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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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    $\begingroup$ win=laser+phone+probability $\endgroup$
    – Pedja
    Jan 14, 2012 at 11:15
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    $\begingroup$ By not playing. $\endgroup$
    – Najib Idrissi
    Jan 14, 2012 at 12:26
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    $\begingroup$ 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. $\endgroup$
    – Matthew Towers
    Jan 14, 2012 at 15:14
  • $\begingroup$ 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. $\endgroup$
    – Peter Shor
    Jan 14, 2012 at 20:20
  • $\begingroup$ @PeterShor Place 100 bets to be sociable and lose the least expected amount of money --> red/black (minimum variance isn't it?) $\endgroup$
    – KingBOB
    Jan 15, 2012 at 12:12

4 Answers 4

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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 expectation 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.

The bold strategy 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".

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    $\begingroup$ 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 $\endgroup$
    – Henry
    Jan 14, 2012 at 15:59
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    $\begingroup$ Of course $\min(a-b,b)$ applies to simple chances (red/black etc.). Does the paper also recommend this over, say, betting $\frac1{35}\min(a-b,b)$ en plein? $\endgroup$
    – Hagen von Eitzen
    Apr 13, 2014 at 14:49
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    $\begingroup$ I think that betting $\frac{1}{35}\min(a-b,b)$ straight is actually even better. It's just like betting red/black $log_2(35)$ times, except that you only have one chance to be hit by the $0$ rather than $log_2(35)$ chances. $\endgroup$
    – Oscar Cunningham
    Apr 13, 2014 at 15:27
  • $\begingroup$ @Oscar Cunningham: Is that Timmy? I adore Timmy. $\endgroup$
    – Vandermonde
    Mar 16, 2015 at 22:50
  • $\begingroup$ Never mind; your gravatar could well be of a different stuffy than I had thought. $\endgroup$
    – Vandermonde
    Mar 18, 2015 at 3:40
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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.

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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.

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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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    $\begingroup$ Classic Martingale system :D $\endgroup$
    – KingBOB
    Jan 14, 2012 at 11:51
  • $\begingroup$ This actually boosts up the house edge. $\endgroup$
    – SarpSTA
    Jan 5, 2016 at 14:15

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