# Roulette tournament?

You have been invited to play in a French Roulette tournament. The tournament-specific rules are as follows:

2. Play lasts for 5 rounds
3. Minimum bets based on color, low/high (1-18 and 19-36), even/odd is set at 2 tokens. Any other type of bet requires a minimum of 1 token.
4. All other rules from standard french roulette apply, but assume a single-zero layout (hence 'french roulette').

What's the best strategy to finish with more tokens than others?

• But welcome anyways, to puzzling SE. Please edit your post, or provide better quality posts in the future. May 1, 2016 at 9:57
• @ghosts_in_the_code :I think one can assume the rules to a standard casino roulette game to be known, one doesn't exlain the rules to chess in a chess riddle either. The key difference between 'french' and 'american' roulette has now been edited in. May 1, 2016 at 11:36
• Do you have to make a bet in every round?
– TTT
May 1, 2016 at 17:07
• @TTT that is indeed implied. My rewording of the question may actually have that made unclear. May 1, 2016 at 17:14

The question seems to imply this is a winner-takes-all tournament, thus having the second biggest stack at the end of the five rounds is worth the same as having the smallest stack in the field. As such how much risk you take on any given round should be a function of the size of the biggest stack in the field, the size of your stack and the number of rounds remaining (at the extreme you would need to bet every chip in your stack on a long shot in the hopes of regaining a competitive stack). I am not sure it is possible to provide a complete optimal strategy in a SE answer.

Well, the game has been around casino's long enough to avoid mathematical long-term methods that win. Theoretical guaranteed win scenario's require pretty much unlimited funds, and it's highly unlikely you'll get away with it a a casino. Given the limit to 100 tokens and 5 rounds, I'll avoid that in my answer, and try and find a 'guaranteed minimized loss' strategy.

In each of the 5 rounds, play two tokens on red, and two tokens on black. In 36/37 cases (all but zero) you'll win on one color, and lose on the other. Given that a 'color' win pays 2x bet, you'll net out at zero, UNLESS a zero pops up 1 or more times. In such cases you lose 2x2=4 tokens.

Expect loss:
5 turns x 4 tokens bet x 1/37 chance you loose = 20/37 chips loss expected, leaving you on average with 99 and 17/37 chips in the end.

• Yup, went a bit too quick there. Edited accordingly, as the rest of the answer doesn't really change. May 1, 2016 at 13:54
• FWIW I agree that this is the correct answer if we play such tournaments ad-infinitum, of course. If it were not winner-take-all, however, then a chip-valuation is required (such as the ICM in tournament poker) May 1, 2016 at 14:10
• There's a local optimization for your solution: You can place a single chip on one of the three columns, winning pays 3x. Three chips on those would leave you losing only three chips on a 0 instead of four. This strategy can never lose against yours.
– Anon
May 1, 2016 at 16:14
• There's also a strategy that actually beats yours most of the time: On the last turn, instead of placing chips on red or black, I can place a chip each on the numbers 1 to 35. I'll very likely end up with one chip more than I started with, so I'll only lose a tournament against your strategy if the chip lands on 0 or 36 in the last turn.
– Anon
May 1, 2016 at 16:17
• Now I'm sure you see the pattern. I could beat this new strategy by placing chips on only 1 to 34. This would be beaten by placing chips on 1 to 33. And so on. Yet placing a chip on only 1 actually gets beaten by your original strategy.
– Anon
May 1, 2016 at 16:20