There are three R-14 slots on the wheel. There is only one R-14 bet square on the table. So yes you get paid if the ball lands in any of the R-14 slots.
Ok, thanks for the clarification. Here are my calculations.
Optimal bet fraction = Kelly = p - (1- p) / b,
where
p = probability of winning
b = odds on the bet
Expectancy = p * b - (1 - p)
My custom score = Expectancy * Kelly
Then,
Kelly(red) = 23/37 - (1 - 23/37) / 1 = 0.24 = 24%
Kelly(black) = 13/37 - (1 - 13/37) / 1 = -0.29 = -29%
Kelly(R14) = 3/37 - (1 - 3/37) / 35 = 0.05 = 5%
Kelly(R16) = 4/37 - (1 - 4/37) / 35 = 0.08 = 8%
Expectancy(red) = 23/37 * 1 - (1 - 23/37) = 0.24
Expectancy(black) = 13/37 * 1 - (1 - 13/37) = -0.29
Expectancy(R14) = 3/37 * 35 - (1 - 3/37) = 1.92
Expectancy(R16) = 4/37 * 35 - (1 - 4/37) = 2.89
Score(red) = 0.24 * 0.24 = 0.058
Score(black) = no need to calculate, negative expectancy and negative Kelly
Score(R14) = 0.05 * 1.92 = 0.096
Score(R16) = 0.08 * 2.89 = 0.231
So, if only a single bet is allowed, the best strategy would be to bet 8% of the bankroll on R16, on every bet. Some sort of combination bet (such as R14 and R16) would probably be even better, but I have not calculated it yet.