# make a green profit in roulette

## Story

You meet a rich drunk CEO at a roulette table in Vegas. She explains to you that she has a way of almost guaranteeing a profit at this game by betting on only black. In the process she gets complimentary drinks from the waiters for just their tips. Her secret (though she constantly brags about it) is that she has 1 million dollars set aside to back up her bets. She complains, however, that the green numbers (0 and 00) on the board mess up the game and are obviously there for the casino's cut saying something about Europe being much better about this.

## Relevant roulette rules

When she bets on black numbers (18 of them are black), and the ball lands on a black number, she is handed back twice her bet value for a net profit equal to her bet. The same would be true for red numbers. If she were to bet on "green", however, and the ball lands on a green number, she is handed back 18 times her bet. She can also land on an individual number (0 and 00 being the only green numbers) and win 36 times her bet if she wins.

At the table she is at, the minimum bet is \$100 dollars and there is not a maximum bet. The deal will accept any whole dollar amount within that range and he is willing to be there for at least 500 more bets. He doesn't expect tips from you considering how much you have at stake. ## The bet The drunk CEO decides that because green numbers are so terrible there is no way to win with them. She, therefore, makes the following bet with you: • She gives you her million dollar stash and you are not allowed to provide any more money. • You are only allowed to bet on green or on one of the two individual green numbers. • So you can't just use random walk, you give a number$N$before starting which is the number of games you will play. She is willing to stay all night (500 games) if she has to. If your strategy requires more games she might be willing to have Jessica run a simulation of it and do pay outs accordingly. You must play every game. • If after game$N$, you have more than \$1 million then you win. If not, you lose.
• If you win, you get to keep all of the money.
• If you lose, she takes what you have left, everything you own, and you become her indentured servant for the rest of your life. (Apparently she thinks that is worth $1 million.) So while Jessica (who apparently the CEO met at a poker table last year) is being forced to write up the contract, figure out your strategy. What number$N$will you pick? How can you maximize the probability you will not be a slave assuming you take the bet? How can you maximize your expected profits (assuming you agree that your life is worth \$1 million)?

• are all the numbers 36? – dmg Jan 19 '15 at 15:05
• Roulette usually goes from 1-36, so I guess we can assume that. And I don't really see how this differs from a normal roulette in this case (negative EV). So I personally wouldn't take that bet, I suppose. Unless I'm feeling lucky or like being a servant. – No. 7892142 Jan 19 '15 at 15:07
• @dmg you get 36 times you bet back if you correctly pick the number... yes. This is the same rules as normal (american) roulette but the difference here is the need to make a profit compared to trying to maximize your profit. – kaine Jan 19 '15 at 15:08
• So there are 38 possibilities? – dmg Jan 19 '15 at 15:10
• @No.7892142 Actually it's roughly equal to the lottery if you play red/black because to turn the minimum bet into a lottery size win by pressing your bet (you would have to hit consecutively some number of times in a row) the house advantage increases to about 50%. – Michael Jan 21 '15 at 4:05

## Preamble

Since you are forced to come up with a number of turns you will play, you can't simply walk out after winning the first bet. In order words, your wins will need to be big enough to cover your subsequent losses.

If you bet on green (higher percentage than 0 or 00), then you have a 1/19 chance of winning a bet, and a 18/19 chance of losing.

The longer you play, the better your odds of having at least one win is. Here is a table that shows this.

Turns  Odds of Losing every game   Odds of Winning at least once
1       18/19      = 94.7%         5.2%
10     (18/19)^10  = 58.2%         42.8%
20     (18/19)^20  = 33.9%         66.1%
30     (18/19)^30  = 19.8%         80.2%
40     (18/19)^40  = 11.5%         89.5%
50     (18/19)^50  =  6.7%         93.3%
60     (18/19)^60  =  3.9%         96.1%
70     (18/19)^70  =  2.3%         97.7%
80     (18/19)^80  =  1.3%         98.7%
90     (18/19)^90  =  0.8%         99.2%
100    (17/18)^100 =  0.4%         99.6%


As you can see, the more games you play, the better your chances of winning at least once.

