The math works out more easily if he bet on the Player. I work out a similar problem in roulette at my mathproblems.info site, problem number 116. For even money bets, the general formula is ((q/p)b-1)/((q/p)g-1), where: b = starting bankroll in units. g = bankroll goal in units. p = probability of winning any given bet, not counting ties. q = probability of losing any given bet, not counting ties. Here the player starts with $12 million, or 60 units of $200,000, and will play until reaches 120 units or goes bust. So in the case of the Player bet the equation values are: b = 60 g = 120 p = 0.493175 q = 0.506825 So the answer is ((0.506825/0.493175)60-1)/(( 0.506825/0.493175)120-1) = 16.27%. It is much more complicated on the Banker bet, because of the 5% commission. That would result in the distinct possibility of the player overshooting his goal. If we add a rule that if a winning bet would cause the player to achieve his goal, he could bet only what was needed to get to $12 million exactly, then I estimate his probability of success at 21.66%. A simpler formula for the probability of doubling a bankroll is 1/[1+(q/p)b]. This question was raised and discussed in the forum of my companion site Wizard of Vegas.