Doomsday statistics worked out...

Assume there is some rate for civilizations to be born A, and each civilization has some chance D to die before a time B at which point they go extra-planetary… Then at any given time T there would have been A*T civilizations born, A*(T-B)*(1-D) extra-planetary civilizations, and at most A*B earth-like civilizations.

Extra planetary earth like Dead Total
A*(T-B)*(1-D) < A*B > A*(T-B)*D A*T

If we assume a constant birth rate for all civilizations then the odds that you are born into an earth-like civilization instead of an extra planetary civilization is at best p = B/(T-(T-B)*D) or a lower bound on D = (pT - B)/(pT-pB)

We can guess D and calculate how lucky we are to be in a young civilization, or we can guess p and estimate how unlucky we probably will be.

If B = 1e4 yr, and t = 1e10 yr and…

If d = 0 then we are 1 in 1,000,000 and if d = .5 then we are 1 in 500,000

On the other hand if we assume that p = .5 then our odds of surviving would be 1 in 500,000

Of course we have not used the fact that our civilization has already been around for Y years … Or how close are civilization is to going extra-planetary or how close Y is to B - and it does not appear the authors of the ​paper considered this either - but as it turns out it can be fairly important if the existential threats are similar to planet killing asteroids that are likely to strike at any time.

Let's assume that lifetimes of civilizations follow some exponential distribution (a*exp(-a(t-t0)) for t < t0+B

Furthermore if (1-D) fraction of civilizations survive for B years, and the lifetime of civilizations that live less than B follows an exponential distribution, then (1-exp(-lambda B)) = D which gives lambda = -ln(1-D)/B. So given that our civilization is Y years old, the odds that we are destroyed before B is (exp(-lambda Y)) - (exp(-lambda B)) = (1-D)(Y/B) - (1-D)

When Y = 0, this gives D consistent with the paper… but obviously when Y = B, this gives 0 as we would be in the process of colonizing other planets…

If we expect to be extra planetary in 100 yrs, then Y = .99 B and if we assume p = .5 which gives D = (1-2e-6) then this gives our odds of us not going extra-planetary at only 2e-7! Which is much better odds than (1-2e-6)…

Of course the value used for D assumed that civilizations that die live for B years, and not the average value from the exponential. We could calculate the average as a function of lambda, and then use p to get lambda, and then use lambda to get D… But given that we are close to B, means that it must happen with some frequency - which means that most civilizations are long lived, and we must just be lucky to have found ourselves in a pre-extra planetary civilization

Thinking about jars of numbered jelly beans

Let's assume we have a jar of numbered jelly beans of unknown size… And we randomly draw a jelly bean with a 1 on it… Does this tell us anything about the jar of jelly beans? Well - we now know the jar had at least 1 jelly bean with absolute certainty… But what are the odds the jar only had 2 jelly beans to choose from - or 3, 4, etc… Or if we were someone gave us even odds on their being more than X jelly beans - should we take it?

Intuitively, we might expect that the odds that there are more than 1,000 jelly beans are slim at best… To actually calculate the odds, we need to assume some distribution for the possible number of jelly beans in the jar… Let's assume that any number of jelly beans is equally likely up to some limit L. Then for each L', the odds of us drawing a jelly bean with the number 1, are just 1/L'. Now to determine the new expected distribution of jelly beans, we can weight each expectation by the relative odds of drawing a 1 compared to the overall odds of drawing a 1. The overall odds are just 1 + ½ + 1/3 + ¼ + 1/5 + … 1/L = HL where H is the harmonic series. The odds of there being L' Jelly beans were just 1/L, but the odds of drawing a 1 with L' jelly beans is 1/L', so the new odds of there being L' jelly beans is 1/L' / HL instead of 1/L. So we can ask ourselves what are the odds of there being only 1 jelly bean? If we thought there could be somewhere between 1 and 100 jelly beans, then the odds are just 1/H100 ~ 20%. The odds that there are either 1 or 2 jelly beans would then be (1/1 + ½) / H100 = H2/H100 ~30 and the odds are 50/50 that there are fewer than 7 Jelly beans…

Now if there could be N jelly beans with equal probability, and we drew a jelly bean numbered 1, what would be the upper limit of beans that gave us 50/50 odds? This is just Hx/HN = .5 Since the harmonic series is sum(1/n) can be approximated by integral(1/x) = ln(x), the harmonic series asymptotes to ln(N), and we have ln(x)/ln(N)=.5 or x ~ sqrt(N). So for L=100, we would estimate 10 beans would be a sufficient upper limit… (It was 7)… But for 10,000 beans, even odds say there are no more than 100 beans… But we can make no useful predictions without having some upper limit on the possible number of jelly beans… We can also ask what the average number of jelly beans we should expect is - as opposed to the 50/50 cutoff… This would just be sum (L' 1/L')/HL = L/HL which assymptotes to L/ln(L) …

What about exponential distibutions instead of linear

So now if the jelly bean jar is flared, and divided into L vertical sections and there are exp(z) jelly beans per section. and each jelly bean is labeled by it's vertical section. Then the odds of drawing a jelly bean from section z' is just exp(z') …

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