Is a losing streak of 10 at 49.5% evidence the game is rigged?

No. In 1,000 Dice bets at 49.5%, the chance of at least one run of 10 or more losses is 41.30%. Long streaks are a property of the odds. Rigging is tested by recomputing bets in the verifier, not by counting losses.

Why a recurrence rather than the usual shortcut formula?

The shortcut 1 − (1 − q^n)^(m − n + 1) treats every window of n bets as independent, which they are not. For the example above it gives 65.69% against the exact 41.30%. The recurrence counts each way the first streak can occur exactly once.

Do my client seed or nonce change these odds?

No. Under the scheme every bet is an independent draw at the published chance. Seeds decide which particular results you get, not how likely each one is.

Losing-streak odds calculator

How likely is a run of n losses somewhere in m bets? Enter the win chance per bet, the streak length and how many bets you will place. The headline probability is computed exactly by a recurrence over every bet, not by the independent-windows approximation most streak calculators use.

inputs

Win chance per bet (%)

Between 0.01 and 99.99.

presets

Losing streak length

Whole number, 1 to 500.

Number of bets

Whole number, 1 to 100,000 (exact recurrence; larger runs slowly).

Independent bets at a fixed chance. Runs in your browser; nothing is sent anywhere.

resultp = 48.65% · n = 6 · m = 200

With a 48.65% win chance, the probability of at least one losing streak of 6 or more somewhere in 200 bets is 84.40%.

P(streak ≥ n)

84.40%

1 in 1.18

expected streaks ≥ n

1.75

separate runs of that length or longer

expected longest streak

7.2 bets

in 200 bets

Any single stretch of 6 bets is all losses with probability 1.83%; the headline figure is higher because there are 200 places for it to start. A streak like this is a property of the odds, not a sign of tampering. If you want to check a specific bet, use the verifier.

See the maths

Loss chance per bet q = 1 − p = 0.5135. Bets are independent with a fixed chance, which is how these games are constructed.

P(at least one run of ≥ n losses in m bets) is computed exactly, not approximated. Let a_j be the probability that the first j bets contain no run of n losses:

a_j = 1                       for j < n
a_n = 1 − q^n
a_j = a_(j−1) − p · q^n · a_(j−n−1)   for j > n

P(streak ≥ n) = 1 − a_m = 0.84401512

The recurrence says: the first run of n losses ends at bet j only if bet j−n was a win, the next n were losses, and nothing before that contained a run. It is evaluated for j up to m = 200.

Expected number of streaks ≥ n counts maximal runs: a run can start at bet 1 unconditionally, or at any of the other m − n positions after a win.

E[runs ≥ n] = q^n · (1 + (m − n) · p) = 1.74866

Expected longest losing streak sums the tail probabilities of the longest run L, each computed with the same recurrence, until the terms fall below 10⁻¹².

E[L] = Σ_(k ≥ 1) P(L ≥ k) = 7.24229

Loading calculator…

questions

Is a losing streak of 10 at 49.5% evidence the game is rigged?: No. In 1,000 Dice bets at 49.5%, the chance of at least one run of 10 or more losses is 41.30%. Long streaks are a property of the odds. Rigging is tested by recomputing bets in the verifier, not by counting losses.
Why a recurrence rather than the usual shortcut formula?: The shortcut 1 − (1 − q^n)^(m − n + 1) treats every window of n bets as independent, which they are not. For the example above it gives 65.69% against the exact 41.30%. The recurrence counts each way the first streak can occur exactly once.
Do my client seed or nonce change these odds?: No. Under the scheme every bet is an independent draw at the published chance. Seeds decide which particular results you get, not how likely each one is.

Every figure on this page is computed in your browser from the inputs shown. The URL carries the inputs, so a copied link reproduces the result. Last reviewed 2026-09-23.

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