There are many things to consider when you decide to take on the subject – craps odds. The experts tend to agree…well, most of them tend to agree, you must first understand craps odds, in order to be knowledgeably equipped to play the game of craps.

In fact, some will stress that you must know the odds before you make a bet, in order to know which bets give the house (casino) a smaller edge over you.

Why does the house edge matter? One can argue that the game of craps cannot be beaten. When considering craps odds, there is mathematical evidence to back this statement. This being true, doesn’t it make good sense to decrease the advantage of the house, thereby hoping to decrease the amount you will ultimately lose?

There is a chance that you may be thinking – Craps cannot be beaten? Heck, I’ve walked away a winner before, so that’s not true. This argument, when not taking craps odds and the house edge into consideration, can hold water under certain conditions.

However, when considering craps odds, the thinking is not that a particular session or series of rolls cannot be beaten. The idea is that craps odds and the house edge are designed to ensure the house cannot be beaten over a long period of time.

Let’s examine this for a moment.

We can begin to understand craps odds by taking a look at the probability (chance, or odds) of rolling a particular number. The first thing for you to do is calculate the number of combinations possible using a pair of dice.

You can see that there are six sides to one die. Each side represents a specific number. The numbers are – 1, 2, 3, 4, 5, and 6.

There are two dice, so you multiply six times six to determine the number of combinations possible. In this case, the number is 36 (6 x 6 = 36).

Next, treating each die separately (die A on the left, and die B on the right), determine how many ways you can roll each of the following numbers – 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.

Here are the results – 2 (1 way), 3 (2 ways), 4 (3 ways), 5 (4 ways), 6 (5 ways), 7 (6 ways), 8 (5 ways), 9 (4 ways), 10 (3 ways), 11 (2 ways), 12 (1 way).

Now, you calculate the probability by dividing the number of ways to roll a number by the number of combinations possible using a pair of dice (36). For example, there is one way to roll the number 2, so you have a 1 in 36 chance of rolling a two. The probability is 1/36 or 2.78%.

Here are the probabilities of rolling each number – 2 (1/36, 2.78%), 3 (2/36, 5.56%), 4 (3/36, 8.33%), 5 (4/36, 11.11%), 6 (5/36, 13.89%), 7 (6/36, 16.67%), 8 (5/36, 13.89%), 9 (4/36, 11.11%), 10 (3/36, 8.33%), 11 (2/36, 5.56%), 12 (1/36, 2.78%).

The probabilities above show what is probable or likely to occur on each independent roll of the dice. Independent because whatever the outcome of the next roll of the dice, it is not dependent on, or affected by previous rolls of the dice.

You may have heard the saying – dice have no memory – well, considering the fact that they are objects without the capacity to think or run calculations, in other words, dice do not have a brain – it is safe to say that dice cannot remember anything, so previous rolls are irrelevant.

Using the same argument, you can say that dice do not know the probabilities, so they are not influenced by probabilities. But, if that is true, couldn’t you also say that dice do not know craps odds, so they cannot be influenced by craps odds? Ooops! Don’t answer that just yet.

Now that you know the probabilities, your next step is to understand how this relates to craps odds.

First off, you cannot establish true craps odds without knowing the probability of rolling a specific number. One definition of odds, according to Merriam-Webster’s Online Dictionary, is as follows — the ratio of the probability of one event to that of an alternative event.

In other words, you need to know the probability of rolling a number in a specific situation, in order to determine the true craps odds.

Here is a simple formula for true craps odds on rolling any number before a 7 on the next roll: P7 divided by PN = true craps odds. The letter P stands for probability, and the letter N stands for the number to roll before seven.

Using this formula you can calculate the true craps odds of rolling a 2 before the 7. P7/P2 = true craps odds, so 16.67% (.1667)/2.78% (.0278) = 6.00. The true craps odds of rolling a 2 before the 7 — is 6 to 1.

This same concept, not necessarily the same formula, is used to mathematically determine the true craps odds of all the bets in the game of craps. However, the house edge is calculated to favor the house, and this is what gives the house the advantage.

For example, the true craps odds of rolling a 6 before a 7 is – P7/P6 =.1667/.1389 = 1.2, or 6/5, or 6 to 5, or 6:5. However, the house pays 7:6 (7 to 6) when you make a place bet on the number 6. The difference between the true craps odds of 6:5 and the actual payout of 7:6 is the house edge, which is 1.52%.

With this in mind, what happens if you bet $12 to place the 6 (make a bet that the 6 shows before a 7), and the shooter rolls a 6?

The true craps odds would be a payout of 6:5 or 6 dollars profit for every 5 dollars you bet, which is about $14.40 profit. However, the house pays you 7:6, instead of the true craps odds, so you only get $14 profit…the difference being 40 cents.

