Hi Bob.

I copied this off another members post. I cannot remember his name else I would give him credit.

On average, how many trials would you need in 38-number roulette before any number is repeated?

Counting the first trial, I show the mean is 8.408797, the median is 8, and the mode is 7.

The probability of two numbers without a repeat is 37/38 = 97.37%.

The probability of three numbers without a repeat is (37/38)×(36/38) = 92.24%.

The probability of four numbers without a repeat is (37/38)×(36/38)×(35/38) = 84.96%.

Following this pattern, the probability of no repeats in 8 numbers is (37/38)×(36/38)×(35/38)×...×(31/38) = 45.35%.

So the probability of a repeat within 8 numbers is 100% - 45.35% = 54.65%.

I think it's the same. Basically the more numbers that come out, the more chance of a repeat.