Easier to beat the odds than house edge

Mar 19, 2002 1:36 AM

Sophisticated gamblers know that if the odds against winning a wager are “x-to-1,” an “x-to-1” payoff would make the bet “fair” ­­”” no advantage for either the house or the player. A lower relative payoff would give the casino an edge. A higher payoff would tip the balance toward the bettor.

A few examples will show the idea. At double-zero roulette, the odds against winning a bet on a single spot are 37-to-1 while the payoff is 35-to-1. Place the four at craps for $10, to win $18; you’re bucking 2-to-1 odds for a 1.8-to-1 payoff. Take insurance when you have a blackjack and the dealer has ace-up; you’re a 2.25-to-1 underdog for a return of 2-to-1.

In these cases, the payoffs are lower than the odds against winning, giving the house an edge. The disparities may seem minor. In fact, few players are aware of ”” let alone notice ”” them. Yet the edges they engender are relatively steep (5.26 percent at roulette, 6.66 percent on the four at craps, and 7.69 percent on the insurance bet at blackjack). Were the odds-payoff gap much larger, the house advantage would be so outrageous that solid citizens couldn’t leave the casino with money in their fanny packs except by lottery-scale strokes of luck.

Games with “payoff schedules,” in which the same bet can win different amounts depending on the outcome, are more complicated. They’re accordingly tougher to assess intuitively with respect to edge, and may leave players thinking they’re being short-changed.

The jackpot for a royal on a jacks-or-better video poker machine offers a good model. The odds against making a royal are about 40,000-to-1. But the payoff with the maximum number of coins bet is only 800-to-1. Likewise, the odds against three-of-a-kind at three-card poker are 424-to-1 while the payoff on the pair-plus bet is only 30-to-1. Yet, house advantage on “9/6” video poker with no bonuses is 0.5 percent and that on the pair-plus bet at three-card poker is 2.3 percent ”” both of which compare quite favorably to the edge on the win-lose bets mentioned previously.

To understand how odds on bets with multiple possible payoffs work, pretend there’s a game at which you can lose on a skunk or win at one of three levels: raccoon, deer, or elk. Probabilities, odds, and payoffs of the various results for one implementation are shown in the accompanying chart. The “expectation” for each result is the amount lost or won per dollar bet, multiplied by the probability. Get the overall expectation for the game by adding the individual components. For this situation, the sum is zero, meaning that players should average $0.64 in wins for every $0.64 lost. This is accordingly a fair game, neither the casino nor the bevy of bettors making a profit in the long run.

Although the payoffs at each level are way below the odds against the results, they don’t stand on their own but are parts of the total expectation. That is, a 9-to-1 payoff for a 99-to-1 shot at a elk only represents an expectation of $0.09 per dollar. But it’s added to the $0.30 and $0.25 for deer and raccoon to yield $0.64. And this sum cancels the $0.64 expected loss on skunk.

Contrast this with a game omitting the raccoon and deer, players still having 1 percent chance of winning on a elk but the full complementary 99 percent chance of losing on the skunk. The odds against winning are 99-to-1. Assume the payoff is 99-to-1. As for expectation, this is a 99 percent chance of losing multiplied by $1 or -$0.99, and 1 percent chance of winning times $99 or +$0.99. The total comes to zero, as it should in a fair game.

Now you know why, when bets can win various amounts depending on the result of a round, house edge can remain reasonable even though payoffs at each level are far less than the odds of the outcome in question. It’s as the poet, Sumner A Ingmark, wrote:

No matter how you slice the pie, You can’t increase its size thereby.