Dear Mark: Maybe this is a dumb question, but it occurs to me that in roulette, if one were to put a dollar on each number, one through 36, that player would loose $35 but win $35 plus their dollar back. This cannot be true because it changes the odds to being on the bettor's side. Can you explain this illusion for me? Pete S.

They put warnings on hair dryers that say, "do not dry hair while standing in a puddle of water” for the same folks who e-mail the real incoherent inquires, and, of course, I answer those too. Remember when Gurth asked if I thought they would put a casino next to Yucca Mountain (Nevada), the nation's first long-term geologic repository for spent nuclear fuel and high-level radioactive waste?

Your question, Pete, stems from layout confusion, plus not knowing how the house does get its edge in roulette.

What you’re missing is that the casino’s advantage of 5.26% in roulette comes from the presence of the 0 and 00 on the layout. You admit to the belief that the casino pays all wagers according to how the odds would be if there were just 36 numbers on the wheel. But It ain’t so, my friend. Include the 0 and 00, and add ‘em up again. 38, right? That’s what pays the light bills. The true odds of hitting your number are 1/38, yet winners are paid only 35-1. 

By playing on a single-zero game, you can cut the house edge in half; the house will still pay you 35-1, even though the true odds are improved for you to 1/37.

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