1. Jun 29, 2010

### KFC

In basic statistic, we consider mean, mode and variance. Take a die as example, there are 6 possible values so the average is (1+2+3+4+5+6)/6= 3.5. For a roulette, besides 1 to 36, there are two special sections 0 and 00. So how do we calculate the average? Take 0 and 00 as numerical ZERO?

In one text, it said that for roulette, in 2000 spins, the average is 52.63 and standard derivation is 7.16. But how come it get an average that large?

2. Jun 29, 2010

### hamster143

You have to specify rules of the game

The average only makes sense if you can add outcomes. In case of a die, if you get $1 for each dot that you roll ($1 for 1, $2 for 2, ...) you can add them and you can say that the average (expected earnings per roll) is$3.5.

52.63% is the chance for the casino to win (and for you to lose) if you put money on a color. Say, you bet on red, if the roulette rolls red, you win, if it rolls black or either zero, you lose. The chance to lose is 20/38 = 52.63%.

3. Jun 29, 2010

### KFC

Thanks for reply. But I don't think the average (52.63) is the payoff odds or (rate to lose) because in the context of the book, it mentioned it is the average (mean) which later central limit theorem will be applied. And the author consider the same game on a biased roulette in which 17's appear at the chance 1/19 instead of the 1/38, and the average now becomes 105.26 instead of 52.63. Following your idea to get 52.63, I don't see how to get 105.26 in this case

4. Jun 29, 2010

### mathman

52.63 is impossible on the face of it. You might want to include more text material from the article or book you are referring to.