Suppose the wheel is spun twice. On the first spin, you place two bets: one on even and one on the first dozen.

Question

Suppose the wheel is spun twice. On the first spin, you place two bets: one on even and one on the first dozen. On the second spin, you also place two bets: one on "high" and one straight up bet on the pocket '13'. What is the probability that you win at least one of your bets on both spins?

Answer 1

I assume that this is a standard American roulette wheel with 2 extra green zeroes and numbers 1-36

1st spin P(at least one bet) = P(even) - P(1st dozen) + P(even and 1st dozen)

P(1st spin at least one) = 18/38 + 12/38 - 6/38 = 24/38

2nd spin P(13 or high) = P(13) + P(high) (These probabilities have no overlap)

P(2nd spin 13 or high) = 1/38 + 16/38 = 17/38

For both of these independent events to occur P(1 and 2) = P(1)P(2) = (17*24)/38²

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