The Basic Math of Craps: Required Understanding for Smart Players

To fully understand the game of craps, you must understand the basic math behind it.  Don’t worry, it’s not rocket science and you don’t need a Ph.D to understand it as you may need if you want to understand this.  In fact, it’s easy as pie.  If you want to be a solid player feared by the casino, then you must understand the relationship among the various values that can appear when you roll two dice.  Without that understanding, you’ll be lost at the table and you’ll look like a fool…and you’ll probably lose every time you play.  So, take the time to read this article.  If you have to read it twice to understand it, then read it twice.  It’s critical to understanding everything else about the game, so just read it, learn it, and practice it.  After a short while, it’ll be as simple as adding 2 + 2.  Afterwards, you can also read  about how to calculate the casino house advantage & about distribution variance “your our weapon against the casino”.

As you know, craps is played with two dice.  (By the way, the little dots on a die are called “pips.”  A pip is simply a dot that symbolizes numerical value.  You find pips on things like dice and dominoes.)  Thirty-six (36) combinations of numbers can be rolled with two dice.  Refer to the table below.

Combinations Using Two Dice

DIE #1DIE #22-DICE TotalDIE #1DIE #22-DICE Total
112415
123426
134437
145448
156459
1674610
213516
224527
235538
246549
2575510
2685611
314617
325628
336639
3476410
3586511
3696612

As you can see in the table above, the possible total values when using two dice range from 2 to 12.  It’s important to know, without even thinking about it, the number of ways to roll each number 2 through 12.  This basic information is used as the basis for determining many aspects of the game, such odds and house advantage.

The table below summarizes the number of ways to roll each number 2 through 12.  For example, how many ways are there to roll the number 12 using two dice?  In the table above, count the number of times the number 12 appears in the “2-Dice Total” columns.  You find that it appears only once; therefore, there’s only one way to roll a 12 using two dice.  The only way as 12 can be rolled is with a 6 on one die and a 6 on the other die.  Comparing the table above to the summary table below, you see that there’s one way to roll a 12 with two dice.

Let’s do another example.  How many ways are there to roll a 5 using two dice?  In the table above, count the number of times the number 5 appears in the “2-Dice Total” columns.  You’ll find that it appears four times; therefore, there are four ways to roll a 5 with two dice.  Comparing the table above to the summary table below, you see that there are four ways to roll a 5 with two dice.

Ways to Roll a Number With Two Dice

2-DICE VALUE# OF WAYS TO ROLL THE 2-DICE VALUE
21
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
121

The table below is a common format for displaying the number of ways to roll each number, and the possible 2-dice combinations for rolling each number.

Ways to Roll a Number and Their Combinations

2-DICE VALUE2-DICE COMBINATIONS FOR ROLLING THE NUMBER# OF WAYS TO ROLL THE NUMBER
2[1-1]1
3[1-2] [2-1]2
4[2-2] [1-3] [3-1]3
5[1-4] [4-1] [2-3] [3-2]4
6[3-3] [1-5] [5-1] [2-4] [4-2]5
7[1-6] [6-1] [2-5] [5-2] [3-4] [4-3]6
8[4-4] [2-6] [6-2] [3-5] [5-3]5
9[3-6] [6-3] [4-5] [5-4]4
10[5-5] [4-6] [6-4]3
11[5-6] [6-5]2
12[6-6]1

To ensure we’re reading the table above correctly, let’s look at an example.  Suppose we want to know how many ways there are to roll an 8 with two dice, and we want to know what two-dice combinations can result in an 8.  We see from the table above that there are five (5) ways to roll an 8, and the two-dice combinations are:

  • 4 on die #1, and 4 on die #2.
  • 2 on die #1, and 6 on die #2.
  • 6 on die #1, and 2 on die #2.
  • 3 on die #1, and 5 on die #2.
  • 5 on die #1, and 3 on die #2.

Memorize the summary table above.  Without having to think about it, know how many ways there are to roll each number 2 through 12.

There’s an easy trick to help memorize it.

As shown in the table above, there are six ways to roll a 7, five ways to roll a 6 or 8, 4 ways to roll 5 or 9, three ways to roll a 4 or 10, two ways to roll a 3 or 11, and one way to roll a 2 or 12.  Notice that all the numbers, except for the number 7, are paired based on how many ways to roll them.  Let’s construct a table to help us memorize the pairings.  We’ll build one column at a time so it’s easy to follow.  Let’s start by putting the pairings in the first column, which are based on the number of ways to roll each number in the pairing.

