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Calculating a Casino's Edge From Odds

Date: 05/30/2000 at 02:00:21
From: Kevin Sheridan
Subject: Calculating a Casino's Edge from Odds

I know that the correct odds for rolling a 7 in craps are 5 to 1. 
There are 6 chances to roll a 7 out of 36 possibilities. This is 1/6, 
or, expressed in odds, 5 unfavorable numbers to 1 favorable. If the 
casino's payout in odds is 4 to 1, I am told that the casino's edge 
is 16.67%. I understand the edge to be the difference between the 
player's chance of winning the bet (correct odds) and the casino's 

I would like to know how to calculate the 16.67%. I am off by a factor 
of 2 when I work out the problem.

Thanks for the help.

Date: 05/30/2000 at 12:32:14
From: Doctor TWE
Subject: Re: Calculating a Casino's Edge from Odds

Hi Kevin - thanks for writing to Dr. Math.

The casino's edge is simply the negative of the expectation (the 
average the player can expect to win per hand). In your example, the 
expectation is -1/6, and the casino's edge is +1/6. In other words, if 
the player bets $1 on every roll, the player can expect to lose $1/6 
(or about 17 cents) per roll over the long run. The casino, therefore, 
can expect to earn $1/6 per roll over the long run.

Expectation is defined as:

     E(X) = Sum[x p(x)] for all x in A

     where E(X) = expectation
             x  = amount won or lost for each possible outcome
           p(x) = probability of that outcome
             A  = set of all possible outcomes

In your example, there are two possible outcomes, rolling a 7 (player 
wins 4) and not rolling a 7 (player loses 1). The probabilities are 
1/6 for rolling a 7 and 5/6 for not rolling a 7. Note that the sum of 
all probabilities must always equal 1. So the expectation is:

     E(X) = (4)*(1/6) + (-1)*(5/6) = 4/6 - 5/6 = -1/6

Thus, the casino's edge is -(-1/6) = 1/6 or 16.67%.

To compute this based on odds and payout, we have to convert the 
expectation formula to those terms. The probability of an outcome is 
defined as:

     p(x) = (chances for)/(total chances)

while odds are defined as:

     (chances for):(chances against)

Total chances is, of course, (chances for)+(chances against).

For an outcome with odds X:Y, let's define O, the odds expressed as a 
single value rather than a pair of numbers, as O = X/Y. Thus for odds 
of 5:1, O = 5; while for odds of 7:2, O = 3.5.

Similarly, for a payout of X:Y, let's define Q (I use Q since p is 
already used for probability) as Q = X/Y.

Now we need to define the equation for expectation in terms of O and 
Q. We'll assume only two possible outcomes, win and loss. The formula 
for expectation above can handle any number of outcomes. So:

     E(X) = (payoff for W)*(prob. of W) + (payoff for L)*(prob. of L)

          = Q*p(W) + (-1)*p(L)

Note that since we converted the payoff to Q, the payout "odds" are 
Q:1, thus the payout for a loss is -1

          = Q*[1/(O+1)] + (-1)*[O/(O+1)]

Similarly, since we converted the 'odds against winning' to O, the 
odds are O:1, thus the probability against winning is O in O+1, and 
the probability for winning is 1 in O+1.

          = Q/(O+1) - O/(O+1)

          = (Q-O)/(O+1)

The casino's edge is -E(X), which is:

     -E(X) = -(Q-O)/(O+1)

           = (O-Q)/(O+1)

To test this formula, let's try your example. The odds are 5:1, thus 
O = 5/1 = 5. The payout is 4:1, thus Q = 4/1 = 4. So the casino's edge 

     -E(X) = (5-4)/(5+1)

           = 1/6  or  16.67%

It seems to work. Perhaps you can test it with other values as well.

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum   
Associated Topics:
High School Probability

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