What is the standard deviation of a $1 bet on any one particular number, e.g., 33, in the game of
American roulette? Notice the payoff is 35:1 for this type of bet.
To determine the standard deviation of a $1 bet on a specific number in American roulette, we need to understand the basic concepts and calculations involved.
In American roulette, there are 38 possible outcomes: the numbers 1 to 36, plus the green 0 and 00. Each number has an equal probability of occurring, which is 1/38.
For a $1 bet on a specific number, such as 33, there are two possible outcomes:
1) The number 33 is selected, resulting in a win with a payout of $35 (since it is a 35:1 payoff).
2) Any other number is selected, resulting in a loss of $1.
To calculate the standard deviation, we need to know the probability of each outcome (winning $35 and losing $1) and the corresponding values.
Probability of winning: Since there is only one winning outcome (33), the probability is 1/38.
Probability of losing: Since there are 37 losing outcomes (all other numbers), the probability is 37/38.
Value of winning: The value is $35 (the payout).
Value of losing: The value is -$1 (the amount lost).
Now, we can calculate the standard deviation using the following formula:
Standard Deviation = √((Probability of Winning * Value of Winning^2) + (Probability of Losing * Value of Losing^2))
Standard Deviation = √(((1/38) * (35^2)) + ((37/38) * (-1^2)))
Simplifying the equation, we get:
Standard Deviation = √(((35^2)/38) + (37/38))
Calculating this, we find that the standard deviation of a $1 bet on the number 33 in American roulette is approximately $5.77.
Therefore, the standard deviation of this type of bet represents the amount of risk or variability associated with it. In this case, the standard deviation indicates that, on average, the amount won or lost on a $1 bet on the number 33 will be around $5.77.