Ways to Roll a Number by Pairings

2-Dice Pairings  
7
6 or 8
5 or 9
4 or 10
3 or 11
2 or 12

Notice that the number 7 stands alone with six ways to roll it.  Also, notice that the first number of each pairing as we go down the column decreases by one (i.e., 6, 5, 4, 3, and 2).  And the second number of each pairing as we go down the column increases by one (i.e., 8, 9, 10, 11, and 12).  It’s important to realize that the number 7 is in the middle of the pack and stands alone.  That is, the number 7 lies directly between the group 6-5-4-3-2 and the group 8-9-10-11-12 (i.e., 2-3-4-5-6—-7—-8-9-10-11-12).

Now, notice that each pairing has a number lower than 7 and a number higher than 7.  The first pairing is 7 minus one (i.e., 6) and 7 plus one (i.e., 8).  So, the first pairing is “6 or 8.”  The second pairing is 7 minus two (i.e., 5) and 7 plus 2 (i.e., 9).  So, the second pairing is “5 or 9.”  Do this technique for the remaining pairings.  To help you memorize the pairings, just think, “One away from 7 in both directions is 6 and 8.  Two away from 7 in both directions is 5 and 9.  Three away from 7 in both directions is 4 and 10.  Four away from 7 in both directions is 3 and 11.  And five away from 7 in both directions is 2 and 12.

Do this technique in your mind a few times and it’ll be imprinted on your brain.  If the pairings do happen to slip your mind, just do the little drill described above and the pairings will quickly come back to you.

Now, we need to figure out how many ways to roll each number.  The little math trick continues.  All you have to do is subtract one (1) from the low number of each pairing.  Let’s fill in the middle column of the table.  Remember, take the low number from each paring and subtract one, as shown in the middle column below.  Although the number 7 stands alone, you still subtract one from it.

Ways to Roll a Number by Pairings

PairingsLow # of the Pairing - 1 
77 - 1 = 6
6 or 86 - 1 = 5
5 or 95 - 1 = 4
4 or 104 - 1 = 3
3 or 113 - 1 = 2
2 or 122 - 1 = 1

Very good!  See how easy this is?  You just calculated the number of ways to roll each number, 2 through 12, with two dice.  See the completed table below.

Ways to Roll a Number by Pairings

PairingsLow # of the Pairing - 1Ways to Roll the Number in Each Pairing
77 - 1 = 66
6 or 86 - 1 = 55
5 or 95 - 1 = 44
4 or 104 - 1 = 33
3 or 113 - 1 = 22
2 or 122 - 1 = 11

Let’s see if you’re paying attention.  Without looking at the table, how many ways are there to roll a 4?  If you can’t memorize it, then do the little math trick.  The pairing is “4 or 10,” and 4 is the low number of the pairing, so 4 – 1 = 3.  Therefore, there are three ways to roll a 4.

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Let’s do one more.  How many ways to roll a 9?  Very good!  You’re a natural at this!  The pairing is “5 or 9,” and 5 is the low number of the pairing, so 5 ‑ 1 = 4.  Therefore, there are four ways to roll a 9.

The basic data in the table above is very important, so memorize it.  If you need help, memorize the pairings and do the simple math trick described above to figure it out.  If you want to win at craps, then you must know and understand this basic relationship among the numbers.

You may be thinking, “Why is this so important?  I hate math.  I use a calculator to add 2 + 2.  Can’t I just play the game without knowing all this gobbledygook?”  No!  It’s easy to see that craps is a game of odds where possible winning two-dice combinations are compared to possible losing two-dice combinations.  Whether numbers are winners or losers depends entirely the type of bets that you make, which we’ll discuss in other articles.  A number may be a winner when making one type of bet, while the same number may be a loser when making a different bet.  Regardless, the game boils down to understanding the relationship among the numbers, particularly how the number 7 relates to all the other numbers.  If you don’t understand it, you’re going to make stupid mistakes and you’re going to lose.  If you want to be a rock-solid player and win consistently, then you must memorize it.

For example, suppose we want to compare the number 7 to the number 10.  Suppose we want to bet on the 10 appearing on a roll before a 7 appears.  Therefore, our bet wins if a 10 shows before a 7, and our bet loses if a 7 shows before a 10.  Assume that all other numbers don’t matter, so we ignore them and keep rolling until either a 10 shows (we win) or a 7 shows (we lose).  Let’s bet $1 and assume it’s an even‑money bet, which means if we lose, we lose the $1, and if we win, we win $1.  The odds for this even‑money bet are expressed as 1:1 (stated as “one to one”).  An even-money bet, or a 1:1 bet, means for each unit we bet and win, we win that exact amount.  For example, if we bet $5 and win, then we win $5; or if we bet $8 and win, then we win $8.  Is betting the number 10 against the 7 for even money a good bet?  No way!  It’s a terrible bet because we have twice as many chances of losing than winning.

From the table above, we see there are six ways to roll a 7, and only three ways to roll a 10.  That means there are twice as many ways for us to lose as there are for us to win.  So, making this even-money bet is not only terrible, it’s stupid.

