@engineer,
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.