The streak that feels wrong
Every dice player eventually sits through a run that feels impossible: eight, nine, ten losses at a bet that wins about half the time. The instinct is that the game has turned, or that a win is now owed. Neither is true, and the cleanest way to see it is to ask a different question. Not “how unlikely was that run?”, which is always a small number, but “how long a run should a session of this length contain?”, which turns out to be a surprisingly large one.
The answer depends on two things only: the chance of losing one roll, and how many rolls are played. Both are known exactly on a provably fair game, so the answer can be priced exactly too.
The numbers
The tables give the longest losing run in a session, by target and by the number of rolls. “Median” is the run half of all sessions reach; “worst 10 %” and “worst 1 %” are the runs that the unluckiest tenth, and the unluckiest hundredth, of sessions reach or pass.
| ROLLS | MEDIAN | WORST 10 % | WORST 1 % |
|---|---|---|---|
| 100 | 6 | 9 | 13 |
| 1,000 | 9 | 13 | 16 |
| 10,000 | 12 | 16 | 20 |
Calculation for this article: exact distribution of the longest run of losses, loss chance 5,050 ÷ 9,999. Each “worst” figure is the first run length that no more than that share of sessions reach.
| TARGET | 100 | 1,000 | 10,000 |
|---|---|---|---|
| ×1.98 | 6 | 9 | 12 |
| ×2 | 6 | 9 | 12 |
| ×3 | 9 | 15 | 21 |
| ×10 | 25 | 47 | 69 |
Calculation for this article; columns are the number of rolls in the session. ×1.98 is over 50.00, ×2 under 49.50, ×3 under 33.00, ×10 under 9.90.
Put as the question players actually ask, the chance of seeing at least ten losses in a row at ×2 is 4.8 % in a session of 100 rolls, 41.3 % in 1,000 and 99.5 % in 10,000. Ten in a row is not rare. It is a normal feature of a long session, and over ten thousand rolls its absence would be the strange thing.
Why it grows so slowly
Look down the ×2 column and the median grows by three each time the session gets ten times longer: 6, 9, 12. That is not a coincidence. A run of k losses has probability qᵏ, where q is the chance of losing one roll, and a session of n rolls offers roughly n × p places for a run to start (p = 1 − q). The longest run is about the k at which those two balance: k ≈ log(n × p) ÷ log(1 ÷ q).
At ×2, 1 ÷ q is about 1.98, so each factor of ten in the session adds log(10) ÷ log(1.98), about 3.4 losses, to the worst run. The approximation gives 5.7, 9.1 and 12.4 for 100, 1,000 and 10,000 rolls against the exact medians of 6, 9 and 12. At ×10 the loss chance is 90.1 %, 1 ÷ q is only 1.11, and each factor of ten adds about 22 losses, which is why a long-shot target produces runs that look absurd and are not.
Two consequences follow. The first is that a long run tells you nothing about the next roll. The engine draws each result from its own seed and nonce, with no memory of the last; the chance on roll 11 after ten losses is the same 49.495 % as on roll 1. The gambler’s fallacy is exactly the belief that it is not. The second is that the run you will meet is set mostly by how long you play, not by luck: a player who plays ten times as many rolls should expect a worst run about three losses longer at ×2, every time.
What a streak costs a doubler
For a flat bettor, a long losing run is unpleasant and nothing more: ten $1 losses cost $10. For anyone who doubles after a loss, the same run is the whole question, because the stake grows as 2ᵏ and the money lost before the next win is 2ᵏ − 1.
| RUN | LOST | NEXT BET | IN 1,000 |
|---|---|---|---|
| 6 | $63 | $64 | > 99.9 % |
| 9 | $511 | $512 | 65.4 % |
| 12 | $4,095 | $4,096 | 12.6 % |
| 15 | $32,767 | $32,768 | 1.7 % |
Calculation for this article. “In 1,000” is the chance that a session of 1,000 rolls contains a losing run at least that long, from the exact distribution above.
This is where the slow growth stops being comforting. Two 1,000-roll sessions in three reach a run of nine, which a $1 doubler survives only with $511 behind the next $512 bet. One session in eight reaches twelve. And every win in between returns exactly the $1 the system was built to make. The Martingale, priced works through the same arithmetic on a roulette wheel; the dice version differs only in the base: at 49.495 % the runs are a touch shorter than at roulette’s 48.6 %, and they still arrive.
Using the number
The practical use of the table is to decide in advance what a session will contain, rather than being surprised by it. If you are about to play 1,000 rolls at ×2, a run of nine losses is the typical case and twelve is one session in eight. If that run would hurt, the stake or the session is too large, and the table says so before the first roll, not during the eighth loss.
The numbers are only this clean because the odds are fixed and published. Every roll on Betkyo dice is drawn from a committed server seed and your own client seed, and the dice odds article lists the chance behind every target. A streak on a game like that is the arithmetic arriving on schedule; there is nothing else in it to read.
Odds you can price in advance9,999 equally likely results, roll-under pays 0.99 ÷ chance, every round drawn from a committed seed and yoursDice →
How many losses in a row are normal at a 50/50 dice bet?
At ×2 on Betkyo dice (49.495 % to win), the typical longest losing run is 6 in 100 rolls, 9 in 1,000 and 12 in 10,000. Fewer than one session in ten of 1,000 rolls reaches 13. Calculations for this article from the engine’s odds.
What is the chance of 10 losses in a row on dice?
At ×2 it is 4.8 % somewhere in 100 rolls, 41.3 % in 1,000 rolls and 99.5 % in 10,000 rolls. The chance of ten losses in the next ten rolls specifically is about 0.1 %, which is why a run feels rarer than it is.
Does a long losing streak make a win more likely?
No. Each roll is drawn independently from its own seed and nonce; the chance on the next roll is the same as on the first. Believing otherwise is the gambler’s fallacy.
Why does the longest streak grow so slowly with more rolls?
The longest run is about log(n × p) ÷ log(1 ÷ q) for n rolls with win chance p and loss chance q. At ×2 that adds about 3.4 losses for every tenfold increase in the number of rolls.
How much does a Martingale need to survive a losing streak on dice?
Doubling from $1, a run of k losses costs 2^k − 1: $511 for nine, $4,095 for twelve, $32,767 for fifteen. In 1,000 rolls at ×2 a run of nine is reached 65.4 % of the time and twelve 12.6 % of the time.
- Betkyo engine source: OriginalBaseDiceService.kt — getResult() (9,999 results 0.01–99.99), getWinChance(), checkAndGetMultiplier() (0.99 ÷ chance, floored), getEarned() (strict comparison)
- Calculation for this article: exact distribution of the longest run of losses by dynamic programme over the run in progress, for loss chances 5,050/9,999 (×2), 5,000/9,999 (×1.98), 6,700/9,999 (×3) and 9,010/9,999 (×10) and sessions of 100, 1,000 and 10,000 rolls; the log(n·p)/log(1/q) approximation for comparison; the 2^k − 1 doubling cost
