Combination bets are usually bad bets

Roulette and craps are very different games, but they have something in common: a large number of betting options, leading to players trying to figure out combinations of wagers that will guarantee success.

They'll try to use one bet to cover up the weakness in another. It doesn't really work that way. No combination of bets with house advantages can add up to a whole with an edge to the player.

That doesn't stop players from trying, even adding weak bets in the hopes of strengthening a system. A roulette player named Ron wrote to me recently to say he was doing just that, incorporating the five-number bet on 0, 00, 1, 2 and 3 into his combination.

"It's pretty simple, really," he wrote. "If I'm at a $5 table, I bet $5 on even. Then I put $5 on the five-number bet. That takes care of the 0 and 00, and it gives me two of the odd numbers. If the 2 comes up, I win twice. The only numbers that can beat me are the odds from 5 on up. So I have 22 of the 38 numbers, and one of them gives me some extra."

The house edge on the bet on even is 5.26%, the same as nearly every bet on the layout. Exception: the five-number bet Ron added to his combo. There, the house edge soars to 7.89%. It's a bet that should never be made, neither alone, nor in a combination.

A better option would be to bet the five numbers individually. Instead of $5 on the five-number bet, he could bet $1 on each on the 0, 00, 1, 2 and 3. Each of those wagers has a 5.26% house edge, and that would leave his system no weaker than any other roulette combination.

But that's not what he as doing, at least not until I showed him the arithmetic. He was betting the evens plus the awful five-number bet.

If Ron were to make these bets 38 times and have each number on the wheel turn up once, he'd risk a total of $380. He'd lose his 10 bucks on each of the 16 odd numbers from 5 through 35. On other numbers, he'd do no worse than break even.

**On the 17 even numbers from 4 through 36, he'd lose his five-number bet, but win his even bet, keeping his $5 wager and claiming $5 in winnings. He'd have a total of $170.

**On 0, 00, 1 and 3, he'd win the five-number bet at 6-1 odds, and lose the even number bet. Each winner would bring him back his $5 wager plus $30 in winnings, for a total of $140.

**On 2, he'd keep both wagers, win $5 on even and win $30 on the five number, for a total of $45.

All told, he'd have $355 at the end of the trial. The house would keep $25. The house edge is 6.58% edge — right between the 5.26% on even and 7.89 on five numbers.

In Ron's attempt to cover weaknesses in the even bet, he takes on the bigger weakness of the five-number wager. It's not a winning combination.

That's the way it works in any game that offers multiple wagers with different house edges. In craps, a friend liked to use a simple system. He made $6 place bets on 6 and 8, and he'd hedge with a $3 place bet on any 7. With a 4-1 payoff, he figured he'd win back the $12 he lost on the place bets when a 7 turned up. When either place bet won, he'd win $7 at 7-6 odds, and just lose $3 on his any 7.

Problem is, the any 7 bet also lost when 2 turned up, or 5, or 9, or any other number, while the place bets just stayed in action. He had to bet a new $3 to keep his hedge 20 times per 36 rolls.

The house edge on the combination is 6.67%, a weighted average that takes into consideration that he had more money on the place bets at a 1.52% house edge than on any 7 at 16.67%.

No combination can have a house edge higher nor lower than that on any individual bet in the system. Be it the five-number bet in roulette or any 7 in craps, a bad bet is a bad bet, and it can't be the key to any winning combination.