Crash odds, priced: the chance of every cash-out target

The number and the clock

Crash looks like a race against a rocket, and mechanically it is two much simpler things. Before the round starts the engine draws one hidden number, the bust point, from a committed hash. Then a clock starts and a multiplier climbs along a fixed curve; when the curve reaches the bust point the round is over, and everyone still riding loses their stake. A cash-out — manual, or an auto target set before the round — locks the multiplier at that moment and pays stake times multiplier. Nothing about the flight is random: the only random thing in the round is the number drawn before it.

Write the random part as a fraction u = h ÷ 2⁵², spread evenly between 0 and 1, and the bust formula becomes floor((100 − u) ÷ (1 − u)) hundredths. At u = 0 that is exactly 100, a bust at ×1.00 before the curve has moved; as u approaches 1 it runs away, and the clamp stops it at 1,000,000 hundredths, ×10,000. Everything below follows from that one line and the strict rule for what counts as a win.

Strictly above: the rule that sets the price

This is the detail the odds hinge on. A target of T hundredths does not win when the bust point equals T; it wins only when the bust point is at least T + 1. From the formula, the chance that the bust point is at least K hundredths is 99 ÷ (K − 1) for any K from 101 up, so the chance that it is at least T + 1 is 99 ÷ T. For a target of ×2, that is 99 ÷ 200 = 49.50%. Multiply by the ×2 payout and the return is 0.99 — and the T cancels the same way at every target.

Win chance = 99 ÷ T with T the target in hundredths, exact under the strict rule. Return = target × win chance = 0.99 at every row. Swing is the standard deviation of one round’s result per unit staked, √(0.99 × target − 0.9801). “Reached at” is 10 × log₂(target) seconds on the doubling curve. The lowest target the bet form accepts is ×1.01; ×10,000.00 is the bust cap itself and a target there can never be strictly below the bust, so ×9,999.99 is the top of the table.

The two middle columns do not move. Crash under this rule is priced exactly like limbo: 1% of every unit staked, whatever target you pick, whether you cash out by hand at ×1.37 or set an auto target at ×500. What the target changes is the shape of the result — a stream of small wins at ×1.10, a drought broken by a windfall at ×1,000 — and the last two columns measure that. At ×2 a round swings by about one stake; at ×1,000, by thirty-one stakes. The ride to ×1,000 also takes a hundred seconds, which is a different kind of cost.

One row deserves a second look. ×1.98 wins exactly half the time, which makes it the median bust point: half of all rounds bust above ×1.98 and half at or below it. The round number ×2.00 is a coin flip with a half-point handicap, 49.50%, and that half point is where the 1% lives at that target.

Seen versus paid: the one-hundredth gap

There are two questions that sound the same and are not. How often does the curve show ×2.00? And how often does a ×2.00 target get paid? The first is the chance that the bust point is at least 200 hundredths, 99 ÷ 199 = 49.75%. The second is the chance that it is at least 201, 99 ÷ 200 = 49.50%. The difference, one round in 402, is the round that busts on exactly ×2.00: the screen reaches the number and the ticket loses.

Reaches = 99 ÷ (T − 1); paid = 99 ÷ T; the difference is 99 ÷ (T (T − 1)), the probability mass on the single hundredth T. The gap matters at low targets and vanishes at high ones.

This gap is the reason the engine draws the bust point as an integer number of hundredths rather than as a decimal that is rounded afterwards. With an integer draw, the strict rule hands exactly one hundredth of the distribution to the house at every target, and that one hundredth is what makes the return come out at 0.99 rather than a little above it. Our earlier comparison of limbo and crash priced crash from the “reaches it” column, 99x ÷ (100x − 1), and concluded that short rides were fractionally cheaper than 99%. That was the wrong column: under the engine’s strict settlement the price is 99.00% at every target, and the article now carries a correction.

Where the bust lands when nobody is aiming

The same formula describes the round itself, regardless of what anyone bets. The chance that the bust point is at least K hundredths is 99 ÷ (K − 1), and that gives the whole distribution of what the history bar shows.

