'What are the probabilities for the player to be ahead in various numbers of trials?'Įverybody can use my probability software SuperFormula, option L: At Least M successes in N trials. To be ahead, means the player has won at least one more roulette spin (number of successes) than the number of losses in N trials. The main thing, mathematically, is the number of player's wins in N trials. But more than 20 spins are NOT recommended the probability (odds) to lose go(es) above 50%! Think about it! Even in even-money bets, the player has a good chance to be ahead by at least one unit after 5, or 10, or even 100 spins. There are moments, however, when the roulette player can be ahead by at least one betting unit. Theoretically, no bankroll will put a player ahead guaranteed, IF flat-betting and playing very long consecutive sessions. I don't agree that it is the best strategy (betting all your bankroll on one spin). On the surface, the best probability for the roulette player to be ahead is in one trial (spin): 48.6% to win (versus 51.4% to lose), as far as even-money betting is concerned. How much cash C would I need to have probability P of succeeding? Let's say I'm betting on a 37 number roulette wheel (18 red 18 black and one green 0)?' Is it possible to generalize this? I want to win W dollars at which point I will quit. To take an extreme example if you had $1000 you could reasonably expect to be up $1 at some point.

'Obviously in roulette betting on in the long run you are going to lose your money but at some point chances are you'll be in profit. This question was posted in mathematical newsgroups (, , ): 'Winning and Quitting on Red/Black in Roulette'. To Be Ahead and Leave the Casino as a Winner By Ion Saliu, Founder of Roulette Mathematics Probability, Odds to Win at Roulette in N Number of Spins ~ Roulette: Odds, Number of Spins to Quit as Winner, Payouts

Roulette expected payout of a red bet