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.
Best Answer
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 $0.50972954$