## General Strategy

So, lets employ the following strategy:

• Pick as many turns as we can.
• Make a medium sized bet initially.
• Every time we lose, increase the bet slightly to cover the losses we just incurred.
• If we win, bet the remaining bets at the minimum \$100 The value of the bet we choose initially needs to cover any losses we'd incur for the duration of number of turns we choose, plus the value of the original bet. Lets say that we choose 10 turns. If we win on the first bet, then we need to cover 9 losses at \$100 each, plus the amount of our initial bet. Since we win $18$ times what we bet, we end up with this equation:

$$18\times bet \ge bet + 100(N-1)$$ $$bet \ge \frac{100(N-1)}{17}$$

If we plug in $N=10$, then we get an initial bet of \$53. But our chances of winning once in 10 turns are only$42.8\%$, and we can do better than that. If we try for$N=500$, then our chances of winning are 99.99999%! But we have an initial bet of \$2,936. The problem with this is that since the bet is increasing after every loss, we'd run out of money after only 53 consecutive turns! This lowers our chances to only 94.3%. Pretty good, but not optimal. In fact, since we chose our initial bet based on 500 turns and we are only getting 53 turns, we can lower our initial bet substantially.

## Solution

After running a simulation with these numbers, it turns out the sweet spot is to have $86$ turns, and an initial bet of exactly \$500. If you win the first bet, you pay \$500 and get \$9,000, for a net profit of \$8,500. This is exactly enough to cover $85$ turns at the minimum bet.

If you lose all your bets, you will still have enough money to make the $86^{th}$ bet with the chance to win it all back.

This gives you a $0.956\%$ chance of losing the game, but a $99.044\%$ chance of winning!

One thing to note is that this strategy assumes that once you win, you don't need to win any more and the amount you made needs to cover the remaining bets at the minimum, plus any previous bets.

The thing is that once you win, and start betting the minimum, you may win some of those bets too. The fact that you can expect to win some of them likely has a big impact on your actual chances of winning.

For example, if we chose $N=500$, but chose \$250 as our initial bet, then if we win on the first bet, we would get \$4,500. The remaining 499 bets would expect to win 1 in every 19. So, we can expect to get approximately \$47,273 in winnings by betting \$49,900 in 499 minimum bets of \$100. The question is, how much does this affect the returns? How many more turns can you add with this knowledge? If$N=86$turns gives you a 99% chance of winning, can you do 100 turns with less of an initial bet with the hopes that you'd make back your shortfall in by winning a few times on the minimum bet too? At what point does the probability of not getting back this money outweigh your overall probability of winning? • I think this is generally correct but not optimized. Thank you for mentioning that multiple wins might work. – kaine Jan 20 '15 at 16:17 • @kaine It is optimized given the parameters. The accepted answer has a lower probability of winning because it starts with the premise of 500 turns, but loses all the money after 53 turns if there is bad luck. This starts with the premise of 86 turns and doesn't run out of money until then. – Trenin Jan 20 '15 at 17:29 • @kaine In the "addendum" I mention that the strategy assumes that you will lose all your minimum value bets, but you will in fact win some of them, which means you could probably extend the main strategy for a couple more turns with a lower initial bet. – Trenin Jan 20 '15 at 17:30 • @kaine This answer doesn't try to go for more than one win. I can see things getting complicated there, but not sure if it would help. – Trenin Jan 20 '15 at 17:34 • This is also know as the Martingale betting system – Kruga Feb 15 '18 at 8:59 ## Preamble Most of the rationale for this answer has already been discussed by Trenin et al. However, there are some errors. In particular, from the original question (my emphasis), "If after game N, you have more than$1 million then you win. If not, you lose."

