The "Bold" strategy of a single large bet is not the best Roulette strategy to double your money

Question

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.

Answer 1

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 $0.50972954$

Answer 2

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.

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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

Answer 3

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.

Answer 4

Just to lay out the principle: If you bet b dollars in any way, then there are 36 numbers where you get a fair payout, and 1 or 2 zeroes where the house takes your money away. The expected value is therefore -b/37 (one zero) or -b/19 (two zeroes).

The best strategy is one that minimises the total amount you bet. I think that means the optimal strategy for doubling your money is independent of the number of zeroes, just the probability of winning changes. So you can find an optimal strategy assuming no zeroes, minimising the expected total amount bet.

A slight modification is needed because the house only takes whole dollars. If you make a bet that could lead to you quitting with 3,005 dollars for example, because your expected amount of money is fixed, your chances of leaving wit 3,005 instead of 3,000 are lower by a factor 3,000 / 3,005.