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This one was too much fun, so I hope you will forgive and indulge me for this installment of What Are The Chances? because it highlights concepts I want you to carry with you always. The video is linked at the bottom.
The first, and really the only, is nothing has a probability! If you learn only one thing from me, I hope and pray it is this. It is the key to understanding all uncertainty, and even all certainty itself. All things are known, unknown, or in between depending on what evidence you assume. You, and none other but you. Change the evidence, change the probability.
It is never wrong, therefore, when somebody asks you for the probability of something that has happened, and often just happened, to say “1”. It happened; therefore, the chance it happened is 1. 100%!
Thus we have answered our titular question: the probability these two great men died on the same day in 1826 is 1, assuming the evidence of historians is true.
I say notation can help as well as hinder. Here it does both (because many are frightened of notation). We can write that same conclusion this way:
Pr(Both died 4 July 1826 | Historians say both died 4 July 1826) = 1.
Simple! Even trivial, so much so that it seems a preposterous waste of time to bother. And yet…
There are more ways to think about the question. Ask this instead: What is the chance I, the Statistician to the Stars!, and you, dear reader, both hand in our dinner pails tomorrow, the day after this fine post greets the world? There is no answer because we have supplied any evidence. So let’s supply some.
I’m in fine fettle, and given that (conditional on that, assuming that, granting that, supposing it is true that), it is unlikely I will crap out tomorrow due to natural causes like a bear attack. However, things happen: tree branches fall, bombs explode, tempers flare. Picking a number out of this, for the sake of picking a number, let’s say the chance of meeting my reward is one in a million. This is, of course, arbitrary, a model, and all models only say what they are told to say. I told mine to say this.
I can’t speak for you, of course, but if you’re reading this, you either have a healthy sense of humor or ache to find—at last!—a mistake in something I have wrote, so you’re in much the same condition I am and want to live. Say it’s the same number for you.
Thus, accepting all this, the chance we both slip under tomorrow is 10^-6 x 10^-6 = 10^-12. In notation:
Pr(Two fine people fizzle out | chance either one does is 1 in a million) = Pr(I die|E) x Pr(You die|E),
where E is that evidence we accepted.
And there is the seed of the answer about Adams and Jefferson. Next consider we are now where we are not but were where we weren’t then. On the very Glorious Fourth, 1776 itself. We ask ourselves then: what are the chances these men die on 5 July 1776? We’d get something like the same answer, depending on what assumptions we make for E.
Now change the question to What are the chances these men die on same day? That day could be 5 July 1776, but it could also be 6 July 1776, or 7 July 1776, and so on down the way. Can you answer that from what we’ve done so far?
DO NOT READ FURTHER UNTIL YOU’VE PONDERED THE SOLUTION
An equivalent way to ask the question is this: What are the chances both men die on 5 July 1776 or 6 July 1776 or 7 July 1776 or, etc.?
The “or” is key. For the 5th alone we have the answer. But what about the 5th or 6th? Supposing the same evidence for the 6th as we did for the 5th (i.e. the one in a million), the answer for just the 6th from the perspective of the 5th is the same: 10^-12. But for the 5th or the 6th, from the perspective of the 4th, the answer is completely different.
Because why? Because to get to the 6th both men had to have survived the 5th; only assuming both made it to the 6th can we ask for the chance of 5th or 6th. The “events”, if you want to call them that, are not the same. In shorthand notation, for just the 5th we want in notation
Pr(DD|E) = Pr(D|E)Pr(D|E),
where D = man dies, and E is our evidence. For just the 6th we want in notation
Pr(ADAD|E) = Pr(AD|E)Pr(AD|E),
where D and E are the same but A means the man lived the one day before he might day on the second day. (We could put subscripts here and keep careful track, but that seems an unnecessary burden here.) Point is: if, say, Adams met his reward on the 5th, and Jefferson lasted at least into the 6th, there as thus no chance after Adams died both could have died on the same day. Obviously!
And so on for any other day. For just those days past the 4th we have
Pr(A^iDA^iD|E) = Pr(A^iD|E)Pr(A^iD|E),
where i = Days past 4th minus 1, or n – 1, if you like, and where there is nothing in E to suggest either man’s death is related to the other; that is, knowledge of ones conveys no information about the other. For the 6th, i = 2 – 1 = 1. That means (and if I did the counting right: don’t trust me), from your perspective on 4 July 1776, the chance both die on 4 July 1826 is
Pr(A^18262 D A^18262 D|E, 4 July 1776) = Pr(A^18262 D|E)Pr(A^18262 D|E, , 4 July 1776),
since (I think) there were 18,263 days from the first to the last 4th.
With E, we assume the chance of death on any day, considered on that day, is one in a million, 10^-6. Of course, that is a bad E as men grow aged. But accept it for the moment; below, we’ll change it. We also deduce
Pr(A|E) = 1 – Pr(D|E).
Thus, given E,
Pr(A^18262 D A^18262 D|E, 4 July 1776) = Pr(A^18262 D A^18262 D|E, , 4 July 1776) = Pr(A^36524D^2|E, , 4 July 1776) = (1-10^-6)^36524 x 10^-12 ~ 0.964 x 10^-12 ~ 10^-12.
Given the large chance of living every day, we haven’t much altered the chance they die together fifty years from today. Recall, and this is crucial, we are assuming our perspective from 4 July 1776. As a bonus we also get the chance these men both live to 3 July 1826, which is 96%. Assuming E.
