At Risk or In Play?

I recently received the following e-mail from one of my readers:

Hi Don,

I've been reading about the "doey-don't" system of placing both a come and don't come (or pass and don't pass) wager. The books claim that this reduces the house edge on the standard line bet. I think 0.92% is cited. The pitfall is 12, which loses on the come but only pushes on the don't come. In all other cases, the wagers cancel each other out. So isn't the house edge 1/36 (or) 2.78%, which is greater than the familiar 1.41% associated with pass?

If you post this please post it under anonymous. Thanks.

Well, Mr. Anonymous, your letter is interesting and certainly raises some curious issues. The 2.78% figure does represent a house edge on a per dollar risked basis but does raise some questions. Why? Well think about it. You are making two wagers each having a house edge of 1.41% and 1.364% respectively (or 1.41% and 1.4026%; I'll discuss this difference below) so wouldn't you expect that by making both bets you would end up with a figure somewhere in between?

To begin with, before reading this article I suggest that you go into the archives, dig out my November 2003 article on Laying the Odds and read it. There I explain how one arrives at the 1.364% figure and also the 1.4026% figure. The difference lies in how ties are interpreted; 12 on the Don't Pass wager is a push. If one thinks of the occurrence of 12 as the end of the game and the wager is removed this leads to the 1.364% figure. On the other hand if the wager remains then the 12 can be interpreted as a non event (like rolling a 5 when the point is 6) and this leads to the 1.4026% figure. These two interpretations will play a role in our analysis of the Doey-Don't system.

What about that 2.78% figure? (Actually it's 2.7777… %) Well if you say that $1 is at risk each roll then in the 36 rolls per game (on average) you would lose 0.02777… x 36 or $1 per game. But there is another way to look at things.

Suppose that in 1980 rolls of the dice we place $1 on the Pass and $1 on the Don't Pass. Instead of using "at risk" as a barometer of the game we use "in play." We form the usual table for computing a house edge but use "in play" as a criterion.

If we consider all 3960 as being in play and we lose 55 then the house edge on a per dollar in play basis is 55/3960 or 1.389%.

Thus we have two figures for house edge and each has a reasonable interpretation. If we figure the $1 is at risk at each roll then every 36 rolls (on average) we have risked $36 and have lost $1. The house edge is then 1/36 or approximately 2.78%. On the other hand we can expect to lose 1.389% of all money that is placed on the table. I actually like the latter figure since, unlike the 2.78% figure, it does not suggest that the Doey-Don't wager is inferior to either of the two wagers that comprise it.

As I pointed out in my Laying the Odds article, the issue of handling ties is an old and contentious issue. Here is an example of why. Suppose that we consider ties as non events. Then the only event is the 12. If we consider $1 at risk then in our 1980 rolls only 55 is at risk and we lose 55 when the bet is settled. The house edge is 100%. Or if we consider that $2 is in play then 110 is in play in our table and the house edge is 55/110 or 50%. Silly? Yes, but correct. When the 12 is rolled (the only event considered in this formulation of the game) we lose our $1 at risk with 100% certainty or 50% of our $2 in play. So these figures are correct but certainly not very enlightening.

As I have said many times in my articles (see May 2009 for example), when you are combining two or more wagers into a combined wager it is important to properly define the precise game your combined wager represents when calculating a house edge. See you next month.

Don Catlin can be reached at 711cat@comcast.net