This is basically @CactusCake's answer, simplified. I think my intuition was the same as theirs: You often get counterintuitive things happening (see: the Birthday Problem, I won't spoil the answer), so we must first get a general idea of "all things being equal, what is the chance?".
But the answer titled "Math" gets bogged down in hard-to-justify specific details ("Take a Boeing 727!" -- "Why?!"), that lead to ballpark numbers (seat occupancy percentages are competitive info, very selectively given out!), that then anyway don't lead to a numerical or algebraic answer: So there was no reason for those numbers anyway (and the comments didn't like them!).
So. Reformulation: Let's assume we have done N different flights (single flights, not ends of connecting flights) where we've always checked 1 suitcase in, and the other passengers always checked in 99 other suitcases altogether. So, what is the probability after N flights that we've NEVER been first to unload?
Very simple, all these flights are independent occurrences (previous flight doesn't influence next flight), so it's the product of these N identical probabilities (see under: IID, Independent & Identically Distributed). And each flight we've got a 99% (or 99-out-of-a-100, or 0.99, whatever you prefer) probability of disappointment.
So the chance of unbroken disappointment is .99^N (using ^ for "power of"; this clearly goes to zero for large N) for N flights, ... But the real question is, Would you have to be
cursed particularly unlucky to be disappointed 100 times in a row??
- 1 flight: 0.99000000 or 0.990
- 2 flights: 0.98010000 or 0.980
- 3 flights: 0.97029900 or 0.970
So I think the OP thought along these lines, "(1- N/100) is a fair approximation, thus after N=100 flights the chance is basically zero", which is just plain wrong (a linear approximation of a power; a nonsensical negative probability past 100; etc). But it's that "tail" past the two most significant digits that grows surprisingly fast!
To save calculating and typing, skip steps by just doubling N so you have the square of the previous: N=2, 4, 8, 16, 32, 64; because A^(2N)= A^(N+N)=A^N * A^N = (A^N)^2. So you must double your flights to halve your chance of unbroken disappointment...
- 1 flight: .99000000
- 2 flights: .99000000 * .99000000 = .98010000
- 4 flights: .98010000 * .98010000 = .96059601
- 8 flights: .96059601 * .96059601 = .92274469
- 16 flights: .92274469 * .92274469 = .85145777
- 32 flights: .85145777 * .85145777 = .72498033
- 64 flights: .72498033 * .72498033 = .52559648
- 96 flights = (64 + 32) flights = .72498033 * .52559648 = .38104711 = about 38%
So if three friends each (independently!) take 100 flights each, you expect one of them to have never been first to unload.
Conclusion: You're unlucky (you missed a 2-in-3 chance of at least once being first), but not particularly unlucky.
And you see the same double-flights-to-halve-chance principle holds for any number of suitcases that are checked in: You're plotting points f(x) for x=N of a function f(x)=a^x with the parameter a close to (but less than) 1, so x going to infinity (basically for b suitcases on each flight, a=1-(1/b) = (b-1)/b -- as said, always less than one, and we have large-ish b, say between 50 and 500?
[[Yes yes I know I'm seemingly horribly rounding in that table, repeatedly multiplying 8-digit-precision and keeping 8-digit-precision... But that was for ease of understanding! They were actually calculated with 20-digit precision, and .381 for N=96 is correct. For 100 flights, it's a 0.36603234 or over-one-in-three chance.]]
So, further confounding factors: (1) Have you actually checked stuff in on all (or vast majority of) hundred flights? (2) Have you actually never been first? Because (2a) if yours appears first but you have poor positioning at luggage belt, you still won't retrieve yours first; and first class/priority have best chance for good positioning, and (2b) on international flights if you don't have "retina scan"/Privium/... fast checking, you may have arrived at the belt with the luggagge already there, exactly the time you were first, natch! Also, (3) practically each flight the first 5items on the belt are pushchairs and child car seats, as they were collected at the gate (after loading checked luggagge, so unloaded first); this may mislead your observations?
With that as a given, on the flights I take you have to pay to check in (always for 'budget' companies, the last 2--3years for more medium like BritishAirways in my case -- the complimentary gin&tonic is a receding memory on shorter flights!!); so I check in only when I'm on a significant trip, say 10+days or 7days in a cold/wet destination; or sports/elegant dinners thus extra clothes. 5-day academic conference = handluggage only. "Significant trips" tends to be further, so larger plane, so 100+ checked in suitcases seems reasonable to me and the 1-in-3 chance (all else being equal) stands: Redo for 50 if you feel 100 suitcases is not justified.
So this is all a priori reasoning, with the other answers (economy = bottom of unloading priority; economy = checkin on intercontinental closes before business checkin; LIFO to an extent; maybe you have a habit of checking in early; ... ) being strong factors on top.