Punter picks a number (1-6) and bets \$1.
Three fair dice are rolled.
If your number shows up once, you get \$2 (stake plus \$1), twice \$3 (stake plus \$2), thrice \$4 (stake plus \$3).
If it doesn't show up at all, you lose your \$1 stake.
In my simple mind the probability of your number showing up just once is 1/6 + 1/6 + 1/6 = 1/2.
If you then add the variable of double or triple winnings when you get your number twice or thrice it would seem the punter has the edge, but surely I've missed something here?
The sample space of rolling 3 dice contains 216 possibilities for the outcome. If you work them out, 91 contain your number showing up at least once, 16 at least twice, and 1 showing up thrice. Note that 75 contain your number showing up exactly once, 15 exactly twice, 1 exactly thrice.
I think you are assuming these are independent events. If they were, you would add them together; however, they are not. The case where the same number shows up twice includes outcomes where the number shows up once, and the case where the same number shows up thrice includes outcomes where the number shows up once or twice. Therefore, you cannot add them.