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Best of Alan Krigman  # How do Edge and Chance of Winning a Bet Differ?

4 April 2011

In casino gambling, edge indicates the fraction of every wager the joint should get to keep. Probability of winning suggests how frequently bets should pay off. Higher or lower edge means that the bosses keep more or less, not that a bet is less or more apt to win. Analogously, higher or lower probability of winning means players win more or fewer rounds, not that they earn less or more on the proposition. The two parameters aren’t interchangeable, although some players consider them essentially equivalent. Adding to the confusion, they’re both theoretical – useful statistically, especially when pictured as averages over extended play by one or many solid citizens – but neither foretells what will happen on any particular coup.

Some casino patrons haven’t a clue about the edge or chance of winning associated with their action. They seems to think “it’s all just gambling; you place your bets and cross your fingers.” This may not be as foolhardy as it seems for occasional players – out for fun and excitement and not wanting to spoil their entertainment by having to work at it. And for single bets or sessions so short as to make invoking averages meaningless, luck plays a bigger role than statistics, anyway.

Edge and chance of winning gain significance as the amount of action goes up. The cumulative impact of bankroll erosion caused by edge then increasingly dominates the state of a player’s fortune, and the frequencies of various outcomes actually encountered start to approach the probabilities of winning. Still, many “regular” gamblers can’t quote the value of either, aren’t sure how they impact their performance, and don’t know how or even if the two are related.
Examples of chance of winning a round and edge in three real cases and one hypothetical situation illustrate how the parameters differ and why the link between them is often obscure.

Bet on a single number at double-zero roulette: chance of winning is 2.63 percent and edge is 5.26 percent. Bet on a 12-number column: chance of winning is 31.6 percent but edge is still 5.26 percent.
Bet on a four at craps: chance of winning is 33.3 percent and edge is 6.67 percent. Bet on a six: chance of winning is 45.4 percent and edge is 1.52 percent.
Bet on a typical “8/5" jacks-or-better video poker machine: chance of pushing and recovering a bet is 21.4 percent, that of winning anything at all is 24.0 percent, and edge is 1.6 percent.
Bet on a fictitious jackpot-only longshot where players pay \$1 to make a draw from a bowl containing one red and 100,000 blue marbles. Red returns \$100,001; blue loses. The chance of winning is one out of 100,001 and edge is zero; on the average the house expects to earn nothing, but the overwhelming majority of people who try will lose.

The factor relating the chance of winning a round and the edge is the payoff. When bettors either lose or win fixed amounts, knowing either parameter, along with the payoff, is sufficient to determine the other. Denote edge by “e,” probability of winning by “p,” and payoff (amount won per dollar bet) by “w.” The formula for edge, knowing probability of winning, is e = (p) x (w) - (1- p). This can be re-arranged to find probability of winning, knowing edge, as p = (1+e)/(1+w).

Here’s how the formula yields the chance of winning a bet on a double-zero roulette column if you know the edge is 5.26 percent and the payoff is 2-to-1. Start with p = (1+e)/(1+w). Using decimals or fractions for the percentages and remembering that edge will be negative when the house is favored, substitute the known values to get p = (1-0.0526)//(1+2/1), and do the math to find p = (0.9474)/(3), which is 0.316 or 31.6 percent. In the other direction, say you’re aware that bets on the six at craps have five ways to win and six to lose so the chance of winning is 5/(5+6) or 5/11 while the payoff is 7-to-6. Start with e = (p) x (w) - (1- p). Substitute the known values to get e = (5/11) x (7/6) - (1 - 5/11); the arithmetic yields 0.0152 or 1.52 percent.

The wicket is stickier when a bet can either lose or win various amounts depending on the specific outcome. Say a game pays w1 = 1-to-1 with a probability p1 = 24.8 percent, w2 = 2-to-1 with a probability p2 = 15.0 percent, and w3 = 3-to-1 with a probability p3 = 0.10 percent. The formula for edge is e = (p1) x (w1) + (p2) x (w2) + (p3) x (w3) - (1 - p1 - p2 - p3). Insert the known probabilities and payoffs to get e = (0.248 x 1) + (0.150 x 2) + (0.001 x 3) - (1 - 0.248 - 0.150 - 0.001); this works out to -0.05 – a house edge of 5 percent. Algebra doesn’t let you go the other way to get the probabilities for the various wins given the edge. For instance, to find p1, you’d have to know not only the edge and all the payoffs, but also p2 and p3 which are two of the three unknown values you wish to determine.

These concepts can be extended to sessions as opposed to individual bets. But, it might be best to stop here for now, because doing so would require introducing the notions of volatility and skew – oops – which I just did. Violating this lesson of the well-loved lyricist, Sumner A Ingmark:
Since information overload is liable to confuse,
Don’t try for more enlightenment than anyone can use. 