Does this mean you lost $.40? Hmmm…You put $12 on the table, won $14 profit, plus you get to keep your $12 bet…would you feel like you lost money at this point?

Do you think the dice know just how much the house edge cost you?

Okay, that’s quite a bit to think about, so let’s dig a little deeper.

You know that the number 6 will be rolled five times in 36 rolls…in theory. You also know that the number 7 will be rolled six times in 36 rolls…in theory.

Let’s alternate the 6 and 7 such that 6 is rolled before 7, then 7 is rolled before 6. Further, let’s do this to reflect the theory that 6 will be rolled five times and 7 will be rolled 6 times. Additionally, we will make a $12 place bet on 6 for each time we alternate the 6 and 7.

By the way, this will represent a total of eleven bets. Five of the bets will be a win for 6, and six of the bets will be a loss due to the 7. This will make more sense as the example progresses.

You start with a $12 place bet on 6 and it wins. This gives you a profit of $14.

Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled before a 6. You lose the $12 place bet, and now have a total profit of $2 ($14 previous profit minus the $12 loss).

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $16 (the previous total profit of $2 plus the $14 profit on this bet).

Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $4 ($16 previous profit minus the $12 loss).

So far you have rolled 6 twice and 7 twice.

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $18 (the previous total profit of $4 plus the $14 profit on this bet).

Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $6 ($18 previous profit minus the $12 loss).

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $20 (the previous total profit of $6 plus the $14 profit on this bet).

Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $8 ($20 previous profit minus the $12 loss).

You have rolled 6 a total of four times and 7 a total of four times. This means you have one more roll of 6 and two more rolls of 7 to go.

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $22 (the previous total profit of $8 plus the $14 profit on this bet).

Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $10 ($22 previous profit minus the $12 loss).

Since you have exhausted the rolls of 6 in our hypothetical scenario, you still have one more roll of 7 to go. This means making one more place bet on 6.

You make your final $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of -$2 ($10 previous profit minus the $12 loss).

Based on the information above, if your bankroll was only the $12 you began with, you just lost 17% of your bankroll. If your bankroll was $100, you just lost 2% of your bankroll.

Here is the real question — Was the loss due to the probability of rolling 6 before 7, or due to the house edge?

By checking out the same scenario, using the true craps odds, we can get a better idea of the impact of the house edge.

You start with a $12 place bet on 6 and it wins. This gives you a profit of $14.40.

Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled before a 6. You lose the $12 place bet, and now have a total profit of $2.40 ($14.40 previous profit minus the $12 loss).

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $16.80 (the previous total profit of $2.40 plus the $14.40 profit on this bet).

Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $4.80 ($16.80 previous profit minus the $12 loss).

So far you have rolled 6 twice and 7 twice.

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $19.20 (the previous total profit of $4.80 plus the $14.40 profit on this bet).

Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $7.20 ($19.20 previous profit minus the $12 loss).

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $21.60 (the previous total profit of $7.20 plus the $14.40 profit on this bet).

Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $9.60 ($21.60 previous profit minus the $12 loss).

You have rolled 6 a total of four times and 7 a total of four times. This means you have one more roll of 6 and two more rolls of 7 to go.

Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $24 (the previous total profit of $9.60 plus the $14.40 profit on this bet).

Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $12 ($24 previous profit minus the $12 loss).

Since you have exhausted the rolls of 6 in our hypothetical scenario, you still have one more roll of 7 to go. This means making one more place bet on 6.

You make your final $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $0 ($12 previous profit minus the $12 loss).

Based on the information above, if your bankroll was only the $12 you began with, you just broke even. If your bankroll was $100, you just broke even.

By examining the two hypothetical scenarios above, it should be plain to see that the house edge is not solely responsible for your losses.

The probability of making a number before 7, and the house edge combined, led to the loss. What would have happened if we disregarded the probabilities, and rolled 6 and 7 five times each?

Looking at the first scenario, with the house edge factored in, you would be ahead, with a profit of $10. Looking at the second scenario, with the true craps odds factored in, you would be ahead, with a profit of $12.

What does this mean? Craps odds are not solely responsible for the long term loss expected in the game of craps.

It takes a combination of the probabilities (the number combinations that will be produced over the long run), plus the odds (actual payouts that factor in the house edge), and in certain cases, the rules of the game (for example, the rule that bars 12 on the come out roll when betting Don’t Pass).

Does this mean that you can make a profit in the short run? Yes! How do you determine what the long run is?

Great question! Maybe you should ask the dice.;-)

Copyright © 2025 by Christian Saul Christian A Saul is the author of a new craps strategy called The Craps Annihilator Find out how you can “make winning a habit.” Be sure to visit his Web site for more information on craps strategy, craps systems and how to play craps. This article is available for reprint in your ezine, your website, or blog, as long as no portion of the article is modified, and this resource box/signature line, with clickable links above are included. Article Source: #