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Best of Donald Catlin

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Calculating Dice Sum Frequencies

17 August 2001

    There are many casino games that depend on summing the faces of dice; Craps and Sic Bo come to mind.  I imagine that there will be new games coming along that depend on dice sums as well.  Although most gamblers are aware of the frequencies for sums with two dice, the frequencies for more than two are probably not familiar numbers to them.  In this article I want to show you how, with just pencil and paper (and not a computer), to compute frequencies for dice sums.

    To begin with, I want you to recall the definition of C(n,k) that I introduced in my article Oh, New York, Bring Back those Big Dippers that I wrote for Rolling Good Times in August of 1999.  C(n,k) is the number of ways of choosing k distinct objects from a set of n objects and is given by the formula

C(n,k) = n!/[k!(n - k)!)] (1)

where
n! = n(n - 1)(n - 2) ... 3 x 2 x 1 (2)

and
0! = 1

For example, the number of ways of choosing 3 objects from a set of 5 is (5 x 4 x 3 x 2 x 1)/(3 x 2 x 1 x 2 x 1), which is 10.  To see that this is correct, suppose that the set of objects is {a, b, c, d, e}.  Then the collection of all three element subsets of this set is:

{a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}
{a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}

They're all there, check it out.  We will also need to observe that the number of ways to arrange n objects in order, say left to right, is n!  The reason is that there are n ways to choose the first element, and for each of these there are n - 1 ways to choose the second, so there are n x (n - 1) ways to choose the first two elements.  For each such choice there are n - 2 ways to choose the third element, so there are n x (n - 1) x (n - 2) ways to choose the first three.  Continuing we get n! for the entire set.  In the above example, for instance, there are 5 x 4 x 3 x 2 x 1 or 120 ways to arrange these five letters in order.  I don't want to write all of them out, so let's look at a simpler example, the set {a, b, c).  According to my contention there should be 3 x 2 x 1 or 6 ways to order these.  Here they are:

abc, acb, bca, bac, cab, cba

    The idea in calculating dice sums is to first see how a particular sum can be obtained by choosing viable numbers on the set of dice in question.  We arrange the numbers in decreasing order from left to right in a table so as not to repeat any choices.  An example will be very useful; let's see how we can obtain a total of 9 using three dice.  Notice, in Table 1, the way I list the dice, slowly reducing the largest number until it cannot be reduced further, then the next until it cannot be further reduced, and so on, making sure that I get all of the combinations:

Highest Next Lowest
6 2 1
5 3 1
5 2 2
4 4 1
4 3 2
3 3 3

Table 1
Ways 3 Numbers can Produce Total 9

    Given Table 1, the next step is to see how many ways we can rearrange the three numbers from left to right.  Think of the dice as being red, white, and blue.  There are three types of patterns: all numbers different, two the same and one different, all of them the same.  If all of them are different, then there are 3! or 6 ways to arrange them.  If all of them are the same, then there is only one way to arrange them.  If two are the same, then there are C(3,2) ways to choose where the matched numbers will go leaving only one spot in which to put the third.  Since C(3,2) = (3 x 2 x 1)/(2 x 1 x 1), there are 3 patterns when two of the dice are the same.  Using this information we can enlarge Table 1 as follows:

Highest Next  Lowest Patterns
6 2 1 6
5 3 1 6
5 2 2 3
4 4 1 3
4 3 2 6
3 3 3 1
Total - 25

Table 2
Ways of Rolling Total of 9 with Three Dice

    Now that you see the idea, let's try a harder problem.  How many ways are there to roll a 12 with four dice?  To begin with, let's see how many different dice patterns there are.  All could be different numbers, we could have a pair with two other distinct numbers, we could have two pair, we could have a triple with another number, or we could have all four the same.  Taking these in order, we know that if all are different there are 4! or 24 different patterns.  If there is a pair with two different numbers then there are C(4,2) or 6 choices to place the pair.  This done, there are two places left for the remaining two distinct numbers and these can be in either order, so there are 2 x 6 or 12 patterns here.  If we have two pair, then again there are C(4,2) choices for the higher pair leaving no choice for the lower pair.  Thus there are 6 patterns in this case.  If we have a triple, there are four choices for the non-repeated number and then no choice in placing the triple, so there are 4 patterns here.  Obviously if all are the same there is only 1 pattern.  With this in mind, here is the table for calculating the number of ways to roll a total of 12 with 4 dice:

Highest Next Next Lowest Patterns
6 4 1 1 12
6 3 2 1 24
6 2 2 2 4
5 5 1 1 6
5 4 2 1 24
5 3 3 1 12
5 3 2 2 12
4 4 3 1 12
4 4 2 2 6
4 3 3 2 12
3 3 3 3 1
Total - 125

Table 3
Ways of Rolling a Total of 12 with Four Dice

    If you try a calculate the frequencies for each of the three dice totals, all of the numbers you calculate should add up to 216 which is the total number of ways one can roll three dice (6 x 6 x 6).  Similarly, the four dice totals should add to 1296.  Five dice totals should add to 7776.  For 6 it's 46,656 (that would keep a body busy for awhile).

    It is unlikely that one would see a game with more than 6 dice.  It is interesting to note, however, that for seven or more dice there will always be at least one pair.  The higher you go the trickier the patterns become.

    If you want to try your hand at calculating dice sums here are some to try.  Yes, I know, it is easier with a computer but what fun is that?  With three dice and a total of 10 you should get 27.  With four dice and totals of 13 and 14 you should get 140 and 146 respectively.  Notice, by the way, that for very small or very large totals the number of dice numbers, and therefore patterns, is small.  Rolling a 5 with four dice, for example, can only be done with a 2 and three 1s.  Now if you really want a challenge here is one to try.  How many ways are there to roll a total of 15 with 5 dice?  You should get an answer of 651.

    If you try some of these dice total problems and things don't seem to work out correctly, you can contact me by e-mail by clicking on the word Technigame in the About the Author box at the end of this article and I'll help you out.  See you next month.

Donald Catlin

Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers