An American style roulette wheel has individual compartments for the numbers 1 to 36, plus 0 and 00. If the payoff for a single-number bet is 35:1, what is the expected value of the winnings on any single-number bet?

Only if you win. The expected value (average return) is 35/38 of the bet.

I assumed that zero is not a bettable number (for the 35:1 payoff), so that there are only 36 possible winning numbers out of 38. The probability of a particular number from 1 to 36 is 1/38.

To calculate the expected value of the winnings on any single-number bet, we need to determine the probability of winning and the associated payout.

In American roulette, there are 38 compartments on the wheel, including the numbers 1 to 36, 0, and 00. Since you're placing a single-number bet, there is only one winning outcome out of the total number of possible outcomes.

The probability of winning on any single-number bet is therefore 1/38.

The payout for a single-number bet is 35:1, which means if you win, you receive 35 times your original bet as winnings.

To find the expected value, we multiply the probability of winning by the amount won and subtract the probability of losing multiplied by the amount lost.

Expected Value = (Probability of Winning * Amount Won) - (Probability of Losing * Amount Lost)

Expected Value = (1/38 * 35) - (37/38 * 1)

Expected Value = (35/38) - (37/38)

Expected Value = -2/38

The expected value of the winnings on any single-number bet in American roulette is -2/38 or approximately -0.0526. Therefore, statistically, you can expect to lose about 5.26% of your bet in the long run.