BoosterHand Lab · simulation

I Tested a '99.999% Win' Roulette Strategy on 10,000 Players

Updated 2026-11-04 · By BoosterHand Lab · Method: Monte Carlo simulation (Python/NumPy, fixed seed)

$99.999% win roulette strategy tested: two dozens plus triple after a loss, simulated on 10,000 players$

Key results

The plan bets $1 on each of two dozens: 24 of 37 numbers, a 64.9% win chance per spin. After a loss the stake triples, so any win repays everything plus $1.
The '99.999%' is the chance of not losing 11 spins in a row. Surviving 11 losses needs $177,146. With $1,000 and a $500 table limit, 6 losses in a row end the plan: 99.81% per win.
10,000 players, one session each, stop at +$10: 98.15% hit the target. The 185 who didn't lost $719 to $728 each.
All together they lost $35,667 of $1,362,006 wagered (−2.62%). The math says −2.70%: 1/37 of every dollar.
Played every day for a year, the plan promised +$3,650. The median player ended at −$1,476 and 81% finished the year down.

There’s a roulette plan that keeps coming back with the same promise: cover most of the table, raise your bet after a loss, take a small profit and leave. “It wins 99.999% of the time.” I worked out the exact math on a single-zero wheel, then gave the plan to 10,000 simulated players with $1,000 each.

98% of them won. Together, they lost $35,667.

How the “99.999%” plan works

The wheel has 37 pockets: 1 to 36 plus a single zero. The plan bets $1 on the 1st dozen and $1 on the 2nd dozen, covering 24 numbers.

Win (1–24): 24 of 37 = 64.9% of spins. One dozen pays 2:1 (+$2), the other loses (−$1): net +$1.
Loss (0 or 25–36): 13 of 37 = 35.1%. Both bets lose: −$2.

After every loss the stake triples: $2, $6, $18, $54, $162, $486. Because a win returns half the stake as profit, the next win always repays every loss in the streak plus $1. Lose $2, $6 and $18, then win on $54: +$27 − $26 = +$1. The player repeats this until the session is up +$10.

Two dozens roulette strategy on a European single-zero table: 24 of 37 numbers covered, 64.9 percent win chance per spin, win +1 dollar, loss −2 dollars — Two dozens cover 24 of 37 numbers: +$1 on a win, −$2 on a loss.

Where “99.999%” comes from

The plan only fails if the losses keep coming. The chance of losing k spins in a row is (13/37)^k:

Losses in a row	Chance	Stake on that spin	Total lost
1	35.1%	$2	$2
3	4.34%	$18	$26
6	0.188% (1 in 532)	$486	$728
7	0.066%	$1,458	$2,186
11	0.001%	$118,098	$177,146

Eleven losses in a row happen about once in 100,000 tries. That’s the 99.999%. It is true, but only for someone who can keep tripling to a $118,098 bet.

Roulette 99.999 percent win claim: chance of losing 1 to 11 spins in a row on a two dozens bet, 0.001 percent for 11 losses — Long losing streaks are rare, and that's where the 99.999% claim comes from.

The wall: bankroll and table limit

With $1,000 the ladder breaks much earlier. After 6 losses the player has lost $728 and has $272 left, but bet #7 is $1,458 ($729 per dozen). That’s more than the bankroll and more than a $500 table limit. So 6 losses in a row end the plan:

per win cycle: 1 − (13/37)^6 = 99.81%, not 99.999%
per session (10 wins for +$10): 0.9981^10 = 98.1%

Roulette martingale-style triple progression from 2 to 118,098 dollars: with a 1,000 dollar bankroll and 500 dollar table limit, bet 7 of 1,458 dollars is impossible — Bet #7 is out of reach, so a 6-loss streak ends the session.

Want to try your own bankroll, base bet and limit? Our Martingale calculator shows how many losses a progression survives.

Before the results: 10,000 players, one session each, 98% win rate. Do they finish up or down overall?

How we ran the simulation

Wheel: single-zero (European), 37 pockets, every spin drawn at random
Bet: $1 on the 1st dozen + $1 on the 2nd dozen, stake tripled after each loss, reset after a win
Bankroll and limit: $1,000 per player, $500 maximum bet
Session rule: stop at +$10, or when the next stake is larger than the bankroll or the limit
Test 1: 10,000 players, one session each
Test 2: the same 10,000 players, one fresh $1,000 session every day for 365 days
Method: exact probability math plus Monte Carlo simulation in Python (NumPy), fixed random seed

10,000 players, one session each

The players spun 152,609 times and won 64.9% of those spins, exactly as the math predicts.

Result	Players	Each
Hit +$10	9,815 (98.15%)	+$10
6 losses in a row	185 (1.85%)	−$719 to −$728

Roulette strategy simulation with 10,000 players: 9,815 hit the 10 dollar target (98.15 percent), 185 busted and lost 719 to 728 dollars each — Each dot is one player: green hit +$10, red hit 6 losses in a row.

