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Take Down the Eight on the Comeout?14 August 2009
I recently received the following letter from one of my readers.
Well, Fred, I'm really happy with your inquiry but I'm not sure you will be happy with my reply. To begin with let me disabuse you of the idea that the total amount at risk is $1. Yes, if the seven is rolled on the comeout you will only lose $1. But suppose you roll a 6 followed by a 7; you would lose $11 so all $11 is at risk.
As I mentioned in my May 2, 2009 article that you reference in the Subject bar in your email (Interpreting House Edge, Part II), when you try to combine two separate wagers into a single "game" you have to be very careful how you define your game. The resulting house edge might not reflect the reality of the situation.
I can think of two reasonable ways to define the game here. One way would be to play until both bets are resolved but not reinstate either bet until both bets are resolved. This leads to lengthy calculations. Let me give you an example. Suppose a 6 is rolled on the comeout. The probability of this is 5/36. Then there are 16 dice totals that affect the rest of the game, five 6s, five 8s, and six 7s. The possible events, probabilities, and payoffs are:
The reason for the 11 in the denominator of the third fraction is that once a point, say 6, is rolled then only the five 8s and six 7s are relevant to settling the bet.
One would have to make similar calculations for each of the points 4, 5, 9, 10, as well. Also one has the craps, 11, 7, and 8 on the comeout to analyze. That's a lot of calculating (32 separate events) and for a game that is rather contrived. Thus, I just didn't do it.
The second game I envision is one in which one keeps reinstating whichever wager is settled and continue playing. I believe this is a more realistic scenario. Similar to what I did in my May article I will play this game for an hour and see what the results are. From my May article we already know that the placed 8 will lose on average $3. What about the pass line wager?
To answer this we need a figure that I calculated in my January 5, 2000 article (How Long is a Craps Roll?). Specifically I showed there that the average number of rolls per pass line decision is the number 557/165. Assuming 108 rolls per hour the number of decisions per hour on the pass line wager is 108 x 165/557 or approximately 31.993 decisions/hr. From my September 1, 2003 article (The Pass Line) the exact house edge for the pass line is 28/1980. Multiplying the number of decisions per hour by 5 we obtain 159.965, which represents the total action for the hour's play. Multiplying this by the house edge give us approximately $2.26 loss. Thus our total loss on the two wagers is, on average, $5.26.
What is at risk? Well, each decision represents money at risk. The figure we calculated above, 159.965, represents the amount at risk on the pass line. From my May article there are, on average 33, decisions per hour for the placed 8 so the amount at risk there is 6 x 33 or 198. Rounding to the nearest penny we then have the total at risk $357.97. Our overall house edge is 5.26/357.97 or approximately 1.469%. Note that this answer is reasonable since it is in between the house edge for the pass line and the house edge for the placed eight.
Well, Fred, I hope this answers part of your questions. What would I do when faced with a placed 8 on the comeout? I would take down the 8 and put my $6 along with my $5 on the pass line. Rather than having $11 riding on a wager with a house edge of 1.469% I would now have $11 riding on a wager with a house edge of 1.414%. You see, rather than looking at the pass line wager as a hedge for the placed 8, as you do in the last line of your email, I look at the placed 8 as increasing the house edge on my pass line bet. Thus I could make my decision to move the $6 to the pass line without any of the above calculations. They were fun though!
See you next month.
Don Catlin can be reached at firstname.lastname@example.org
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