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9 July 2016
By Jerry Stickman
I shoot craps quite a lot, but I tend to be “mathematically challenged” or so it seems!
I understand that on average you will roll a 7 more than any other number. I’ve seen the “list” posted many, many times. But, if you take, let’s say, the 6 and 8 "together" on the dice, when you actually roll those babies down the felt don’t you have a better chance of making one of those two numbers than a 7?
My feeble brain assumes if there are 5 ways to make a 6 and 5 ways to make an 8, why don’t you have 10 chances to make either of those numbers and only 6 chances to make a 7? If this is true it would therefore obviously give you the advantage.
I would love to hear an explanation I can understand!
Thanks. I enjoy your articles.
Your question is an excellent one as a great many supposedly knowledgeable players also get it wrong.
There are actually two pieces that make up the entire picture.
First of all – yes, it is true that you will throw a 6 or an 8 more times than you will throw a 7.
There are 36 possible combinations with a pair of six sided dice. With this being the case, on an average 36 throws of the dice that have a random distribution, you will throw five 6s and five 8s – a total of 10 6s and 8s combined.
Over the same average 36 rolls of the dice, you will throw six 7s. You throw 10 6s and 8s but only six 7s. It certainly seems like you should be able to win.
So why can you not make money by betting the 6 and 8? After all, you will hit one or the other more often that you hit the 7.
Well, as Paul Harvey used to say, “Now, the rest of the story.”
If you hit a 6, you win on the 6 only – just one number. You will have just one win of 7 to 6. When you hit the 8, you win on the 8 only – just one number. You will have just one win of 7 to 6. But when you throw a 7 you lose both the 6 and the 8. You lose a total of 12 – six on the 6 and six on the 8.
Let’s look at the totals. Out of the 36 possible combinations you will win 7 to 6 on 10 of the rolls (five 6s and five 8s) for a total 70 units. On the same 36 possible combinations you will lose 12 units six times for a total of 72 units lost. In the normal random distribution of dice throws you will win 70 units and lose 72 units. This is not a winning proposition.
Your logic would work only if you won on both the 6 and 8 if you hit one of them but, alas, that is not the way the game works. If it did, the casinos would not offer craps – or they would change how they paid the winning numbers.
I hope I was able to explain it in a manner that you can understand. If not, let me know and I will give it another shot.
Follow up from Holly
Thanks so much for your reply. I do understand your explanation, and it’s only true in the context of craps and payouts, etc. For example, if you and I were just sitting on the floor rolling the dice, the 6’s and 8’s – lumped together – would come up more than the 7. In other words, if I bet you I would roll more 6s and 8s combined than I would roll 7s, I would be correct, right? By George, I think I’ve got it!! (Unless you tell me I’m wrong again!) (LOL)
Yes, you are exactly correct with your “sitting on the floor” analogy.
May all your wins be swift and large and all your losses slow and tiny.
Jerry “Stickman” authored the video poker section of "Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker!" You can contact Jerry “Stickman” at firstname.lastname@example.org
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