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 &lt;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?
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".
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:


first round: bet 1 on red (total loss: 1, win: 2)
second round: bet 2 on red (total loss: 3, win: 4)
3rd round: bet 4 on red (total loss: 7, win: 8)
....


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: $P(&quot;n-not-red&quot;) =(\frac{19}{37})^n$ , so with $n = 5$ round you have 96% of probability of win 1 (but also 4% probability of loose 31)