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Best of Donald Catlin
New bet in Detroit1 July 2007
Thanks for your letter Nick. The good news is that your simulation is producing numbers that are in the right ball park. Let me show you how to do an exact analysis of this wager. Since, as you correctly point out, wagering more than $1 on this bet produces a progressively worse game for the player, I will use the $1 wager in my calculations. You can easily adapt my analysis to larger wagers.
To begin with, the difficulty in doing an exact analysis is not so much in the conceptual procedures but rather in the numerical accuracy. The probabilities for the larger payoffs are so small that one is in danger of losing accuracy due to truncation and round off errors. Therefore we want to carry as many places as possible in our decimals and minimize the number of operations we perform.
To achieve the latter I will proceed as follows. Suppose that p represents the probability of making a pass (here defined as rolling 7 or 11 on the comeout or rolling a point and making it). Then q = 1 – p is the probability of a seven out.
If we make k passes and then seven out the probability of doing so is pkq since these are all independent events. Consider the sum
Sk = 1 + p + p2 + p3 + ... + pk (1)
It is well known that this sum can be written in closed form as
Sk = ( 1 - pk + 1)/(1 – p) = ( 1 - pk + 1)/q (2)
It follows that if we define Pk = qSk then we have the formula
Pk = 1 – pk + 1 (3)
So why bother with this? Consider the quantity P4. This is the same as setting k equal to 4 in (1) and multiplying the result through by q. This is the probability that the shooter sevens out on the first roll or makes one pass and then sevens out or makes two passes and then sevens out or makes three passes and then sevens out or makes four passes and then sevens out. In other words this is the probability that the shooter makes four or fewer passes before sevening out. Note that with (3) this is just 1 – p5. Similarly the probability that the shooter makes 5 or 6 passes and then sevens out is just P6 – P4. This trick can be used for all of the classes of events in this wager. Finally, we want to obtain the probability that the shooter makes 20 or more passes. This is just the probability that the player does not seven out in 19 or fewer passes and is 1 – P19. But if we solve (3) for this quantity the answer is simply p20. Neat!
Now let's determine p. To this end refer to the archives on this site and look up my September 2003 article entitled The Pass Line. There you'll find that in 1980 rolls there are 976 winners and 1004 losers. However, for the wager we are analyzing the 220 craps losers don't count so there are only 1760 relevant outcomes and of these only 784 are losers (miss outs). Anyway, from these numbers we can easily determine that p is 976/1760 or 61/110. Using this figure and the techniques above we can construct the following table.
There you have it Nick; the house edge in this game is a whopping 27.189% just as you figured. See you all next month.
Don Catlin can be reached at firstname.lastname@example.org
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Best of Donald Catlin