Calculating Average for Roulette

[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?

 

[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%.

 

[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

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%.

 

[mathman]

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?

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.

 

[CellCoree]

I would first like to clarify that the concept of average in statistics is typically referred to as the mean, which is calculated by adding all the values and dividing by the number of values. The mode refers to the most frequently occurring value, and the variance is a measure of how spread out the data is from the mean.

In the context of roulette, the average or mean can be calculated by considering all the possible outcomes and their probabilities. In this case, there are 38 possible outcomes (numbers 1-36, 0, and 00) and each has an equal probability of occurring, making the average or mean value (1+2+3+...+36+0+00)/38 = 19.5. This means that over a large number of spins, the average number that will come up is 19.5.

It is important to note that the average value of 52.63 mentioned in the text is not referring to the number that will come up on the roulette wheel, but rather the average amount of money that a player can expect to win or lose over 2000 spins. This is calculated by taking into account the payouts and the probabilities of winning for each bet on the roulette table.

Regarding the standard deviation of 7.16, this is a measure of how much the data varies from the mean. In the case of roulette, it indicates that there is a significant amount of variability in the amount of money a player can expect to win or lose over 2000 spins.

In summary, the average or mean value in roulette is 19.5, which takes into account all the possible outcomes and their probabilities. The average value of 52.63 mentioned in the text refers to the expected amount of money a player can win or lose over a large number of spins, and the standard deviation of 7.16 indicates the variability in these outcomes.