But suppose 10 is our favorite number and we want to bet it, so is there any circumstance where betting the 10 against the 7 is a good bet?  Yes, of course.  When betting the 10 against the 7, we take a much greater risk because there are twice as many ways to lose as there are to win, so we want to be compensated for taking that risk.  We’re compensated by getting odds on the 10, which means if we bet $1 and win, then we expect to win more than just even money.  But how much more?  It’s simple, so don’t fear the math.  Again, there are six ways to roll a 7 and three ways to roll a 4.  The comparison of those outcomes is expressed as 6:3 (i.e., 6 ways to roll a 7 to 3 ways to roll a 10).  This expression is like a fraction, so we reduce the expression to 2:1.  Therefore, we expect to get 2:1 odds when betting the 10 against the 7.  If our $1 bet wins (i.e., a 10 shows before a 7), we expect to win two times our $1 bet, which is $2 (i.e., $1 x 2 = $2).  If, for example, we bet $3 on the 10 against the 7 and win, at 2:1 odds, we expect to win two times our $3 bet, which is $6 (i.e., $3 x 2 = $6).

To illustrate this further, let’s roll the dice 36 times and assume the results are distributed exactly according to the number of ways to roll each number (i.e., a perfect distribution).  After 36 rolls with a perfect distribution, we would see the following results:

  • On 6 of the rolls, a 7 will show.
  • On 5 of the rolls, a 6 will show.
  • On 5 of the rolls, an 8 will show.
  • On 4 of the rolls, a 5 will show.
  • On 4 of the rolls, a 9 will show.
  • On 3 of the rolls, a 4 will show.
  • On 3 of the rolls, a 10 will show.
  • On 2 of the rolls, a 3 will show.
  • On 2 of the rolls, an 11 will show.
  • On 1 roll, a 2 will appear.
  • And on one roll, a 12 will appear.

Note: The above outcome over 36 rolls occurs only with a perfect distribution, which is unlikely, but useful for illustrating how and why odds are important in the game of craps.

We know there are six ways to roll a 7 and three ways to roll a 10.  For a 1:1 even-money bet, if we bet $1 on the 10 against the 7 on each of the 36 rolls with a perfect distribution, we expect to win $1 three times (when the 10 shows) and lose $1 six times (when the 7 shows).  So, for an even‑money bet, our net result is a $3 loss ($6 – $3 = $3).  But with a perfect distribution, we expect to break even, not lose.

Let’s use the same example except this time we get the 2:1 odds that we expect to get when we bet the 10 against the 7 (i.e., instead of getting only 1:1 even money).  If we again bet $1 on each of the 36 rolls, we expect to win $2 three times (when the 10 shows) and lose $1 six times (when the 7 shows).  Therefore, for a 2:1 odds bet, our net result is that we break even, as we expect (i.e., we win $2 three times, or $2 x 3 = $6; and we lose $1 six times, or $1 x 6 = $6).  So, with the 2:1 odds bets, we win $6 and we lose $6 (i.e., we break even as we expect).

In the world of statistics, if everything balances out after a long period of time with a large quantity of dice rolls, how does the casino make money?  Why don’t the player and casino break even over time?  The answer is simple…they screw us!  And they don’t feel the slightest bit guilty about it.  There’s something called “house advantage” that we’ll discuss in detail in another article.  “House advantage” is a nice way to say, “If you don’t know what you’re doing, we’re going to screw you and take your money.”  The house (i.e., the casino) takes a set percentage out of the payout of every possible bet when you win a bet (except the Free Odds bet, which is discussed in another article).  The easiest way to illustrate this concept is to compare the results of making a Place bet on the number 4 or 10.  (We’ll look at Place bets in another article.)

Since 10 is our favorite number, let’s look at Place betting the 10 against the 7.  As we know from the table above, there are three ways to roll a 10.  If we Place bet $5 on the 10 against the 7, we expect to win $10 (remember, there are six ways to roll a 7 and 3 ways to roll a 10, so we expect to get 2:1 odds on the 10, so we expect to get $5 x 2 = $10 when we win a $5 bet).  Ready to find out how the casino screws us?

Instead of giving 2:1 odds for a Place bet on the 10 against the 7, the house gives odds of only 9:5.  Ouch!  That means, when we bet $5 and win, we receive only $9 instead of the $10 we expect.  They’ve screwed us out of that extra dollar that we should get because of the true odds of 2:1.

