Roulette cover all.

On a board where there are 36 numbers and two zeroes for a total of 38 spots, is there a way to calculate how many spins are needed to have a 50% chance of hitting every number once or more?

I don't think "expected number" is the correct term in this case. Do you mean, the minimum number of spins such that the probability of getting all 38 numbers at least once in that many spins (or fewer) is at least 1/2?

Assuming that is what you want, some simulating shows that the probability of getting all 38 numbers in 151 or fewer spins is just less than 50%, and in 152 or fewer spins is just greater than 50%.

It's very easy to make a Markov chain for this since there are only 38 states.

0.491968759 chance of hitting all 38 in <152 spins
0.501599171 chance of hitting all 38 in <153 spins

Not that you asked, but the expected value to hit all 38 is 160.6602765052 spins, which can be calculated exactly via calculus and Markov chain, and can be closely approximated by (Ln(38) + c)*38 where c is the Euler–Mascheroni constant

I guess what I'm looking for is the number of spins that it would be a fair bet for hitting all the numbers. 100 spins was obviously too little, and 200 seemed too many. I was thinking 150 and that was surprisingly close.
Thanks for your replies.

An over/under of 153.5 would be pretty close to a fair bet. Assuming even money, betting the under would give you a slight (0.32%) edge