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Building a Proof


For the previous page, we saw the probability function for various scores if three dice were thrown. You may have noticed a symmetry to the probabilities.


If we drew a line between 10 and 11, we could say that each side of the line was a mirror reflection of the other.

Let's assume we want to develop some sort of dice gambling game, throwing five dice. Two immediate concerns come to mind:
  1. If the chance of the customer winning are too low, the gamblers will stop playing, and we won't make any money.
  2. If we don't understand the probability function, we may overlook a way that gamblers could have winning odds, and then we will lose money.
It is necessary to understand all the characteristics of the probability function to prevent either problem from happening.

For three dice, it was possible to determine the probability function "by hand" (as opposed to using some computer program), since there were only 216 combinations possible. However, if our gambling game involves throwing five dice, then there will be 7776 possible combinations.

We start with the high and the low score.
  • 1,1,1,1,1 yields a score of 5, and is the only possible throw that results in 5.
  • 6,6,6,6,6 yields a score of 30, and is the only possible throw that results in 30.

    There are five possible combinations that will yield a score of 6.
    1. 2,1,1,1,1
    2. 1,2,1,1,1
    3. 1,1,2,1,1
    4. 1,1,1,2,1
    5. 1,1,1,1,2
    There are five possible combinations that will yield a score of 29.
    1. 5,6,6,6,6
    2. 6,5,6,6,6
    3. 6,6,5,6,6
    4. 6,6,6,5,6
    5. 6,6,6,6,5
    At this point, you may say that you sense, intuitively, that we could continue this process until we have examined all the possible scores, and that for each possible low score, there would be a corresponding high score with the same probability.

    But we want to prove it. What to do...

    Algebra was developed, to replace tables of specific information: with a general formula:

    A collection of numerical data corresponding to one scenario is called a specific solution. Above, the point (3,6) only tells you that y=6 when x=3.
    An algebraic formula that corresponds to all possible scenarios is called a general solution. The formula (x,2x) can be used to determine any y value, if the x value is known.


    In the above problem, we might say that we proved a correspondence between the scores of 5 and 30 when we showed that each score could only occur from 1 combination.
    We might then say that we proved a correspondence between the scores of 6 and 29 when we showed that each score could occur from 5 combinations.

    But to prove the correspondence for all possible scores, we need to create formulas, using algebra, not specific numbers, that cover every possible scenario.

    We are going to need variables for the scores.

    Let's use the variable 'i' for low scores: 5, 6, 7, ...

    and the variable 'j' for high scores: 30, 29, 28, ...

    We need to determine the math for the correspondence between i and j. When i=5, j=30. When i=6, j=29.

    j = 35 - i.

    Now, how do we prove that there are as many scores for an i value as there are for the j value?

    There may be more than one approach to do this, but currently, I can only think of one: proving a correspondence between scores.

    Let's say we proved there were 'x' combinations that produce a particular i. How would we prove that there were 'x' combinations for j?

    Here's the plan. We prove there is a "mirror" symmetry between the combinations that yield i, and the combinations that yield j. Each i combination matches to an 'equivalent' j combination, and each j combination matches to an 'equivalent' i combination.

    I would like a color cartoon here with seven of those
    M&M characters looking into a mirror, and
    it is easy to see that there are seven reflections
    of M&M characters in the mirror.

    We might then find some technical way of saying that if
    a mirror image exists for every "real" object, the
    number of mirror image objects must be equal to the
    number of "real" objects.


    Assume that we could prove:
    1. There is a j combination for each i combination,
    2. There is an i combination for each j-combination.
    What do we mean by this? Let's make i=7, and write several possible combinations. Since i=7, j = 35-i = 28. Look at the combinations below: You may notice that every 1 in the 'i' combinations corresponds to a 6 in the 'j' combinations.
    Likewise, every 2 in the 'i' combinations corresponds to a 5 in the 'j' combinations.



    Here's the important part. We want to prove, that for any possible combination that results in i, there is a combination that results in j.

  • We need to define more variables. We will use 'a', 'b', 'c', 'd', and 'e' for the i combination.
  • We will use 'v', 'w', 'x', 'y', and 'z' for the j combination.

    We are going to need a constant, n=7. The dice give outputs from 1 to 6. If we had dice that gave outputs from 1 to 4, then n would be 5. More about this later.

    Finally, to prove the correspondence between the i combination and the j combination, we need to express the relation using only the variables above.

  • v=n-a
  • w=n-b
  • x=n-c
  • y=n-d
  • z=n-e

    Feel free to use the above cominations of numbers to verify the algebra above.

    The final question is this:



    The score of a + b + c + d + e = i

    What is the score of v + w + x + y + z?

    v + w + x + y + z
    (n-a) + (n-b) + (n-c) + (n-d) + (n-e)
    + n - a + n - b + n - c + n - d + n - e
    + 5n - a - b - c - d - e
    + 5n - (a + b + c + d + e)
    + 5n - i

    Since n=7, 5n - i = 35 - i = j

    What we have proven:

    For every possible combination a,b,c,d,e resulting in a score of 'i', there is a combination v,w,x,y,z with a score of j, where j = 35 - i.

    It took all this work to prove that the left side of the function was the mirror image of the right side of the function.

    However, the next question is, can we determine what probability factors correspond to each score i?

    If we know how many times out of 100 the score will be 9 or less, we can decide how much to charge per roll of the dice, and how much the prize will be for a roll with a score of 9 or less.

    To advertize this game, we might put up posters showing all the possible scores that give a score of 9 or less (but not tell how many even more there are that don't qualify.)

    Another option: We might call this game "Highs and Lows", and advertise that both scores of 9 or below and 26 or above, qualify as winning scores. If we know how many combinations give 9 or less, we know how many give 26 or more.

    Now: Let's see if we can find some way of determining the probability factor for each score.




    When writing the proof, I found myself writing the number 7 into the proof. Yet previously I had said we would not use numbers.

    As I understand it, it is acceptable to add constants to the math, as long as you explain what they represent.

    By using the constant 'n', rather than writing in 7, we can apply the math to situations where the dice have any number of sides, and n will be one greater than the number of dice sides.

    There was one other constant. In defining j in terms of i, we wrote j = 35 - i. Can you guess where 35 comes from? (Hint: What if we decided to throw 4 dice? Would that have some effect on the math?)




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    Last Revised 02/03/98.
    Copyright ©1998 by William L. Dechent. All rights reserved.