The opening bet of \$500 is thus not sufficient, as if this were to be followed by 85 losses, we would be back to precisely \$1 million, which counts as a loss. In terms of the puzzle design this has to count as a loss, as otherwise, a winning strategy would be N = 0, just give the million dollars straight back.

In addition Trenin's answer stated N = 86, but when I followed the same methods I was able to do N = 87, which I've now reproduced by a different method below.

A critical point is that all losing conditions are equivalent - whether the CEO is down by a single dollar or you've gambled away the whole million, so once the value of N is chosen, after N-1 losing bets the Nth bet should be for the entire remaining balance.

However, a dollar that is left until the last bet has almost no effect on the expected value - the earlier we bet it, the more the potential winnings increase our expected profit given success. (consider the possibility of adding an extra dollar to the first bet, or to the second bet... if both lose, you end up in exactly the same position for subsequent bets, but the first one should win 1/19 times, but the second wins only 18/(19*19) times, so the expected value from a dollar bet as early as possible is important).

Thus, when in a losing position each bet should be for the largest amount possible that still leaves enough funds to make all remaining bets from the million dollar float... and each of THOSE bets must be enough to win back all earlier losses and also ensure a float to make \$100 bets for the remainder of the games. Also, as already discussed in other answers, we're better off with fewer 18:1 bets rather than more 36:1 bets (I ran this both ways, and agreed), and we also need to maximise the value of N, as each possibility to avoid indentured servitude greatly outweighs the expected gain from betting larger sums. ## Construction of optimal solution Consider the very final bet. We will bet all remaining funds in an attempt to win back all the earlier losses, with anything in excess of \$1 million being our profit, which is incidental. The next multiple of 18 above 1 million is 1000008, so we should bet 1/18 of that, which is \$55,556. This means that for the most efficient solution, the total of all preceding bets (the amount by which we are down before placing this final bet) will be \$944,444.

Similarly for the penultimate bet, we know that if we after placing this bet we'll be down \$944,444 to put us in the correct position for the final bet. The winning amount also needs to cover (at minimum) a \$100 stake for the final bet, and \$1 profit, so we need winnings of at least \$944,545 and therefore a bet of \$52,475. Thus just before the penultimate bet we should be down by \$891,969, which defines the winnings for the 2nd before last bet as at least \$892,170, etc... Continuing this iteratively, we find that this implies that the 86th before last bet should be for \$507 at a position where we're down by \$1. In this case our winnings would be \$9,109 which covers the \$507 stake, 86 further bets of \$100, and \$2 winnings. However, this cannot be iterated further so instead we make the first bet for \$508 from the starting position.

## Solution

This means we can last for N = 87 games, and the chance of losing that many times in a row is just 0.906%, so the chance of winning is 99.094%

The increasing series of stakes to use is \$508 \$530 \$556 \$583 \$611 \$641 \$673 \$707 \$742 \$780 \$820 \$862 \$907 \$955 \$1,005 \$1,058 \$1,115 \$1,174 \$1,237 \$1,304 \$1,375 \$1,450 \$1,530 \$1,614 \$1,703 \$1,797 \$1,897 \$2,003 \$2,115 \$2,233 \$2,359 \$2,491 \$2,632 \$2,781 \$2,939 \$3,106 \$3,283 \$3,470 \$3,668 \$3,878 \$4,100 \$4,335 \$4,585 \$4,848 \$5,128 \$5,423 \$5,736 \$6,068 \$6,419 \$6,791 \$7,184 \$7,601 \$8,042 \$8,509 \$9,004 \$9,528 \$10,083 \$10,670 \$11,291 \$11,950 \$12,647 \$13,385 \$14,166 \$14,994 \$15,870 \$16,798 \$17,780 \$18,820 \$19,921 \$21,087 \$22,321 \$23,629 \$25,013 \$26,478 \$28,030 \$29,673 \$31,412 \$33,254 \$35,204 \$37,269 \$39,456 \$41,771 \$44,222 \$46,817 \$49,565 \$52,475 \$55,556 Note that these add up to precisely \$1,000,000

