Is the “Big Six” better or worse than roulette?

Question: Last time I gambled, I played the "Big Six," betting $1 on Joker. It hit on the third try and paid $45, so I quit $43 ahead. A friend called it a miracle, saying this was a sucker bet. But he couldn't tell me why it was worse than betting on one number at roulette, which pays only $35. Can you explain this?

Answer: The usual criterion for the quality of a bet is the house edge. This figure indicates the rate at which casinos use the laws of probability to earn their money. Edge depends on chances of winning and payoffs. It's usually given as a percent of the initial bet. Here's how to picture it for the games in question.

Big Six has 54 positions, one being the Joker. On the average, if you make 54 bets on Joker, you'll lose 53 times and win once. Say you bet $1 each time – $54 in all. Statistically, you expect to lose $53 and collect $45. Expected net loss on $54 is therefore $8. Edge is 8/54 = 14.8 percent, almost $0.15 per dollar bet.

"Double-zero" roulette has 38 positions, and bets on a single number pay 35-to-1. On the average, if you make 38 bets on any number, you'll lose 37 times and win once. At $1 a spin, you expect to lose $37 and collect $35. Expected net loss on $38 bet is $2. Edge is 2/38 = 5.26 percent, over $0.05 per dollar bet.

Edge works on averages. Some players will exceed the average, some fall below it. Several factors influence individual games, one of which is that the lower the edge, the longer bettors can keep trying for a stroke or run of luck to put them over the top.