving both snaps is given by $P(A \wedge B) = 0.5\cdot 0.5 = 0.25$.

chickenNinja123
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  • I am not convinced by the independence of the surviving events here. But the probability of surviving both is 0.25. – Snoop Mar 24 '22 at 23:49
  • Makes sense, clean and simple. And to extrapolate the mortality rate, you simply subtract survival rate from all events, i.e. 1 - 0.25 = 0.75 – Enic Mar 24 '22 at 23:51
  • @Enic If you indicate with $S_j$ the surviving event at snap $j$ and $D_j$ the death event at snap $j$: $$P(S_2)=P(S_2|S_1)P(S_1)+P(S_2|D_1)P(D_1)=P(S_2|S_1)P(S_1)=0.25$$ $$P(S_2,S_1)=P(S_2|S_1)P(S_1)=0.25 \neq P(S_2)P(S_1)=0.5\cdot 0.25$$ – Snoop Mar 24 '22 at 23:53
  • If $P(S_2,S_1)\neq P(S_2)P(S_1)$ they are not independent @Jean-ClaudeArbaut – Snoop Mar 25 '22 at 00:19
  • @Jean-ClaudeArbaut Careful. I was dissatisfied with my original answer, and thought it best to make no assumption about the independence of events. I realized, per my (new) answer, that no assumption is necessary. The tagged analogy is (also) plausible, but leads to convoluted (and unnecessary) analysis. – user2661923 Mar 25 '22 at 00:23
  • "Now, the probability of surviving a snap is 0.5 independent of when the snap happens." That's not what "independent" means. "Independent" doesn't mean "same probability for both events", it means "probability for one event doesn't depend on the result of another event". Whether the events are independent depends on how you define them. From one point of view, the people obliterated in the first snap have a 0% chance of being obliterated in the second snap. – Acccumulation Mar 25 '22 at 07:49
  • But we could also say that people obliterated in the first snap still get assigned into snapped/not snapped categories for the second snap, even though it's moot, in which case the events are independent. But as user2661923 says in their answer, since we definitely have $P(S_2|S_1)$, and $P(S_2)$ is both open to interpretation and not directly used (we would use it only to find $P(S_2|S_1)$, which we already have), we should just use $P(S_2|S_1)$. – Acccumulation Mar 25 '22 at 07:49