Calculating expected value at American roulette. MAking 2 side bets at the same time.

If I understand the US roulette wheel correctly, there are 36 numbers and 2 zeros. The chance of winning your first twelve bet is 12/38. The chance of winning your red bet is 18/38. The chance of winning either of them is (12+18-6)/38. (There are 12 wins between 1 and 12, 18 red slots and 6 that are in both so you need to subtract out the double count.) The chance of winning both of them is 6/38.

Wrote engineer: "The chance of winning either of them is (12+18-6)/38."

This can't be right. Even if we only consider cases where one of the two bets (but not both) wins, thats 24/38 = 12/19, which is better than even odds of winning, thus giving a long term advantage to the better, not the house ,-- and in roulette! Winning only the Red bet would net $10; and winning the number bet only would also net about $10, assuming that the $10 was spread equally among 12 numbers and that winning a number bet pays 36 to 1.

The odds of winning a number bet are not 12/38 since half of those are red and we're concerned with an exclusively numeric win. The odds of that are 6/38. The odds of an exclusively Red win are 12/38 since there are 18 reds but six of them are number winners. So the total odds of winning either Red or a number but not both, are (6+12)/38 = 18/38 which is the same as the odds of winning Red by itself without making any number (or other) bets. It's also less than even odds, giving the House an advantage.

The expected value of the $10 bet on the first 12 is:

(12/38) x $20 win and (26/38) x $10 loss = +6.31 - 6.84 = -$0.53

The expected value of the $20 bet on red is:

(18/38) x $20 win and (20/38) x $20 loss = +9.47 - 10.53 = -$1.06

The losses of 5.3% in both instances arise because, more generally, the 2 zeroes out of 38 numbers generate losses of (2/38) ie. 5.3%.

Just using logic, all bets favor the house. Otherwise, they eouldn't be able to pay the workers and overhead.

Yeah, that's what clued me into examining the odds myself.

There are some games where the better can get an advantage over the house, e.g. blackjack where card-counting is used, which is why casinos don't like card counters, and also why they have introduced more decks more often.

That problem with card counting was identified many years ago. If one is lucky, they can win big on machines like one arm bandits. The odds are about the same as lotteries of winning big.

The advantage the house enjoys has nothing to do with whether or not the probability of winning is above or below 50%. The advantage is strictly due to the fact that they pay out based on 36 numbers when there are actually 37 or 38 numbers.