After the first win from any of these, we switch to making \$100 bets, and have ensured sufficient funds to do so without dipping back into the original million, thereby guaranteeing we finish in profit even if there are no other wins (which will be the case with non-negligible probability). If my other calculations are correct (2 different methods disagreed by a few cents, and I couldn't trace any more errors), the weighted expected profit is \$-2,590.63 if we value indentured servitude at \$-1,000,000 as in the question. Considering only the winning cases I get a weighted average profit given non-loss of \$6469.71

I believe this is now fully optimal.

## Towards proof of optimality (or an even better solution?)

The above construction demonstrates that this is the optimal solution under the presupposition that an optimal solution is one that requires only 1 win in N rolls of the roulette table.

However, for a complete proof of optimaility, this presupposition would also need to be proven.

If it is not in fact optimal, then the optimal solution must require more than 1 win.

For strategies requiring only a single win, we fail with probability (18/19)^N - i.e. only in the case where there was not even a single win.

For strategies requiring at least 2 wins, we also fail in at least the cases where there was precisely 1 win, so the probability of failure is (18/19)^N + (18/19)^(N-1)*N/19. In order for this to beat the single-win strategy, it would need to work for N >= 126. Far higher values of N would be required for strategies always requiring 3 or more wins.

The iterative construction for the single-win strategy can be constructed beyond 87 iterations (... and we can stick with the calculated \$507 bet rather than \$508 on that iteration), and gives the "target" that a multi-win strategy must reach. For example, when there are 87 rounds remaining with only a single win required, the pot must have at least $999,999 in it. If a strategy requiring 2 wins exists, it needs to ensure that after the first win, we have sufficient funds to make the corresponding bet sequence from the single-win strategy. This can be extended beyond 87, for example, with 125 rounds remaining (i.e. after the first bet of a hypothetical 2 win strategy with N=126), we would need to be "down" by -\$9,921 (i.e. have at least \$1,009,921 in the pot). i.e. in a multi-win strategy, the first bets will necessarily be LARGER than the initial bets in the single-win strategy (this is compensated by the later bets being much smaller). In order to check if a strategy requiring precisely 2 wins is feasible, we can construct one in a similar way as the 1-win strategy, i.e. If we've only got 2 rounds left with no wins so far, we would need to win both of them. There's no point saving any money for a final bet if we lose, as we're going to be enslaved whatever happens, so we bet all remaining funds hoping to get the \$55,556 required for the final bet. That requires a bet of \$3,087. With only 3 rounds left with no wins so far, we need to win 2 of the next 3. If we lose this one we need to leave \$3,087 for the last ditch attempt as above, but if we win we need an extra \$104,944 for the last 2 bets of the single-win strategy, so a bet of \$5,831 is required, meaning the pot must have at least \$8,918 in it before making this bet. This can similarly be iteratively extended, and we find that this 2-win strategy works up to N = 120, at which point the probability of failure is 1.014%, higher than the optimal single-win strategy, implying a probability of success of 98.986%. Beyond 120 rounds remaining, an initial pot of more than \$1 million is required. In particular this strategy cannot work for N = 126 as required.

