A checked finding

Which primes get unlucky making pairs — 3 vs 5 vs 7 — as it grows?

The question, in your words: “which prime pairs happen to have unluckiness making pairs — like 3 vs 5 vs 7 — as it grows; take the percent chance vs the pairs and show the probability of finding as it grows, because right now it just seems like luck.” So I took each small prime p, measured how often N − p is also prime, and watched it grow.

Grow N up to

drag (200,000)

up to 20,000up to 100,000up to 300,000up to 600,000

shared success rate

17.98%

spread (luckiest − unluckiest)

0.003%

≈ 2 / ln N (predicted)

16.39%

the revealDown every column below the primes are the same — no prime is luckier. Across every row the chance shrinks as N grows. The (un)luckiness isn't in the prime; it averages out to a dead heat.

prime p	up to 1k	up to 10k	up to 100k	up to 200k
3	33.53%	24.57%	19.18%	17.98%
5	33.40%	24.57%	19.18%	17.98%
7	33.47%	24.58%	19.18%	17.98%
11	33.40%	24.59%	19.18%	17.98%
13	33.47%	24.59%	19.18%	17.98%
17	33.60%	24.60%	19.18%	17.98%
19	33.47%	24.61%	19.18%	17.98%
23	33.61%	24.62%	19.18%	17.98%
29	33.61%	24.61%	19.18%	17.98%
31	33.47%	24.62%	19.18%	17.98%

Success rate of each prime as a Goldbach partner, up to 200,000 — the bars are the same length on purpose:

317.98%

517.98%

717.98%

1117.98%

1317.98%

1717.98%

1917.98%

2317.98%

2917.98%

3117.98%

For each small prime p, this is the fraction of even numbers N (up to the cap) for which N − p is also prime — i.e. how often p works as one half of a Goldbach pair. They come out equal because as N sweeps the evens, N − p lands on primes at the same average density no matter which small p you fixed.

◆The answer: it's not luck, and no prime is special

It really looks like luck — but the numbers say otherwise, in a surprising way:

no lucky primeAveraged over all even numbers, every small prime is an equally good Goldbach partner. At a million, 3, 5, 7, 11, … all succeed 15.70% of the time — identical to two decimals. The tiny wobble you see at small ranges is just sample noise and vanishes as N grows.

the chance shrinksEach individual prime's success rate falls as N grows (~24% → ~19% → ~16%, tracking 2 / ln N) — yet the total number of Goldbach pairs still grows, because you get more and more primes to try. Fewer wins per prime, but far more primes.

So where does the real (un)luckiness live? Not in the prime — in the even number itself. Numbers with more small factors (especially 3) get many more pairs. That's the banding you can see in the comet, and the growth story is on the pattern page.

01The odds a pair works, as it grows

Your algorithm, drawn out: there are C = π(N/3) candidate pairs, each works with w ≈ 1/ln N. So the chance all miss is (1−w)^C, and the chance at least one lands — Goldbach holds — is 1 − (1−w)^C.

Grow N up to

drag (100,000)

up to 20,000up to 60,000up to 150,000up to 300,000

expected working pairs (C·w) actual Goldbach pairs chance it works = 1 − (1−w)^C

candidates C = π(N/3)

3,569

one pair works, w ≈ 1/ln N

8.7%

expected pairs C·w

310

chance it works

99.9999…%

chance it fails (1−w)^C

1 in 10^141

Your algorithm, growing: there are C = π(N/3) candidate pairs, each works with w ≈ 1/ln N. The chance all miss is (1−w)^C, so the chance at least one lands is 1 − (1−w)^C. The gold line (pairs) climbs; the cyan line (chance it works) slams into ~100% almost immediately and never leaves — and the failure odds fall to 1 in 10^141. Not luck — an avalanche.

02So how far must we check? (the luck number)

Add up every even number's tiny failure chance and ask: how far do you have to check before you're near-certain nothing's out there? That threshold is the “luck number.”

Compute using evens up to

drag (200,000)

up to 40,000up to 100,000up to 250,000up to 400,000

the luck numberCheck every even number up to about 1,104 and you're already 99.9% likely that no counterexample exists anywhere above it. That's the number you asked for.

to be this sure nothing's out there…	…check up to about
90%	456
99%	762
99.9%	1,104
99.9999%	2,244

Why so small? The cumulative expected counterexamples climb at the very start, then flatline — each new even number adds almost nothing:

Reading it: the running total climbs to ≈ 11.2 then stops. Almost all of that comes from the first few even numbers, where this crude estimate is loose — and those are directly checked by hand and all hold. The honest quantity is the tail past the luck number, which is already under 0.001 and keeps collapsing toward zero, so a counterexample is never statistically “due.” The wild part: a few thousand checks make you near-certain, yet computers have verified Goldbach to 4×10¹⁸ — astronomically past the luck number — and found nothing, exactly as predicted. (Still a heuristic, not a proof — being 99.9999% sure isn't certain, which is why it stays open.)

03Drag it to the astronomical

Now go past what any computer could check — up to 10^10000. Watch each number's fail chance get added onto the running cumulative total, and watch that total refuse to budge.

N = 10^12.0 = 1,000,000,000,000

10^0.510^310^610^1210^1810^10010^100010^10000

cumulative chance of fail — every number so far, added up

Σ ≈ 11.2029

This is the running sum of every even number's fail chance from 4 up to 1,000,000,000,000. The step at N = 1,000,000,000,000 adds only +1 / 10^2.01×10^8 — so past the first few numbers the total barely moves, and it converges to ≈ 11.20 and stops.

this number's own fail chance

1 in 10^2.01×10^8

cumulative Σ so far

11.2029

converges to (all N)

≈ 11.20

luck number (99.9% clear)

≈ 1,104

Each ×10 block: what it adds, and the running cumulative total — add this step's chance to all the previous ones:

block up to	this step adds	cumulative Σ
10^1	+1.2810	1.2810
10^2	+6.9548	8.2359
10^3	+2.9650	11.2008
10^4	+0.0021	11.2029
10^5	+0.0000	11.2029
10^6 … 10^12	+0.0000… (each < 10^−300)	11.2029 (unchanged)

so it never breaksThe cumulative fail chance is dominated entirely by the first handful of even numbers (where this crude estimate runs high — and those are all directly verified to hold). Every step after adds a smaller number than the one before, so the running total flatlines by the luck number ≈ 1,104. Drag all the way to 10^10000: that number's own fail chance is 1 in 10^10^9,990, and it adds essentially nothing. Add up every fail chance to infinity and you still get a finite ≈11.2 — so a counterexample is never expected.

The sum Σ P(fail) is the expected number of counterexamples. Because each term shrinks faster than the count of numbers grows, the sum converges instead of running to infinity — that convergence is exactly why Goldbach is believed. (A heuristic, not a proof: a finite tiny expected count can't rule out one freak N, which is the wall the problem still sits behind.)