Gaming Guru

Interpreting House Edge, Part I

In general terms the house edge is a statistic that tells you how much of the money that you wager is going to be retained by the casino. There are some technical issues that mathematicians love to squabble about, that is, how should one handle ties and should the figure be based on the ante or the actual amount wagered. These issues are not what this article is about so I am going to avoid these issues by considering wagers that don't have ties and where the initial amount wagered is the total amount wagered.

From the player's point of view the house edge is a statistic that ostensibly tells the player which games are better. If game A has a lower house edge than game B then game A is the better of the two. Not always. I want to consider two wagers at craps: The place bet on 5 and the field bet with a 3:1 payout on the 12.

Let's look at the field bet first. We can make a table to calculate the house edge as follows.

There you have it; the house edge for the field bet is 1/36 or 2.778%.

If we place the 5 then the only numbers that matter are the 5 and 7. Payoff for the 5 is 7 to 5 and the 7 loses. There are six ways to make the 7 and four ways to make the 5 so there are ten outcomes that determine this wager. Here is the table.

So the house edge for placing the 5 is 1/25 or 4.00%. So it appears that playing the field is a better bet than placing the 5. It isn't so.

I will need a small piece of mathematics for what follows. Suppose that we have a series of Bernoulli trials. This means that we have a series of independent trials with two outcomes at each trial, one favorable and one unfavorable. If the probability of a favorable outcome is p and an unfavorable outcome is q at each step, what is the expected number of steps to reach a favorable outcome? Let's call this number E. Then at the first step we reach a favorable outcome with probability p or an unfavorable outcome with probability q. Before the first step the expected number of steps to reach a favorable outcome is E. This number is either 1 with probability p or 1 + E with probability q. Why 1 + E? After the first unfavorable outcome the expected number of remaining steps is still E so the total number of steps from the start is 1 + E. Thus we have the relationship

E = p + q(1 + E)

and since q = 1 - p we have

E = p + (1 - p)(1 + E)

Solving this last expression for E we obtain

E = 1/p

If we consider a favorable outcome as rolling a 5 or a 7, then the probability of resolving a place 5 wager is 10/36. The expected number of rolls to resolve the place 5 wager is therefore 36/10.

Let us suppose that in one hour of play that 108 rolls occur. I use this figure because it makes the arithmetic cleaner and it is probably in the right ball park. Let us assume that we are betting $5 on the place 5 wager. Since this wager has a 4% house edge we would expect to lose 0.04 x 5 or 20 cents on each resolution on average. If we divide 108 by the expected number of rolls to resolution we will have the expected number of resolutions per hour. This figure is 30. Thus we would expect to lose, on average, 30 x 0.20 or $6 per hour.

Now let's look at the field bet. If we bet $5 on the field on each of the 108 rolls we are wagering a total of $540. Our expected loss here is .02778 x 540 or about $15. So which wager looks better to you?

Before I leave you this month I want to clear up something that occurred in my February 6th article on Three Card Poker Strategy. In that article I referred to hands of 12-6-3 and 12-6-4. Several readers wrote to me and wondered what that 12 was? Good question. When I write poker programs I use the integers 2 through 14 to encode the cards so the 12 represents the Queen. I should have written Q-6-3 and Q-6-4. I am so used to thinking in terms of my code that I simply let it slip by me and into the article. Sorry if it confused you; I should be more careful. See you next month.

Don Catlin can be reached at 7121cat@comcast.net