(House Edge continued)

Calculating

     

Calculating the house edge in most casino games can be done by comparing the true odds to the pay out odds and representing the difference in terms of a percentage. Or, simply calculate the expected loss and divide it by the average wager. These two methods are the same thing from two slightly different angles. (The first of the two methods I have abandoned. The second method is easier and much more common.)

A simple example of calculating edge could be a coin toss wager. Heads can only come up one way of two possible ways. Therefore, the true odds that heads will come on any coin toss is 1 divided by 2, or, expressed in typical gaming fashion, 1 chance against (tails) and one chance for (heads) 1:1. If winning a coin toss paid the true odds, there is no advantage or disadvantage to the house or the player. The difference in the payout odds to the true odds (1:1 - 1:1) is 0/2 = 0%. If the payout odds were 2:1, then this would work out to 3-2= 1, 1/2 = .50 = 50% advantage for the player. Or, if the payout was 50 cents on the dollar, then this would be a payout of 1:2 which would be 2-3/2 = -.50 = 50% to the house.

More simply, we can imagine that we bet twice on the coin toss and win one and lose one. Allowing the average wager to be $1, we can see that when the payout is 2:1, we have a net of $2 - $1 = $1/2 = .50 per wager, our expected gain. .50/1 = .50 = 50% advantage for the player. Conversely, a 1:2 payout produces a net loss of $1 - $2 = -1. Divide -1 by 2 here to get the expected loss of -.50. -.50/1 (expected loss divided by the average wager of $1) = -.50 = 50% to the house.

Some practical examples of calculating edge

House Edge = Expected loss divided by average amount wagered

Example 1: Horn Bet in craps

First find the expected loss.
Horn bet pays 30:1 if the next roll produces a 11 and 15:1 if it produces craps (2,3,12).
But if you win one you lose the others because you split the wagers between the 4 numbers. If we let x be the amount we are betting on each number, then we have a net payout of 30x-3x for the 11 and 15x-3x for craps. So, 27x for 11 and 12x for craps. How many times will we be paid each? Well, there are 36 total combinations of the dice and since we are betting on the 11 (5-6 or 6-5) and craps (1-1, 2-1, 1-2, 6-6), we have a 6/36 chance of winning. Meaning we will win 16.67% of the time. Part of the 16.67% winnings will be paid at 27x and part will be paid at 12x. Of our 6/36 chances of winning, craps makes up 4 of these chances and 11 makes up the other 2. So we will be paid 12x 4/6 times (of the 6/36) and be paid 27x the other 2/6 times. Supposing x = $1, then our 11 will yield (2/6 * 6/36) * (27 * $1)= 1.5012 and (4/6 * 6/36) * (12 * $1) = 1.3332. Adding the two winnings together, we know that 6/36 times we will have a return of 2.8344.

Now if we win 6/36 times that means we are losing 30/36 times. So our losses will be 30/36 * 4x, because for this case we lose the wager we had on each 2,3,11, and 12. Again, with x = $1 this comes to 30/36 * (4 * $1) = 3.3332. Subtracting our winnings from our losses we have 3.3332 - 2.8344 = a expected loss of .4988.

So what about the average bet? The average bet for this example is simple. Every wager was $1 on 2,3,11 and 12. So average wager is $4 since we bet $4 every time. Dividing Expected Loss by Average Wager Amount gives us .4988/4 = 12.47%

Example 2: Insurance bet in blackjack

I beat this into the ground in the blackjack section, but, as Letterman always says, "Anything worth doing is worth over-doing". For this example I have to use a very generic situation where I do not consider the fluctuating composition of the decks or the number of decks etc. (although, logically, I would think that this would work out to be the average case). An insurance bet is a wager that the dealer has a 10 in the hole on an ace and pays 2:1 if he/she does. There are 4 ten value cards for every 13 cards in a suit. Therefore we will win this bet 4/13 times and it will pay us 2:1. So, with x as the amount wagered, we will get 4/13 * 2x. If x = $2 then this comes to 1.23076. The other 9/13 times we lose the $2; so, 9/13 * 2 = 1.3846. Our expected loss will be 1.3846-1.23076 = .15384. Our average bet is $2 and therefore, the house edge on insurance is .15384/2 = 7.692%.

Example 3: Big 6 Wheel on the $1 spot

Example 4: Single Zero Roulette Bet on Red/Black