Editing Roulette (section)

==Mathematical model==
As an example, the European roulette model, that is, roulette with only one zero, can be examined. Since this roulette has 37 cells with equal odds of hitting, this is a final model of field probability <math display="inline">\left(\Omega, 2^\Omega, \mathbb{P}\right)</math>, where <math>\Omega = \{0, \ldots, 36\}</math>, <math display="inline">\mathbb{P}(A) = \frac{|A|}{37}</math> for all <math>A \in 2^\Omega</math>.

Call the bet <math>S</math> a triple <math>(A, r, \xi)</math>, where <math>A</math> is the set of chosen numbers, <math>r \in \mathbb{R}_+</math> is the size of the bet, and <math>\xi: \Omega \to \mathbb{R}</math> determines the return of the bet.<ref>{{cite book |last1=Barboianu |first1=Catalin |title=Roulette Odds and Profits: The Mathematics of Complex Bets |year=2008 |publisher=Infarom |isbn=9789738752078 |page=23}}</ref>

The rules of European roulette have 10 types of bets. First the 'Straight Up' bet can be imagined. In this case, <math>S = (\{\omega_{0}\}, r, \xi)</math>, for some <math>\omega_{0} \in \Omega</math>, and <math>\xi</math> is determined by
: <math>\xi(\omega) = \begin{cases}
          -r, &\omega \ne \omega_0\\
  35 \cdot r, &\omega = \omega_0
\end{cases}</math>

The bet's expected net return, or profitability, is equal to
:<math>M[\xi] =
  \frac{1}{37} \sum_{\omega \in \Omega} \xi(\omega) =
  \frac{1}{37} \left(\xi(\omega_0) +
    \sum_{\omega \ne \omega_0} \xi(\omega)\right) =
  \frac{1}{37} \left(35 \cdot r - 36 \cdot r \right) =
  -\frac{r}{37} \approx -0.027r.
</math>

Without details, for a bet, black (or red), the rule is determined as
:<math>\xi(\omega) = \begin{cases}
  -r, &\omega \text{ is red} \\
  -r, &\omega = 0 \\
   r, &\omega \text{ is black}
\end{cases},</math>

and the profitability
:<math>M[\xi] = \frac{1}{37}(18 \cdot r - 18 \cdot r - r) = -\frac{r}{37}</math>.

For similar reasons it is simple to see that the profitability is also equal for all remaining types of bets. <math display="inline">-\frac{r}{37}</math>.<ref>[[Wikibooks:Roulette/Math|Roulette Math]], en.wikibooks.org</ref>

In reality this means that, the more bets a player makes, the more they are going to lose independent of the strategies (combinations of bet types or size of bets) that they employ:
:<math>\sum_{n = 1}^{\infty}M[\xi_n] = -\frac{1}{37}\sum_{n = 1}^{\infty}r_n \to -\infty.</math>
Here, the profit margin for the roulette owner is equal to approximately 2.7%. Nevertheless, several roulette strategy systems have been developed despite the losing odds. These systems can not change the odds of the game in favor of the player.

The odds for the player in American roulette are even worse, as the bet profitability is at worst <math display="inline">-\frac{3}{38}r \approx -0.0789r</math>, and never better than <math display="inline">-\frac{r}{19} \approx -0.0526r</math>.{{cn|date=October 2024}}

===Simplified mathematical model===
For a roulette wheel with <math>n</math> green numbers and 36 other unique numbers, the chance of the ball landing on a given number is <math display="inline">\frac{1}{(36+n)}</math>. For a betting option with <math>p</math> numbers defining a win, the chance of winning a bet is <math display="inline">\frac{p}{(36+n)}</math>

For example, if a player bets on red, there are 18 red numbers, <math>p = 18</math>, so the chance of winning is <math display="inline">\frac{18}{(36+n)}</math>.

The payout given by the casino for a win is based on the roulette wheel having 36 outcomes, and the payout for a bet is given by <math display="inline">\frac{36}{p}</math>.

For example, betting on 1-12 there are 12 numbers that define a win, <math>p = 12</math>, the payout is <math display="inline">\frac{36}{12} = 3</math>, so the bettor wins 3 times their bet.

The average return on a player's bet is given by <math display="inline">\frac{p}{(36+n)} \cdot \frac{36}{p} = \frac{36}{(36+n)}</math>

For <math>n > 0</math>, the average return is always lower than 1, so on average a player will lose money.

With 1 green number, <math>n = 1</math>, the average return is <math display="inline">\frac{36}{37}</math>, that is, after a bet the player will on average have <math display="inline">\frac{36}{37}</math> of their original bet returned to them. With 2 green numbers, <math>n = 2</math>, the average return is <math display="inline">\frac{36}{38}</math>. With 3 green numbers, <math>n = 3</math>, the average return is <math display="inline">\frac{36}{39}</math>.

This shows that the expected return is independent of the choice of bet.