Using the 36-roll perfect-distribution example again, we find that the casino makes tons of money off suckers who don’t know what they’re doing.  For each of the 36 rolls, suppose we bet $5 on the 10 against the 7, and the odds are only 9:5 instead of the 10:5 true odds that we expect to get (i.e., the expression 10:5 reduces down to 2:1).  That means, for each losing roll, we lose our $5 bet; and for each winning roll, we win $9.  After 36 rolls with a perfect distribution, we expect to lose six times for a total of $30 (6 x $5 = $30), and we expect to win three times for a total of $27 (3 x $9 = $27).  The net result is that we come out a $3 loser overall, even with a perfect distribution.  You may be thinking, “I’m still a little confused, so how did we come out losing $3 at the end of the 36 rolls?”

Remember, there are six ways to roll a 7, and 3 ways to roll a 10.  In a perfect distribution over 36 rolls, we expect a 7 to appear six times, and a 10 to appear three times.  On the six rolls when a 7 appears, we lose our $5 bet.  On the three rolls when a 10 appears, we win the bet.  Because there are twice as many ways to roll a 7 as there are to roll a 10 (i.e., the odds are expressed as 2:1), we expect to get two times our $5 bet when we win, or $10.  So, for all three winning bets, we expect to win a total of $30 (i.e., with a $5 bet at 2:1 odds, and with the 10 appearing three times in 36 rolls in a perfect distribution, we win $5 (our bet amount) x 2 (for the 2:1 odds) x 3 (for the three times the 10 appears in 36 rolls) = $30.  Now, let’s look at how much we win when the casino only gives us 9:5 odds instead of the full 10:5 odds.  In this case, for all three winning bets, we expect to win a total of $27 (i.e., with a $5 bet at 9:5 odds, and with the 10 appearing three times in 36 rolls in a perfect distribution, we win $5 (our bet amount) x 9 (for the 9:5 odds) x 3 (for the three times the 10 appears in 36 rolls) = $27.  If the casino paid true odds, then over time, everything would even out and no one would make a profit.  However, the casino is in business to make money, so they only give us 9:5 odds (instead of 10:5), so over time when everything evens out, they end up making a $3 profit from us.

Think about all the people playing craps 24 hours a day, seven days a week, 52 weeks a year.  And consider that every bet on the craps table has a built-in house advantage except the Free Odds bet.  By not paying us the true amount based on the true odds, they make a small profit on every bet made by every person at the table, every minute of every day.

The house advantage varies among the many different types of possible craps bets.  We’ll discuss them all later and you’ll learn which bets have high house advantages and which have relatively small house advantages.  Obviously, you want to avoid the bets with the higher house advantages and focus on those with the smallest.

I know what you’re thinking.  “If the casino has a built-in advantage on every bet which means I’m supposed to lose over time, why should I bother playing at all?”  Great question!  We play because there’s an important thing called “variance” that allows us to win.  We’ll go into detail about variance in another article, but variance is our friend at the craps table.  Variance means you likely won’t see a perfect distribution of the results when rolling the dice.  Instead, distribution variance is what gives us the hot and cold streaks that are so common.  Knowing how to recognize those variances (i.e., hot and cold streaks) and knowing how to adapt your play to them are what enables us to win.  And winning is what it’s all about.  Don’t be a stupid player.  Be a rock-solid, smart player.  Make the casino fear you, hate you.  It’s fun asking the pit boss for a wheelbarrow to carry all your chips to the cage.

Here we list some more Craps Math and Statistics related posts:

You can now head over to the table of contents to find more great content.

Author
Written by John Nelsen in partnership with the team of craps pros at crapspit.org.
  • Jesse Champion

    The “bag of marbles” example was perfect. And I get it that the same odds are in play on every throw, however the likelihood of missing or choosing the same marbles, while infinitely possible, is extremely improbable. The possibility of throwing sevens each toss is also infinite but highly improbable given the other numbers also exist in their relative ratios; as do the marble colors.

    No argument here – just seeking another viewpoint on that logic.

    ….. My bad – obsolete website. Go figure – by the time anyone sees this I’ll have my answer.

    • crapspit

      Hi Jesse, thanks for your comments.

      We’re unsure of what you’re asking, so if you’d like to post a clarifying comment, we’ll be happy to try again. You wrote, “…However, the likelihood of missing or choosing the same marbles, while infinitely possible, is extremely improbable. The possibility of throwing sevens each toss is also infinite but highly improbable given the other numbers also exist in their relative ratios; as do the marble colors.”

      We’re unsure of the point you’re trying to make. The example of the bag of marbles was intended simply to explain to our readers who have difficulty understanding the concept that outcomes associated with previous dice roles have no effect on the probabilities associated with the outcomes of future dice roles in a legal game of craps. We agree with your logic that throwing sevens each toss over infinity is highly improbable, but we fail to understand what that has to do with the basic subject about the law of probabilities that future outcomes in a legal craps game are not influenced by previous outcomes; therefore, when you state, “…just seeking another viewpoint on that logic…,” we’re unable to provide any viewpoint without first getting clarification of the point you’re trying to make.