Roulette, being a negative expectation game, means you can expect to lose money. Even the so called "bold" strategy here will, on average lose less than making multiple, similar bets. But is a single bet on even odds (black/red or even/odd) the best approach to doubling your money. Clearly it isn't.

For the following example I use the single zero, American style Roulette as played in Barona and a few Vegas casinos but it applies with worse odds to games with two zeros.

Take, for example, a starting bankroll of $1,500.

A single, even odds, bet of 1,500 will win P=18/37 or .4865 of the time. Not bad, and way better than making multiple bets of 150 until you lose it all or double your pot. But this is slightly better:

Bet 750 on a 2 to 1 payoff (effectively 12 numbers). If you win go home. If you lose bet the remaining 750 evenly across 9 numbers. If you win, go home.

Let's look at the odds.

The odds of winning following the above strategy is: 12/37 + 25/37 * 9/37. The 12/37 is the probability of winning on the initial bet of 750 on a 2:1 payoff while the 25/37 * 9/37 is the probability of winning when the first bet fails.

So we have 18/37 v (12/37 + 25/37 * 9/37) or .4865 v .4887

The expected house take with this simple approach decreases over 15% and it's a bit more fun.

So the question is would making a series of bets on a single number (35:1 payoff) but taking care not to overrun the goal produce an even better improvement in the probability of doubling the bankroll? Since the only rational reason to gamble is the fun factor clearly it would be a lot more fun to make many smaller bets and if the probabilities of achieving doubling one's money improve, so much the better.

Added results from Monte Carlo simulations: I ran a simulation of the decreasing, high odds, bet strategy to double or nothing using Barona Roulette rules. 10 minimum, no fractional bets. The simulation placed the minimum dollar bets on odds ranging from 35:1 17:1, 11:1, 8:1, 5:1, 2:1, and 1:1, based on the current bankroll such that a win would produce exactly a doubling of the initial stake of 1,500. It was adjusted if the bankroll dropped below the required bets to win. In that case only bets on the 35:1 was made. If necessary, a remaining bankroll of 10 was required, all of it being potentially bet next.

The results (probability that a game sequence that either doubled or busted, doubled):

P(simple, even odds bet of 1,500): .486486
P(two step 3:1 bet followed, if lost, by a 4:1 bet): .488678
P(Barona Roulette Rules): .490027
P(Theoretical, fractional bets, no mins): .490270

Interestingly, the average number of roulette spins per completed game for the Barona version was 19. So that's a reasonable "fun factor" and the total expected loss in that double or nothing game is just under \$30 as opposed to just over \$40 for the simple, single bet of \$1500 that provides only one spin of entertainment.

Just ran another simulation but using only 35:1 bets. This requires that wins will usually exceed the goal. Nonetheless, the average win not only occurs with a higher likelihood (.4874) than the "Bold," single bet of \$1,500 on 1:1 but on average each win is leaves you with \$3,017.28. Very cool. More fun, less loss. It's the most practical way to gamble as it doesn't require complicated calculations to make sure each bet doesn't go over the \$3,000 goal.


Your strategy of betting on single numbers so as to double your initial stake would work for the first $24$ bets, and has a probability of $1-\left(1-\frac{1}{37}\right)^{24} \approx 0.481893977$ of succeeding in at least one of these.

The $25$th round hits the problem that you cannot reach double as you have less than $\$\frac{3000}{36}\approx \$83.33$ left. So you might say bet everything that remains on a single number, which you will probably lose, ending the game. This has an overall probability about $\left(1-\frac{1}{37}\right)^{25} \approx 0.50410316$.

But there is a probability of $\frac{1}{37}\left(1-\frac{1}{37}\right)^{24} \approx 0.014002865$ still to deal with if you are still playing after $25$ rounds. What happens to this depends on exactly how much you have left at this stage, and that depends on the casino betting rules.

  • If you must bet whole numbers of dollars, so rounding up your earlier bets, you will have $\$34$ left after losing the first $24$ rounds and $\$1224$ left after winning the $25$th round. Following the same strategy as before will lead to an overall probability of winning of about $0.48745373$ and of losing of about $0.51254627$

  • If you can bet amounts including arbitrary fractions of dollars, so not rounding up your earlier bets, you will have about $\$50.70501$ left after losing the first $24$ rounds and $\$1825.38036$ left after winning the $25$th round. Following the same strategy as before will lead to an overall probability of winning of about $0.49027046$ and of losing of about $0.50972954$

