Using zero on the right side of the equation would closely approximate x were x not exact. At that point, we would determine how many of those trials lost would put us even using zero as the target result and x as the number of trials lost (multiplied by probability of losing) and overall number of trials, then -number of trials - x -multiplied by gain on wins as the number of trials won. We would start with a fixed number of trials. This is possible for Roulette because amounts to be won (assuming a flat bet amount), unlike BJ and UTH (for two examples), are not variable. This would be a fairly trivial calculation that could easily be done with simple algebra solving for x where x is zero.*** The only added step would be to determine (based on the number of attempts) x, where x is the number of trials you’d have to lose in order to be down money. The bet amount has to remain constant for obvious reasons. The reason why is because, the greater the number of trials, the more likely you are to finish down anyway.

This is important because being able to bust out increases the bust rate substantially, though this effect would be diminished the greater the number of trials. I’d also like to add that what I have detailed above would work for literally ANY isolated Roulette bet, provided: You asked me not to attempt to do this until Wizard posts in this thread, so I won’t. The Binomial Distribution calculator can give you the probability of losing x or more where x is one more than half of all attempts. If you lose more trials than you win, then you are down money. The Wizard answered my PM indicating that there is a way to figure this out although there may not be a handy formulaįor this one, you could either use the Beating Bonuses simulator or a binomial distribution calculator the fact that these are flat bets at Even Money makes this quite easy to do if you are specifying a particular number of trials. If a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero RouletteĪfter 1,000 spins - 98% probability the player will be downĪfter 600 spins - 92% probability the player will be downĪfter 100 spins - 80% probability the player will be downīut I don't know how to do the math to figure this out correctly Please don't attempt an answer until after the Wizard has posted The question has not been considered here (to my knowledge) or anywhere else that I know about By PM I posed a question to the Wizard and he indicated if I posted it he would respond

Simple roulette wheel