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Best of Donald Catlin  New bet in Detroit

1 July 2007

I recently received a letter from Nick in Ann Arbor asking about a new bet at the MGM Grand in Detroit. Here is his letter:

First off let me state that I enjoy reading your articles at scoblete.com.

Now on to the question. MGM Grand in Detroit has created a new wager to compete with the "fire bet" found at a competing casino.

Here is the bet in a nutshell. Prior to a new shooter's first roll of the dice, you can make a "challenge" wager between \$1 and \$100. The "challenge" is the number of pass line winners that will be thrown by the shooter before the eventual seven-out. For this wager 7s and 11s count as winners on the come-out (craps do not count against this bet).

The pay table is as follows:

 Passes Payout 5 or 6 5 to 1 7 or 8 10 to 1 9 or 10 25 to 1 11 or 12 50 to 1 13 or 14 100 to 1 15 or 16 1,000 to 1 17, 18, or 19 2,000 to 1 20 or more 5,000 to 1

The maximum payout is \$5,000 per player.

I've been running some computer simulations to get an estimate on the house edge for this bet. Currently a \$1 wager is approx. -27.34% (40 million trials), a \$5 wager is approx. -34.18% (40 million trials), and a \$100 wager is approx. -49.42% (10 million trials). Obviously the edge itself is a function of the money wagered on the bet.

How does one mathematically compute the house edge on a bet that seems to get progressively worse the more money you wager on it?

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.

 Passes Probability Payoff Product 0 thru 4 0.947557214 -1 -0.9475572 5 or 6 0.036315546 5 0.1815773 7 or 8 0.011167780 10 0.1116778 9 or 10 0.003434323 25 0.0858508 11 or 12 0.001056125 50 0.0528062 13 or 14 0.000324780 100 0.0324780 15 or 16 0.000099877 1000 0.0998770 17, 18, or 19 0.000036790 2000 0.0735800 20 or more 0.000007565 5000 0.0378250 Totals - 1.000000000 --- -0.2718851

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 711cat@comcast.net

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Best of Donald Catlin
Donald Catlin Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers