Easier to beat the odds than house edge

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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

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:

matter how you slice the pie, You can’t increase its size thereby.


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