New bet in Detroit

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 p^kq since these are all independent events. Consider the sum

S_k = 1 + p + p² + p³ + ... + p^k (1)

It is well known that this sum can be written in closed form as

S_k = ( 1 - p^{k + 1})/(1 – p) = ( 1 - p^{k + 1})/q (2)

It follows that if we define P_k = qS_k then we have the formula

P_k = 1 – p^{k + 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 711cat@comcast.net