  • $\begingroup$ Nice! .49027.. sure beats .486486.... from a single, even odds bet. That's like a 28% reduction in the expected house take and way more fun because one makes a lot more bets on average. Of course no casino allows that so we are limited to round numbers. Additionally, the Barona casino has a minimum bet of 10 which is an additional constraint that eats into this. $\endgroup$ – doug Nov 1 '16 at 23:18
  • $\begingroup$ There cannot be any reduction in the house take. The house take should not be based on your initial bankroll but on the expected number of dollars staked during the round. Whatever strategy you follow, in the long run the house take is always 2.7\% of any dollar you bet! $\endgroup$ – user164118 Nov 3 '16 at 14:28
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    $\begingroup$ @user164118: so the aim is to minimise the (expected) amount you bet before stopping. $\endgroup$ – Henry Nov 3 '16 at 15:01
  • $\begingroup$ I don't understand your point. Let me explain my point with the following example. Suppose your initial bankroll is \$1500 and your goal is to raise your bankroll to \$3000. You bet each time \$100 on red (payoff odds are 1 to 1 and the win probability is 18/37). Using the gambler's ruin model, the expected number of bets until you are gone broke or have reached your goal is 213.48 and so the expected amount you have staked is \$21348. The probability of reaching your goal is 0.3077 and so the expected amount you have lost is 0.3077. 3000-1500=-10.2. The ratio of 10.2 and 21348 is 0.027. $\endgroup$ – user164118 Nov 3 '16 at 17:42
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    $\begingroup$ @user164118 - Of course you can't change the house take per dollar bet. It is fixed. But that was not the question and the answer did not purport to modify this. The question was whether a betting scheme could reduce the expected overall house take on a game sequence when the goal was double or nothing. The simple approach is to bet the stake, 1500, on an even odds bet. The answer provides an approach that requires on average, much less than placing 1,500 bets hence reducing the expected loss when the gaming sequence is double or nothing. $\endgroup$ – doug Nov 3 '16 at 19:05

Thanks to Henry's work using a fractional algorithm, I wanted to see how close to that optimum strategy a workable, defined, single number betting sequence would be. It turns out to be essentially identical!

Any single number strategy will produce some variation in the pot size of a completed game since any win will produce multiples of 35. However, one can still produce a series of bets that will produce an average win that is as close to doubling the initial stake as desired.

This was done by calculating the expected win size at each step and adjusting the bet size by +/- 1 as required to more closely approximate average winning game doubling. This yields the following bet sequence for an initial stake of \$3,500:

43 44 45 47 48 49 51 52 54 55 57 58 60 62 64 65 67 70 71 73 75 78 79 82   51  
34 35 36 37 38 39 41 42 43 44 45 47 48 49 51 51 52 55 55 58 59 60 63 64 66 68 70 71 74 76 78 81 82   24  
62 64 66 68 69 71 73 76 78 80 157

The first bet is 43 on any single number. If the selected number is hit your total pot size is 1,500 + 35*43 or 3,005. If that number was not hit, which leaves 1,457, then the next bet would be for 44 for a winning pot size of 2,997 (1457 + 44*35), and so on.

Unlike preceding bets, the last bet in each row is insufficient to constitute a win that concludes the game. Rather, it serves to increase the gambler's bankroll if won or terminate the game if lost.

This sequence has probabilities that are easy to calculate by summing the probabilities of winning for each step using these bet sizes as outlined in Henry's fractional betting answer and are for a starting stake of 1,500:

Probability of the sequence winning: .49027
Average winning pot total: 3,000.01:
Average casino's edge for each sequence: 29.19
Average number of spins per completed game: 19

For comparison purposes, this is the same statistic for the single "Bold" bet:

Probability of the sequence winning: .48685
Average winning pot total: 3,000.00:
Average casino's edge for each single bet: 40.54
Average number of spins per completed game: 1

These numbers were calculated, then verified by Monti Carlo simulation.

And, this is the statistic for an all too common betting pattern. Hoping for a hot streak a person bets 250 at a time on even odds until bust or double. A good way to lose much more money over time.

Probability of the sequence winning: .44608
Average winning pot total: 3,000.00:
Average casino's edge for each completed sequence: 161.74
Average number of spins per completed game: 16

I strongly doubt whether a series of bets on a single number gives a better value for the probability of doubling your bankroll. It is certainly not the case when each stake has the same size. This follows directly from the famous gambler's ruin formula. In your specific example with an initial bankroll of 1,500 dollars, the maximum probability of doubling your bankroll is even slightly higher than 0.4887. Using the optimality equation from dynamic programming, the maximum probability can be calculated as 0.4890.

  • $\begingroup$ No. I strongly believe that a sequence of bets on a single number produce more optimum results for an overall goal of doubling or busting. Secondly, I have no idea where the .4890 probability number you give comes from. The probability of the two step sequence is quite simply the sum of the two possible paths to a win which is 669/1369=.48867.... $\endgroup$ – doug Nov 1 '16 at 18:24
  • $\begingroup$ Note that a series of bets on a single number would result in gradually increasing bets as one's bankroll decreases after each bet. For instance the initial bet would be 42 followed by 44 if the first bet fails, etc., down to busting out or doubling the initial bankroll. There would have to be some adjustment since casinos don't allow fractional, no minimum, bets. A simple, but probably sub optimal, approach would be to use even odds bets when close enough to the goal as well as adjustments on the other side to prevent a residual bankroll dropping below the casino minimum bet size. $\endgroup$ – doug Nov 1 '16 at 18:38
  • $\begingroup$ @doug Your initial bet has to be at least $\$42.857143$ (or $\$43$ if using fixed numbers of dollars) $\endgroup$ – Henry Nov 1 '16 at 22:12
  • $\begingroup$ I suppose the goal could be loosened to allow a win to exceed the goal of doubling. That would simplify things and make it easier to play but I'm curious as to what an optimal strategy would be if one always wound up either doubling the 1500 or zeroing out. I suspect such a strategy would produce near .49 probability of success but it appears difficult to compute though a Monte Carlo approach could be used to test various approaches. $\endgroup$ – doug Nov 1 '16 at 23:23

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