# How to win at roulette?

I know that the answer to my attention-grabbing question is : "You can't win at roulette, it's a negative Expected Value game".

Yes, you're right, long term speaking.

Let's imagine a medium-short term situation (just an after-dinner at the casinò) where there will be <100 bets.

What's the best strategy to use to win some money?

I imagine a hypothetical formula that considers our "money target" and percentage of success as inversely proportional (the more we want to win the less likely we are going to succeed).

What do you think? Are those "red-black double if lose" systems useful?

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win=laser+phone+probability –  pedja Jan 14 '12 at 11:15
By not playing. –  nik Jan 14 '12 at 12:26
The book "How to gamble if you must" by Dubins and Savage might be of interest. It's not a practical book about gambling: the subtitle "Inequalities for stochastic processes" gives more of a clue to what it contains. But it sets out to solve the problem "what should you do if you need $10000 by midnight?". Naive Decision Making by Körner (another one with a slightly misleading title) is also worth a read, it is more elementary than How To Gamble. – mt_ Jan 14 '12 at 15:14 What is your objective? Do you want to maximize your probability of winning? Place 100 bets to be sociable and lose the least expected amount of money? Maximize the probably of attaining some given amount of money (for example, the sum you have to pay tomorrow to avoid having your legs broken by the mob)? All of these questions have different answers. – Peter Shor Jan 14 '12 at 20:20 @PeterShor Place 100 bets to be sociable and lose the least expected amount of money --> red/black (minimum variance isn't it?) – Tom Dwan Jan 15 '12 at 12:12 show 2 more comments ## 4 Answers If you want to maximise your probability of increasing your fortune to some target amount of money$\$a$ (greater than the amount you already have) then the correct strategy is the "bold strategy". If you have an amount of money $\$b$, then you should bet$\min\left(a-b,b\right)$(i.e. bet everything you have unless you are near to your target, in which case bet just what you need to reach the target). This is because roulette is a negative sum game, every time you bet you are on average going to lose money. So you want to achieve your target as quickly as you can. The Martingale strategy (double-if-you-lose) is bad because you make lots of bets, each one of which is negative expectation. This is proved to be optimal in this paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC223086/pdf/pnas00211-0067.pdf. Of course, while this strategy does maximise the chance of hitting the target, it also has a large probability of losing all your money. In real life, "the only winning move is not to play". - In fact this strategy is essentially a quick Martingale strategy, particularly if$b$is much bigger than$a$, where you double your bet every time you lose and walk away if you win. If$a$is bigger than$b$it becomes a "let it ride" strategy – Henry Jan 14 '12 at 15:59 add comment The negative expected value applies to short term as well. You simply do not have the odds. Read up on Martingale, it is the system you mention at the end. Even that is not going to help you. If you lose too many times in a row and do not have enough cash, you lose big time. So it is true, it is a negative EV game. Play for fun, not for sure profits. Edit: If I had to, I would play a game where expected win is around zero and does not require great capital. That is e.g. betting a fixed amount on the same color over and over (or switch colors, does not matter). You should be about even (zero disrupts it), but if you get lucky, you might win/lose something. - add comment If the expected number of plays is less than 100 use this strategy: play until you are in the positive, even if only$1. Then quit.

You might never get over $1, but if you do, you will quit as a winner. - add comment I think, best strategy is this: Choose an event with double choice, let's say red/black. Bet in this way: always on the same color, doubling bet at each round if you do not win. I mean: 1. first round: bet 1 on red (total loss: 1, win: 2) 2. second round: bet 2 on red (total loss: 3, win: 4) 3. 3rd round: bet 4 on red (total loss: 7, win: 8) 4. .... When you win, you start again from beting 1. At the n-round you can loose only if$n$not-red (black or zeros) are consecutively drawn:$P("n-not-red") =(\frac{19}{37})^n$, so with$n = 5\$ round you have 96% of probability of win 1 (but also 4% probability of loose 31)

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Classic Martingale system :D –  Tom Dwan Jan 14 '12 at 11:51