Is there a best win chance to play Dice at?

Not in expectation: every chance returns 99% of the stake on average, because the multiplier is 99 divided by the chance. What changes is the shape. At 95% you win almost every roll for 1.0421× and a single loss costs about twenty wins; at 2% you win 1 in 50 rolls for 49.50×. Pick the shape you can stand; the average is fixed.

Does martingale work on Dice?

It changes what losing looks like, not how much you lose on average. With a 1,000 bankroll and 10 base bet at 49.5%, the plan covers 6 straight losses; the 7th happens 0.84% of the time per streak and at least once in a 200-roll session 57% of the time. Each completed cycle wins 10; the wall loses everything. Average them and the answer is still 1% of turnover.

Do roll-over and roll-under have different odds?

No. Over 50.50 and under 49.50 both cover 49.5% of the 10,001 possible rolls and both pay 2×. Switching between them after a loss changes nothing either: each roll is an independent HMAC-SHA256 float and has no memory of the last one.

Dice guidePublished 2026-09-27

Dice strategy: what the maths actually says

Dice strategy comes down to one formula. The multiplier is 99 ÷ win chance, so every setting returns 99% of the stake on average and nothing you choose changes that. What your choices do change is the shape of the ride: how often you win, how long the losing streaks run, and how fast a staking plan hits its wall. This guide computes all of it for Stake-style Dice, the same formula the free Dice game here runs on, with a simulator link for every plan so you can watch it happen over thousands of sessions.

How Dice pays

The roll is a number from 0.00 to 100.00 with 10,001 equally likely values, read from one HMAC-SHA256 float. You pick a target and a direction; your win chance is the share of the range you cover. The payout is floor(99 ÷ chance × 10,000) ÷ 10,000. A fair game would use 100. The missing 1 is the house edge, and because it is applied as a factor, it is exactly 1% at every chance.

What the win chance does

Lowering the chance raises the multiplier and lengthens the losing streaks in lockstep. Ten losses in a row at 49.5% is a 1 in 927 event per attempt; at 10% it is the normal state of affairs.

Dice: multiplier, losing-streak odds and expected wait at each win chance
Win chance	Multiplier	P(10 losses in a row)	Ten losses, one in	Expected rolls per win	Return
2%	49.5000×	81.71%	1 in 1.2	50.0	99%
5%	19.8000×	59.87%	1 in 1.7	20.0	99%
10%	9.9000×	34.87%	1 in 2.9	10.0	99%
25%	3.9600×	5.63%	1 in 18	4.0	99%
49.5%	2.0000×	0.11%	1 in 927	2.0	99%
66%	1.5000×	0.002064%	1 in 48,441	1.5	99%
75%	1.3200×	0.000095%	1 in 1.0 × 10^6	1.3	99%
90%	1.1000×	1.0e-8%	1 in 1.0 × 10^10	1.1	99%
95%	1.0421×	9.8e-12%	1 in 1.0 × 10^13	1.1	99%
98%	1.0102×	1.0e-15%	1 in 9.8 × 10^16	1.0	99%

The return column is chance × multiplier and reads 99% on every line. Everything to its left is variance: how often, not how much.

What martingale does

Martingale doubles the stake after every loss and returns to base after a win, so that any win recovers all previous losses plus one base bet. The catch is the wall: the streak that outruns the bankroll. With base b and bankroll B, the plan survives k losses where b × (2^k − 1) ≤ B. Here is that wall computed for a 49.5% roll, and how often a session runs into it.

Martingale at 49.5% (pays 2×): where the wall is and how often a session hits it
Bankroll	Base bet	Rolls	Losses covered	Stake at the wall	P(wall) per streak	P(hit wall in session)	Win per cycle
100	10	200	3	80	6.50%	99.9%	10
1,000	10	200	6	640	0.84%	56.6%	10
1,000	10	1,000	6	640	0.84%	98.6%	10
10,000	10	1,000	9	5,120	0.11%	41.3%	10

“P(hit wall in session)” is the exact probability, from a Markov chain on streak length, of at least one run of 7 losses somewhere in the session. With 1,000 and a 10 base, 57% of 200-roll sessions hit it; over 1,000 rolls it is 99%. Ten times the bankroll buys only 3 more doublings.

Run it yourself: martingale, 49.5%, 1,000 bankroll, 10 base, 200 rolls, 2,000 sessions. The simulator prints the ruin rate, the median session (which looks healthy) and the realised return on turnover (which sits near −1%). Compare flat betting on the same settings: fewer ruins, the same expectation. At a lower chance the doublings run out faster: martingale at 10% pays 9.9× per win but nine losses in a row is a 1 in 2.6 event, and 100% of 200-roll sessions contain a run of seven.

What d'Alembert, Fibonacci and reverse martingale do

d'Alembert adds one base unit after a loss and removes one after a win. It climbs slowly, so it survives longer than martingale and recovers slower; a long losing run still leaves you staking many units into a game that returns 99% of each. Run d'Alembert. Fibonacci steps up the sequence after a loss and back two after a win; it sits between the two. Run Fibonacci. Reverse martingale doubles after wins instead, turning a stream of small losses into a rare large win; the same trade in the other direction. Run reverse martingale.

Every one of those runs reports the same figure in the corner: net result divided by total turnover, hovering around −1%. That is the house edge showing through whatever pattern the stakes made.

The conclusion is arithmetic, not opinion. Each roll returns 99% of its stake in expectation. A staking plan decides the size of the next stake before the roll is known, so the expected loss of a session is 1% of everything staked, whatever plan produced the stakes. Martingale, d'Alembert and Fibonacci reshape the distribution of outcomes, mostly by trading many small wins for rare large losses. None of them moves the −1%.

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Questions

Is there a best win chance to play Dice at?: Not in expectation: every chance returns 99% of the stake on average, because the multiplier is 99 divided by the chance. What changes is the shape. At 95% you win almost every roll for 1.0421× and a single loss costs about twenty wins; at 2% you win 1 in 50 rolls for 49.50×. Pick the shape you can stand; the average is fixed.
Does martingale work on Dice?: It changes what losing looks like, not how much you lose on average. With a 1,000 bankroll and 10 base bet at 49.5%, the plan covers 6 straight losses; the 7th happens 0.84% of the time per streak and at least once in a 200-roll session 57% of the time. Each completed cycle wins 10; the wall loses everything. Average them and the answer is still 1% of turnover.
Do roll-over and roll-under have different odds?: No. Over 50.50 and under 49.50 both cover 49.5% of the 10,001 possible rolls and both pay 2×. Switching between them after a loss changes nothing either: each roll is an independent HMAC-SHA256 float and has no memory of the last one.

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Last reviewed 2026-09-27