Similarly, a strategy requiring 3 wins can be constructed, which only works up to N = 149 with an initial pot of \$1 million, etc... Intuitively I did not expect any strategy to outperform the optimal "single win" strategy, because every bet made reduces our expected return by 1/19 of the amount bet (which is how the casino makes its money!), and multi-win strategies would almost certainly be making more bets and/or initial bets for larger amounts. Whilst the above does not definitively rule out every other possible alternative strategy, it hopefully makes a good start towards a more rigourous proof. ## To reconstruct my "experimentation" spreadsheet In excel, enter 1000001 in cell B1. This represents the amount needed to guarantee success with 0 wins and 0 games remaining. Cells A1 and E1 will contain zero (or can be left blank). Enter the following on the second row: A2: =A1+1 B2: =B1+100 C2: =B1-E1 D2: =IF(C2=0,0,MAX(CEILING(C2/18,1),100)) E2: =D2+E1  Replicate these downwards as far as you are interested in investigating. Column A contains the number of bets that must be made, column B the amount that must be in the pot to guarantee success with 0 wins, column C the minumum amount that must be won in the current bet, column D the bet that must be made to achieve that (0 means that if we reach this point we're already doomed to failure, otherwise the minimum bet of 100 is enforced), and column E the amount that must be in the pot to guarantee success with 1 win. I also added in D1: =A1+1 as a marker for how many bets need to be won. Columns C to E can then be replicated to the right as required to investigate the condition for 2 or more wins required to guarantee success. e.g. column G gives the bet that must be made if 2 wins are required, and column H the amount of money needed to guarantee success with 2 wins. Both show zero when the current position is a guaranteed loss. The first few cells in column F (and I, L, etc.) can be deleted to avoid the highly improbable end conditions - rather than saving some money right until the end so we can still win if we get two greens in a row, we can delete a few rows so that the strategy only works if the first win is before a given point. This can allow us to start from a slightly higher value of N with the same money, but gives more opportunities for the system to fail. For any value of N, requiring at least K wins where the Jth win is within M[J] games (N, K and all M[J] being integers that define the strategy), this can easily be evaluated to find out the minimum amount of money required to guarantee being able to perform the strategy with N games remaining... One combination that this can confirm works for N = 500 is "at least 16 wins, of which the first is within the first 100 games, the second is within 150 games, the third within 175, ... the 15th within 475 games.". This has some scope for minor optimisations, as the initial balance only needs to be 999756 with that precise combination. However, even without the side-conditions that are necessary to make it affordable from a starting pot of \$1,000,000, and which increase the probability of failure, "at least 16 wins out of 500" seems to fail more than 1% of the time. As such I did not attempt the more complex probability calculation for the overall strategy's chance of failure.

Following these and various other similar experiments with specific cases, I am convinced (albeit without a full proof, so it is possible that I am mistaken) that no strategy can outperform the "1 win in 87 games" strategy.

