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Mensa Mystery4 May 2003
One of this site's readers, Ray Foley, recently wrote to Frank Scoblete with a question about a combination bet at craps. Frank forwarded Ray's letter to me. I wrote to Ray and asked him if I could quote his letter in my article and he said that would be fine. Here it is:
I have a question for you on something I read in American Mensa Guide to Casino Gambling. Specifically, the author states that if you place the 5 for $5 (4% edge), the 6 & 8 for $6 (1.52%) and the field for $5 (2.78% assuming 3x on 12), that the house edge is 1.136%. I have read many books on craps, including yours, and they all state without exception that you cannot hedge a low house edge bet with a high house edge bet. Furthermore, they state that the house edge of the combination of bets should fall somewhere in the middle of all of the bets. I have worked out the odds for the author's claim, and I keep coming up with the same percentage, even though I intuitively know that it can't be right. The combination of all of these bets is lower than the lowest bet by itself? I have attached my calculations on an Excel spreadsheet. Can you please let me know if you agree with them? I really hope you disagree, and have a valid explanation, otherwise all of my beliefs about probability are shot. Additionally, I realize that with this strategy you will have money at risk on every roll, which leads to more exposure, and most likely a higher monetary loss.
Thanks in advance,
Here are the calculations that Ray sent. I am going to compress
them a bit to save space but this table accurately reflects what he sent
To explain the table look at the row for rolling the 8. There are 5 ways to roll an 8 and on each of these there is $22 at risk so the total at risk is $110. The player wins $7 each on his Place 8 bet (total returned 65) and loses $5 each on his Field bet (total returned 0). He neither loses nor wins on the other two wagers (total returned 25 & 30, respectively) so the return to him is 25 + 30 + 65 or 120. Other rows are similar. The difference between the total at risk and the total returned is 9. 9/792 expressed as a percentage is 1.136%. So there you have it, the same number that appears in the book Ray cited.
Well, Ray, you certainly got my attention. This is an interesting and thought provoking calculation. Likewise, the remarks in your letter certainly express your skepticism and concern about the calculation. I agree. I would put it this way: It doesn't pass the "smell test."
I looked up the passage you referred to in the American Mensa Guide to Casino Gambling and sure enough it is there on pages 139-140. This guide was written by Andrew Brisman, who is a gambling writer and (obviously) author. In fairness to Mr. Brisman, this particular wager is listed as "Faulty Craps System." He doesn't recommend it nor is it listed in his Appendix of casino wagers and their edges. On the other hand, I read it very carefully and he does say that the house edge for the bet is 1.136%, he provides a table with the calculation (somewhat different from the table above), and nowhere did I see any mention of the fact that the calculation doesn't make sense. Well, actually one can make sense out of it but not with respect to any craps game that I ever played. More on that later.
Ray, to help straighten out this mystery, I want to back up a bit and look at a single wager, namely placing the 5. Let's see how the house edge is usually calculated. There are ten ways to settle this bet, four 5s and six 7s. The probability of winning this wager is, therefore, 4/10 or 2/5 and the probability of losing is 6/10 or 3/5. The expected return per game is, therefore,
exp = (2/5) x 7 + (3/5) x (-5) = - 1/5 (1)
There is $5 at risk so the expected return per dollar risked is -1/5 divided by 5 or - 1/25. Expressed as a percentage this is 4%, the usual figure stated for this wager.
Let's try to calculate this wager using the Mensa method. Here's
So the money at risk is 180 and the return is 178 for a loss of $2. 2/180 represents a 1.111...% house edge. Hmmm! Not 4%, is it? Here's the problem.
This calculation represents a game in which you would place the 5, then every roll except 5 and 7 you would say to the dealer "Take down my bet" and as soon as he does you would chalk that up as a return and say to him "Put it back up again" and consider it a new wager. If you kept score this way you would indeed lose only 1.111% of all such wagers you made in this fashion. You would also probably be kicked off the table for harassing the dealer.
So what is the correct way to do the calculation? Sometimes the
words we use influence the way we do things. In particular the words
"At Risk" are misleading. Certainly any wager on a craps table is at
risk on every roll but that doesn't mean that it is a new wager on each
roll. I would prefer to use the words "In Play." Specifically, I mean
that the wager is in play only on those rolls where a decision occurs.
Here is what my table for placing the 5 would look like.
So we have 50 in play and get back 48 for a loss of 2. If we divide 2
by 50 we get 1/25 or 4% as the house edge per dollar wagered. Now let's
try the combination bet while handling the money in play in this same
So 362 was in play and 353 was returned for a loss of 9. 9/362 represents a house edge of approximately 2.4862% which is how I would rate this wager.
Now I think I can explain my earlier remark about making sense of the 1.136% figure. If on every roll you told the dealer to remove the place wagers that were not resolved, count them as a return to the player, then tell him to immediately put them back up along with a new field bet and wager on whatever place bet, if any, was resolved, count all these as new bets, then it is true that you would only lose 1.136% of the total wagers so counted. This time, however, you would definitely be kicked off the table if not kicked out of the casino.
By the way, Ray, the table in the Mensa book uses probabilities to calculate the expected profit per game (-1/4) and then divides this figure by 22. The correct figure to divide by is not 22 but the average money in play which is 10.0555... . Try it. I prefer the above calculations because I think they make the interpretations clearer.
Stating house edges in terms of per dollar risked is a useful but tricky business. In the book Finding the Edge edited by Vancura, Cornelius, and Eadington, published by the Institute for the Study of Gambling and Commercial Gaming, University of Nevada, Reno, I have an article entitled Using Overall Expected Return per Dollar Risked to Determine Strategy Decisions in Gambling Games. In it I discuss some of the pitfalls in using this as a statistic for evaluating games. The points made are not directly related to the above discussion, but if you found the above discussion interesting I think you will find my article interesting as well. It contains some seemingly paradoxical examples that should pique your interest.
Ray, thanks for your interesting letter. Because of your healthy skepticism I'll bet a few readers learned a some things. See you all next month.
This article is provided by the Frank Scoblete Network. Melissa A. Kaplan is the network's managing editor. If you would like to use this article on your website, please contact Casino City Press, the exclusive web syndication outlet for the Frank Scoblete Network. To contact Frank, please e-mail him at email@example.com.
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