Counting the Pass Line And Cows

22 September 2012

By Donald Catlin

I recently received a letter from Peter Chao, one of my readers. Here is part of Peter’s letter:

“Instead, could we say [in craps] that there are two separate bets based on an event (setting the point): Pre-Point Bet of $10 and Post-Point of $40-$60 (depending on point)? Both have a Pass Line component, and Post-Point adds the odds.

“The reason I would put forth this assertion is that these games do not end at the same time nor are they even played at the same time. Pre-Point Bet (33.3% player advantage by my undergraduate level calculations) ends once a point is set with a “push.” Post Point bet (-3.64% house advantage … same calculations) begins with a point and ends once a point is hit or seven out. Assuming my calculation of Post Point house edge is correct; wouldn’t we converge to -3.64% house edge (I assume the point is set on the first come out roll)?”

Well Peter, first of all let me thank you for writing to me. It is always interesting to hear what players are musing about. I am not sure how the word “converge” in your last sentence is used but from what follows I’m not sure that it matters

Peter you certainly have a knack for taking a simple calculation and making it challenging. It reminds me of that old joke about counting cows by counting their legs and then dividing by four.

Let me illustrate this to my readers by taking Peter’s ideas and using them to calculate the house edge for the Pass Line. In deference to my readers I am not going to use any odds bets in the calculation; I’ll address this later.

Let me define two games using dice. Both games use a unit bet. Game One uses the numbers 2, 3, 7, 11, 12 using the usual payoffs for a craps come-out roll. Game One ends when any of the other numbers are rolled. This game has a player advantage of 1/3 percent. (I will use fractions to keep things precise. This will be important as the calculation proceeds.)

Game Two is played with the numbers 4, 5, 6, 7, 8, 9, 10; all other numbers are ignored. In what I’ll call the come out roll if any number other than 7 is rolled this number becomes the point, as in craps, and must be re-rolled before 7 in order to win, a 7 loses.

Let us calculate the expected return for Game Two. If 4 is the point then the expected return, which I’ll write as E(4) is (3/9)x1 + (6/9) x (-1) or -1/3. Similarly E(5) is (4/10) x 1 + (6/10) x(-1) or -1/5. I’ll leave it for you to calculate E(6) as -1/11. Now all of these are conditional expectations. To get the overall expectation for Game Two we have to multiply each of these numbers by the probability that that point is established and then add them up. If P(k) is the probability of k being established then since there are 24 ways to roll a point P(4) is 3/24, P(5) is 4/24 and P(6) is 5/24. Thus the overall expectation for Game 2 is:

E = 2[(-1/3) x (3/24) + (-1/5) x (4/24) + (-1/11) x (5/24)]

where the multiplier 2 is present because the calculations for 8, 9 and 10 are identical to those for 4, 5, and 6. If you add up these and clean up the resulting fraction the answer is -124/660. Incidentally, this number expressed as a decimal is approximately 18.79%.

We now see that a craps Pass Line game can be described as follows. The player plays Game One until a point is rolled. Then the player plays Game Two and considers the last roll of Game One to be the come-out roll of Game Two. Game Two is played to the end.

What is the overall expected return for the Pass Line at craps? We have to determine how frequently each game is played. This is easy because 1/3 of the time we play Game One and 2/3 of the time we play Game Two. Hence the overall return for the Pass Line is

Return = (1/3) x (1/3) + (2/3) x (-124/660) = -14/990

14/990 expressed as a decimal is -0.014141414… which is the usual house edge of 1.41414141…%.

What about odds? Using calculations similar to those we just did all we have to do is calculate the expected bet for the game For Game One this will be 1 and occurs 1/3 of the time; for Game Two it will be a weighted average of the odds placed on the various points and occurs 2/3 of the time. Whatever the result the expected return will be -14/990 divided by the expected bet.

Peter I hope this helps. Obviously I don’t recommend analyzing things as is done above. I would like you to check out my September, October, and November 2003 articles in the archives to see the various house edge interpretations for the Pass and Don’t Pass lines in craps. See you in a couple of months.

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