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Best of Alan Krigman

Gaming Guru

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How do the casinos set the payoffs on their games?

18 June 2012

Most casino aficionados have no idea how the bosses set the payoffs on the games. It’s a blend of science and art. The science is based on the math that links the probabilities associated with the results of a coup, the payoffs, and the edge or house advantage. The art involves judging what it takes to earn a profit and how much pain the traffic will bear.

The starting point is a formula that says the payoff per dollar bet (PO) times the probability of winning (PW), minus the probability of losing (PL), equals the edge (E). You’d write the equation as (PO x PW) - (PL) = E. If you knew the payoff and the probabilities, you could do the multiplication and subtraction to find the edge. Alternately, if you knew what edge you wanted along with the probabilities, you could “solve” for payoff in the form PO = (E + PL)/PW.

Pretend you’re a casino bigwig considering a game you think will be a hit. The dealer and the players each get one card, face down, from a single freshly-shuffled deck. Players bet whether their card ranks will be over, under, or equal to the dealer’s. The composition of the deck dictates the probabilities, although you may have to do some calculations to find the values. When you have these figures, payoffs and edge are the variables. Choose either to find the other.

To determine the probability of a match, assume the dealer draws any rank – say it’s a nine; the chance of this is four out of 52. For a match, the player must draw one of the three remaining nines from the 51 cards still in the deck; the outlook is three out of 51. The prospect of both draws together is (4/52) x (3/51). This can happen for any of the13 ranks from deuce through ace. So the overall probability of a match with two unknown cards is 13 x (4/52) x (3/51). This equals 1/17 or 5.88 percent. The probability of not matching is the other 16/17 or 94.12 percent.

The probability the player will be higher depends on what the dealer draws. The chance of any particular dealer’s card is four out of 52. Say ace is high. If the dealer has a deuce, 12 ranks, each comprising four cards, will be higher. The joint probability – a dealer’s deuce and a player being higher – is therefore (4/52) x (4x12/51), which equals 7.2389 percent. If the dealer has a three, the likelihood also being 4/52, 4x11 = 44 of 51 cards will be greater so the joint probability of being higher is (4/52) x (4x11/51), which equals 6.6365 percent. Continuing this reasoning to a king, 4 x 1 out of 51 will be higher. And, for an ace, it’s zero out of 51. The overall probability with two unknown cards is accordingly (4/52) x (48 + 44 + 40 + 36 + 32 + 28 + 24 + 20 + 16 + 12 + 8 + 4 + 0)/51, which equals (4/52) x (312/51). This reduces to 8/17 or 47.06 percent to win and the complementary 9/17 or 52.94 percent to lose. Identical logic applies to the player having the lower of the two cards, except that the figures are reversed with the dealer’s ace having 4x12 = 48 of 51 being lower, down to zero out of 51 for a dealer’s deuce.

Having the probabilities, you could assign a trial payoff for each bet, calculate the edge, then decide if it’s satisfactory. For a game such as this, it’s simpler to begin with edge equal to zero and solving the formula for payoff in the form PO = PL/PW. You’d then decrease the zero-edge payoff you got and calculate the new associated edge, stopping when you have a figure you like.
For the bet on match, the zero edge situation gives you PO = (16/17)/(1/17), which equals 16. If you drop this to 15, you can solve to find an edge of 15 x (1/17) - (16/17) = -1/17 or -5.88 percent. Were you to cut the payoff to 14, edge would be 14 x (1/17) - (16/17) = -2/17 or -11.76 percent. Here, a value judgement is necessary. You’d prefer to earn 11.76 rather than 5.88 percent of the gross wager on this proposition. But you don’t want to hammer your patrons heavily enough that they avoid this bet, or your casino, altogether.

For the greater or less than bets, the no-edge payoff would be PO = (9/17)/(8/17). This gives $1.25 per dollar bet as your initial value. If you drop this to even money, edge becomes 1 x (8/17) - (9/17) = -1/17 or -5.88 percent. Basic strategy blackjack buffs and dice devotees who bet and take or lay Odds on Pass, Come, Don’t Pass, and Don’t Come would consider 5.88 percent usurious, but it’s in line with most other casino games so you could get away with it, especially if the pace of the game isn’t too fast.

To be sure, the calculations can get much more complex. In some situations, you can vary the probabilities as well as the payoff. Or, a game might call for different payoffs depending on the result of a coup. And games that are easy for solid citizens to play don’t always lend themselves to analysis using the laws of probability directly so simulations might be needed to determine the chances of various outcomes. But the underlying concepts are always the same and once you have the idea, well, you have the idea. As the irrepressible inkster, Sumner A Ingmark, wrote:
When based on solid principle,
Your logic is invincible.

Alan Krigman

Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.
Alan Krigman
Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.