Each row is 99 ÷ (lower − 1) minus 99 ÷ upper, in hundredths. The first row is the chance that (100 − u) ÷ (1 − u) falls below 101, u < 1 ÷ 100. The last row is every draw the clamp pins to 1,000,000 hundredths. Rows sum to 100%. Durations are 10 × log₂(multiplier) seconds.

Two rows explain most of what a crash session feels like. Exactly one round in a hundred busts at ×1.00, before the curve has moved a single hundredth, and no target — not even ×1.01 — is paid on it. And about one round in ten reaches ×10, which is often enough to be seen several times an hour and to make a high target feel reachable. It is reachable; the table says how often, and “often enough to remember” is a different number from often. The median round busts at ×1.98, just under ten seconds in, and the average round lasts 14.3 seconds because the rare long flights pull the mean up.

The rounds you sat out are drawn from this same distribution, and every one of them is a fresh HMAC of a fresh chain link. A ×500 in the history bar makes the next ×500 exactly as likely as it was before, 0.198%. That is the gambler’s fallacy in its purest form — here independence is not a statistical assumption but a property of the hash.

What chasing a big multiplier costs

Because the edge is the same everywhere, the choice of target is a choice about variance, and variance has a price that is easy to state: how many rounds you should expect to wait, and how likely a long wait is. With win chance p per round, the chance of at least one win in N rounds is 1 − (1 − p)ᴺ.

1 − (1 − 99 ÷ T)ᴺ, identical to limbo because the win chance is the same. Time assumes the average flight of 14.3 s plus the ten-second betting window, 24.3 s per round; a session at ×1,000 is dominated by the wait, not by the flight.

Put a stake on it and the arithmetic turns cold. A hundred one-unit bets at ×100 cost 100 units to place and return 99 on average — the 1% again — but the average is made of a 63% chance of getting back roughly 100 and a 37% chance of getting back nothing. Nothing about the target changes the 1% you pay per unit. Everything about it changes how the results are distributed among players, and in crash it also changes how long you sit watching a curve.

The low-target version of the same trap is the martingale: cash out at ×2, double after each loss, and collect one unit per win. The win chance at ×2 is 49.50%, a losing run of ten happens 0.11% of the time, and the eleventh bet is 1,024 units. The martingale, priced works that through on a roulette wheel; the crash figures are within a tenth of a point of it. Risk of ruin gives the general formula for how long any fixed-stake plan lasts against a 1% edge.

Cashing out by hand

A manual cash-out is paid at whatever hundredth the live multiplier shows when the server receives it, provided that hundredth is still strictly below the bust point. The curve does not tick evenly: floor(100 × e^(0.0693 t)) advances by one hundredth every 143 ms near ×1.01, every 72 ms around ×2, every 14 ms around ×10, and every 1.4 ms around ×100. A hand on the button at ×10 is choosing a number to within a few hundredths, at ×100 to within a few tenths, and the hundredth that finally lands is decided by network latency as much as by intent. None of that changes the price — each hundredth is paid at exactly 99% in expectation — but it does mean a manual target above ×10 is a looser instrument than an auto target, which fires at exactly the number you typed.

Checking a round yourself

1. Before the round, the panel shows the commit hash of the chain and the salt. After the round it reveals the chain link for that round; hashing the link must give the previous round’s link, which is how the chain proves the order was fixed in advance.
2. Compute HMAC-SHA256 with the revealed link as the key and the salt as the message. Take the first 52 bits of the digest — the first thirteen hex characters — as h.
3. Compute floor((100 × 2⁵² − h) ÷ (2⁵² − h)) in integer arithmetic and clamp it to 100–1,000,000. That is the bust point in hundredths, and it should match the number the round crashed on.

Any HMAC-SHA256 tool and a big-integer calculator will do; the open-source verifier’s shared HMAC helper is the same primitive. The integer division is not optional: a floating-point version disagrees with the server at the floor boundaries, which is exactly the hundredth the strict rule is about. One rounding, not many explains why the verifier reproduces the server’s arithmetic rather than approximating it.