Now add it all up:

	Amount
9,815 small wins	+$98,150
185 busts	−$133,817
Net, all players	−$35,667
Median player	+$10
Average player	−$3.57

One bust erases about 72 winning sessions. To break even the plan would need about 98.6% winning sessions; it gets 98.1%. That small gap is the house edge.

99 percent win roulette strategy: 98,150 dollars in small wins vs 133,817 dollars in busts, net loss 35,667 dollars, median player +10, average −3.57 — The typical player wins; the average player loses.

The players wagered $1,362,006 in total and lost 2.62% of it. The expected loss on any bet on this wheel is 1/37 = 2.70% per $1. Tripling doesn’t change that number. It only changes how much is bet.

”+$10 a day” for a year

$10 a day sounds like $3,650 a year. So every one of the 10,000 players ran the plan once a day for 365 days, with a fresh $1,000 each day.

After a year	Result
Promised	+$3,650
Busts per player (average)	6.8
Players who never busted	12 of 10,000
Median result	−$1,476
Average result	−$1,343
Finished the year down	81%

Roulette strategy 10 dollars a day for one year: 40 sample players and the average, 6.8 busts per year, median loss 1,476 dollars vs promised 3,650 — A slow staircase up, then a cliff: about seven times a year.

What a real 99.999% would cost

You can push the per-win success rate higher, but only with a much bigger bankroll and no table limit:

Plan	Steps survived	Per win cycle	Per +$10 session	At risk	Per $1 bet
$1,000, $500 limit	6	99.81%	98.13%	$728	−2.7%
$2,186, no limit	7	99.93%	99.34%	$2,186	−2.7%
$177,146, no limit	11	99.999%	99.99%	$177,146	−2.7%

The same plan with a +$100 target ($1,000, $500 limit) hits its goal only 82.8% of the time (83.35% in a separate 10,000-player run). And even the $177,146 version still loses 2.7% of every dollar on average: it just risks $177,146 to win $10.

99.999 percent roulette strategy: bankroll needed 177,146 dollars for 11 steps, every version keeps a 2.7 percent house edge — Every version has the same expected loss per dollar.

Limits of this model

This is one generic single-zero wheel with a $500 maximum bet and our exact stop rules. Real tables set their own limits, and some players would keep betting after a broken streak instead of stopping. None of that changes the 1/37 edge on each spin. On a double-zero wheel the edge is higher; see our European vs American roulette test.

What this means for you

Covering more numbers buys win rate, not edge. 24 of 37 numbers won 64.9% of spins and still lost 2.7% per $1.
Progressions trade many small wins for one huge loss. 98% of sessions won $10; the other 2% lost about $723 each.
Any plan on this wheel loses 2.7% on average. Our 10,000 players lost 2.62% of $1,362,006, right where the math says.

Roulette is a game of chance, and no betting pattern turns the edge around. Set a budget you can afford to lose and stop when it’s gone. If gambling stops being fun, our responsible gambling page has tools and support.

Did you guess up or down? For the classic double-after-a-loss version, see our Martingale strategy test on 10,000 players.

FAQ

Is there a roulette strategy that wins 99% of the time?

Yes, you can win most sessions: covering 24 numbers and tripling after a loss hit a +$10 target in 98.15% of our 10,000 simulated sessions. But the rare failure costs about $723, so the plan still loses on average. Winning often is not the same as winning money.

Where does the 99.999% number come from?

On a single-zero wheel, a two-dozens bet loses with probability 13/37. Losing 11 spins in a row has a chance of (13/37)^11, about 0.001%, so '99.999%' of win cycles end before that. The catch is the 11th bet: with a $2 start it is $118,098, and the whole ladder costs $177,146.

What is the two-dozens roulette strategy?

You bet the same amount on two of the three dozens, covering 24 numbers out of 37. A win (64.9% of spins) pays one dozen at 2:1 and loses the other, for a net profit of half the stake. A zero or a number from the third dozen loses both bets.

Does tripling the bet after a loss beat the house edge?

No. Every bet on a single-zero wheel loses 1/37 (2.70%) of the stake on average, and a progression is just a series of such bets. Our 10,000 players wagered $1,362,006 and lost 2.62%, in line with the expected 2.70%.

How much bankroll do you need for a 99.9% roulette system?

With $1 on each dozen, 99.9% per win cycle needs 7 steps of tripling: $2,186 and a table that accepts $729 per dozen. 99.999% needs 11 steps and $177,146. Even then the expected loss stays 2.7% of everything you bet.

Simulations use generic models, not any real game or casino. Results show long-run math, not predictions for one session. 18+ only. If gambling stops being fun, take a break: responsible gambling.

Sources

Martingale (betting system) — Wikipedia
Roulette — Wikipedia
Gambler's ruin — Wikipedia
Expected value — Wikipedia

Next experiment: How Slot Machines Work: The Hidden Reel Strip (5 Billion Spins Tested)