If you choose the perspective of 3 July 1826, then you are right back to Pr(A^2*i D^2|E) where i = 0, i.e. Pr(A^2*i D^2|E) = Pr(D^2|E), and where to keep readability I’ve suppressed our new date perspective as evidence we consider with E.
So the answer, as it always does, depends on our perspective!
None of this yet answers the question what are the chances of the 5th or 6th or 7th or etc., from the perspective of 4 July 1776. We want, using our A/D notation, the probability of this “event”:
DD or ADAD or AADAAD or AAADAAAD or etc. up to 4 July 1826.
To do that, you might think we need the inclusion-exclusion formula, which was featured in the world-gripping article On The Probability Of Large Rocks From Space! Recall NASA calculated the chance of each individual object hitting the earth. But I thought a better question was What Are The Chances ANY Object Hits Us, Year By Year? That is object 1 hits or object 2 hits or etc. Without going into all that again, which you can review in that article, we there had to care for double counting and the like, because multiple objects could hit. We could not simply add the individual probabilities.
But here we don’t need it. Because multiple events cannot happen. Once either man dies, it’s over. The simplest inclusion-exclusion is for ORing two events. Our first two events are DD and ADAD. With inclusion-exclusion, the probability is Pr(DD|E) + Pr(ADAD|E) – Pr(DD & ADAD|E), but the probability of “DD & ADAD” (given E) is precisely 0, because if the men die on 5 July, they cannot re-die on the 6th. If you review the NASA article, you’ll see that holds for any number of events we OR together.
That makes life easy! Finally, for any n days out, we can calculate, given E (with its constant daily death chance), the probability of the event DD or ADAD or AADAAD or … etc. That is
$$\sum_{i=o}^{n} \Pr((A^2)^i|E)\Pr(D^2|E) = \Pr(D^2|E)\sum_{i=o}^{n} \Pr((A^2)^i|E)= q\frac{1-p^{n+1}}{1-p}$$,
where $q=\Pr(D^2|E)$ and $p = \Pr(A^2|E)$, because (as you might recognize) the sum is a geometric series. Neat.
We did the chance of the 5th, which was 10^-12. For the 5th or 6th, using this formula, gives 2 x 10^-12. Plugging in n = 18,263 (i.e. 50 years hence), we get about 1.79 x 10^-8. That’s a huge increase from the single day, in relative terms, i.e. from 10^-12 to 10^-8, or 10,000 times higher.
If you stare at the formula a bit you realize it converges for large n (suppose both men are Methuselan) to $q/(1-p) = \Pr(D|E)/(2-\Pr(D|E))$, which is about 5 x 10^-7. This works for any Pr(D|E) you pick, as long as it lives between 0 and 1.
That, given E, is the answer for the chance both men die on 4 July 1826 from the perspective of 4 July 1776. For today’s perspective, the answer is 1. For other perspectives before 4 July 1826, simple change the n. We’re done!
Except that this E doesn’t seem especially realistic as men age. From the perspective of 3 July 1826, and if you knew both men then, you’d know they were creaking along and not far from Final Exit. The evidence E won’t do. Of course, you could adjust E at that day to something more reasonable, say 0.9. But something close to one in a million better fit 4 July 1776. What we need is to adjust E as a function of time. We’ll lose our simple series if we do that, but gain realism.
What say you to this guess? We’ll call it F, which is E by time:
Small daily chances, rising rapidly toward the end (a tweaked logistic function, which you don’t need to know about, but if you want it, the code is below). The formula is now dependent on where you are on the curve, and it still of course assumes both men have the same chances on any given day, which is reasonable given they were only 8 years apart in age. This starts at around 10^-6 per day and ends around 0.9 for individual daily chances.
If you plug this curve into the sum above, conditional now on F, you get, from the perspective of 4 July 1776, 5.5 x 10^-7. From the perspective of 3 July 1826 you get 0.9 x 0.9 = 0.81 for both.
And that is the final answer. It depends on the evidence you assume. All probabilities do.
Thanks to JRob from the comment section who suggested this topic. He had a clever analysis in the same spirit as ours, counting days from 4 July 1776, assuming that perspective. He also guessed the men would live a long time after that date, with a potential life of 120 years old each. Then he counted the pairs of days on which the two deaths could happen, and divided the two numbers, coming to about 10^-9. This is not a bad first guess, though it leaves out the times in which one man dies before the other.
JRob also pointed to a Forbes article which had a different analysis. Must have been from a frequentist, because he started by calculating probabilities the men lived longer than they did! Which, from our perspective, is 0. Very p-value in flavor. His answer, which was only from the perspective of one particular day, was 6.6 x 10^-5.
VIDEO
CODE
# the daily death curve, or F
m = 18263
n = seq(-10,3, length.out = m)
k = 2
q = 10^-6
f = .9/(1+exp(-k*n)) + q
# the probability from the perspective of m days out
q = f^2
p = (1-f)^2
sum( p^(m-1)*q ) # this is the anwerSubscribe or donate to support this site and its wholly independent host using credit card click here. Or use the paid subscription at Substack. Cash App: \$WilliamMBriggs. For Zelle, use my email: matt@wmbriggs.com, and please include yours so I know who to thank. BUY ME A COFFEE.
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“…a mistake in something I have wrote…” Genius.