• This does look like a slight improvement on the accepted answer, and the numbers do indeed add up to 1000000. However, there's still no guarantee that this is the most optimal solution - Trenin's addendum still applies. There may be a way to get a better chance of winning by not assuming that all bets after winning once will be $100 bets. – Rob Watts Feb 14 '18 at 20:55 • @RobWatts : any bet larger than \$100 once we're "in profit" will reduce our expected winnings - just from the odds of roulette, each bet will expect to return just 18/19 of the amount on average, and smaller bets than \$100 are not allowed. We need to make the smallest bets possible in order to maximise the expected return. There seemed to be some suggestion that there may be a strategy that requires more than 1 win could work. I've not worked out a formal proof of impossibility, but based on the iterative construction in my answer, it seems highly implausible. – Steve Feb 15 '18 at 9:27 • @Steve But we can expect to win some of those \$100 bets. So, that means we have "over-engineered" the test. We could bet a slightly smaller amount initially and make $N$ slightly larger as a result because we can expect that we will win a few games late to make up for our shortfalls. That is why I think (I could be wrong) that there is a better strategy that allows you a higher winning percentage by being able to count on some more late wins. But not necessarily. – Trenin Feb 15 '18 at 17:10
• @Steve To be clear, neither Rob nor I are suggesting you bet more than \$100 after your first win. But you can expect that some of those \$100 bets will win, which may let you make $N$ larger as a result. I expect the impact to be minimal at best. – Trenin Feb 15 '18 at 17:12
• @Trenin the point is that, if the first win is not sufficient to GUARANTEE success without any further wins, then we're absolutely reliant on (at least) another win, and effectively in our starting position. The analysis for the base case applies - with no further wins we might as well bet everything on the final game, and it's more efficient to bet any "spare" earlier. The full table I constructed includes many interesting side-results. e.g. with 100 games remaining, \$1,010,001 guarantees success with 0 wins, \$1,004,922 with 1 win (next bet \$293), \$983,800 with 2 wins (bet \$1,226), etc. – Steve Feb 16 '18 at 10:24 Prelude: This is not the final, optimised solution, this is just the idea of it. It lacks mathematical conciseness and can be improved in various ways - for example, on your nth attempt, you only need to bet the amount that turns you profit into 50.000-100*n. This is not accounted for. Hence, this solution doesn't maximise the profit. It does however maximise the chance of winning the game, I believe. The idea is to just bet the amount that almost ensures you can't lose the game, even if you lose all the remaining games at the minimal stake. So our general goal is to achieve a profit of 50.000 US Dollar, which ensures we can bet 100 US Dollar 500 times, lose everytime and still end up in profit. We will bet on green, because it mostly doesn't matter and it keeps the stakes smaller. For this purpose, we calculate the amount we need to bet to gain 50.000 US Dollar (pure profit): 50.000/17 = 2.942 US Dollar. If we lose, we need to win 52.942 US Dollar (actually less, see the prelude), so we bet 52.942/17 = 3.115 US Dollar. We keep doing this until we finally get a winner - for the remaining bets until we arrive at 500, we bet the minimum amount, 100 US Dollar - it is then irrelevant whether we win or lose. The bad outcomes for us in this situation are running out of money or losing 500 times in a row. Running out of money on 500 tries means the average stake would have to be 2.000 US Dollar - considering our starting stake of about 3.000 US Dollar, this will happen eventually. For the purpose of this solution, I assume this happens after 100 tries, but could be totally off. (In fact, I am, see comments below.) The chance of us losing 100 times are as follows: (36/38)^100 = 0.45%. So we're fairly certain to win. Considering all this, this turns into a maximisation problem: We want to have to win as little as possible to be secure (so we want to bet rarely), but we want to have as many tries to win as possible (so we want to bet often). There is an ideal N out there for this problem, I'm not the one to produce it. And with our million we free Jessica from her shackles - if she looks good - and live happily ever after. The End. • I don't get it. How can there be any bet which ensures I don't lose the game? – A E Jan 19 '15 at 17:58 • In this case, does 50.000 mean fifty thousand or fifty? I am American, and this means just fifty dollars to me - I'm just wondering whether or not this is the case with you. – mdc32 Jan 19 '15 at 18:04 • You did! :) "The idea is to just bet the amount that ensures you can't lose the game", you said. – A E Jan 19 '15 at 18:15 • @AE That's like solving a linear equation with a zero vector, of course. Also added an almost - again I've never been one for concise expressions. – No. 7892142 Jan 19 '15 at 18:16 • I see a problem. When you talk about the bad outcomes, you "assume this happens after 100 tries" when talking about running out of money. The only problem is it runs out far before 50 tries. Even at 50 tries, this would bring your 0.45% chance up to a 6.7% chance. – mdc32 Jan 19 '15 at 18:22 This is a simple answer intended only to show the concept clearly. If I set$N=105$and bet \$1000 each time, I expect to lose \$5250. There is, however, a greater than$50\%$chance I will make a profit. This is because if I win I am most likely to win \$3000 ($22\%$ of the time total) while if I lose I lose either \$105000, \$69000, or \$33000. While I on average lose money, I win more often than I lose. If I do this during the game, I will have a$52.4\%$chance of getting over \$1,000,000 and keeping that money. I will then have a $47.6\%$ of becoming an indentured servant which was valued (probably too high) at \$1,000,000. This means my expected actual winnings is roughly \$50,000.

There are much better solutions than this but this should help counter concerns that taking the bet is a loss with regard to profit.