As an example, we can examine the European roulette model (roulette with one zero). Since this roulette has 37 cells with equal odds of hitting, it is clear that this is a final model of field probability , where , for all . We'll call the bet S a three (A,r,ξ), where A for a certain random event|event , and = random size. The event A naturally leads to a winning event, r to the size of the bet (in dollars, for example), ξ to the bet rule, and the mathematical expectation M[ξ] relates to the bet profitability.
The rules of European roulette have 10 types of bets. First we can examine the 'Straight Up' bet. It's clear that in this case, , where , and ξ is determined by this law
The bet's probability is equal to
Without details, for a bet, red or black, the rule is determined as
and the profitability . For similar reasons it is simple to see that the profitability if also equal for all remaining types of bets. .[6][7]
In reality this means that, the more bets a player makes, the more he is going to lose independent of the strategies (combinations of bet types or size of bets) that he employs:
Here, the profit margin for the roulette owner is equal to approximately 2.7%. Nevethless, several roulette strategy systems have been developed despite the losing odds.[8]
It's worth noting that the odds for the player in American roulette are even worse, as the bet profitability is , and the rest are .
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