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Math Help - Rolling Dice

  1. #1
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    Rolling Dice

    Hello. I am not a math major, just a math tutor. I haven't taken anything beyond Calculus I, but even then my calculus is rusty. Enough about me, now for my problem.

    I hope this belongs here and not in the pre-university threads. It doesn't seem elementary to me, but I am hoping I can be proven wrong.

    Recently, a friend provided me with this formula:



    With it, I can derive the probability, P, of reaching a sum, p, when rolling a certain number of dice, n, each with the same number of sides, s.

    For example, the odds of reaching a sum p=19 for n=6 dice each with s=6 sides are 3,906 out of 46,656.

    It's the next step with which I am having trouble (and is actually my question).

    How do I calculate the odds of reaching a sum of 19 rolling six 6-sided dice when I disregard the die showing the lowest value? the highest value? the two lowest values? etc.

    How do I calculate the odds of reaching any sum, p, rolling n s-sided dice when I disregard one or more dice based on their comparative values?

    Thanks in advance for the help. I hope this doesn't prove impossible.

    Oh. I don't want to write out all the combinations by hand. I want to use a formula perhaps like the one above, not a computer program which does the calculation for me.
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  2. #2
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    If you are willing to use a computer, you can try every combination. There are a managable number of them(46656)

    Edit sorry i misread the last line of your post, i thought you said you were willign to use a PC
    Last edited by SpringFan25; June 18th 2010 at 02:07 AM.
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  3. #3
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    Alternative

    Alternatively, can anyone produce a function or equation for me which fits the following data set?

    f(x)=0 when x=3
    f(x)=1 when x=4
    f(x)=4 when x=5
    f(x)=11 when x=6
    f(x)=23 when x=7
    f(x)=41 when x=8
    f(x)=66 when x=9
    f(x)=95 when x=10
    f(x)=121 when x=11
    f(x)=142 when x=12
    f(x)=151 when x=13
    f(x)=145 when x=14
    f(x)=121 when x=15
    f(x)=88 when x=16
    f(x)=51 when x=17
    f(x)=20 when x=18

    The values for f(x) are the differences between the total combinations which result in a sum of x between "4d6 sans lowest" and "3d6".

    But wait, you might ask, how did I derive this information if this thread is asking for "how many combinations are there for a sum of x for 4d6 sans lowest"? Aren't I missing the "4d6 sans lowest" data? Isn't that the point of this thread? How could I be using it here if I don't know it?

    Many months ago--possibly a year--I wrote all the combinations for it out by hand. I also did so for "5d6 sans the two lowest values". I don't want to have to do that again for any other calculation I might want to do. That is why I am seeking a formula to assist me.

    Thanks and good luck.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Mild View Post
    Alternatively, can anyone produce a function or equation for me which fits the following data set?

    f(x)=0 when x=3
    f(x)=1 when x=4
    f(x)=4 when x=5
    f(x)=11 when x=6
    f(x)=23 when x=7
    f(x)=41 when x=8
    f(x)=66 when x=9
    f(x)=95 when x=10
    f(x)=121 when x=11
    f(x)=142 when x=12
    f(x)=151 when x=13
    f(x)=145 when x=14
    f(x)=121 when x=15
    f(x)=88 when x=16
    f(x)=51 when x=17
    f(x)=20 when x=18

    The values for f(x) are the differences between the total combinations which result in a sum of x between "4d6 sans lowest" and "3d6".

    But wait, you might ask, how did I derive this information if this thread is asking for "how many combinations are there for a sum of x for 4d6 sans lowest"? Aren't I missing the "4d6 sans lowest" data? Isn't that the point of this thread? How could I be using it here if I don't know it?

    Many months ago--possibly a year--I wrote all the combinations for it out by hand. I also did so for "5d6 sans the two lowest values". I don't want to have to do that again for any other calculation I might want to do. That is why I am seeking a formula to assist me.

    Thanks and good luck.
    There is a similar thread here. I don't know of a formula, but the algorithm(s) I discuss on the other thread could be done on paper if one desired (although it would be pretty tedious).
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  5. #5
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    I took a look at the other thread, but you lost me. Can you dumb it down some? Matrices were never my strong point.
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Mild View Post
    I took a look at the other thread, but you lost me. Can you dumb it down some? Matrices were never my strong point.
    I made an image to illustrate the algorithm: (click twice for full size)

    Rolling Dice-dice_distr_algo.png

    This image is for four dice and counting every die (including the lowest). The row is the total face value of the dice, and the column is how many dice are rolled, and the cell is the number of ways to roll it. So for the column labeled 2, we can think of this information stored in an array L where

    L[0] = 0
    L[1] = 0
    L[2] = 1
    L[3] = 2
    L[4] = 3
    L[5] = 4
    L[6] = 3
    L[7] = 2
    L[8] = 1

    and this array gives us all the information we need to get the column labeled 3.

    In order to omit the lowest-valued die, my suggestion in the other thread was to use a two dimensional array to keep track of both the lowest die, and the total value. So for four-sided dice we would have (I'll use A instead of L, which will allow me to naturally specify B... bear with me)

    0 dice

    A[4][0] = 1

    1 die

    A[1][1] = 1
    A[2][2] = 1
    A[3][3] = 1
    A[4][4] = 1

    all other values 0

    2 dice

    Here's where it gets very tedious (on paper).

    B is initialized with all values set to 0.

    Look at each non-zero entry of A, and then consider what happens when you roll a 1, 2, 3, and 4 from there, and update B accordingly.

    So for A[1][1] = 1

    We update

    B[1][2] += 1
    B[1][3] += 1
    B[1][4] += 1
    B[1][5] += 1

    where += means increment by.

    For A[2][2] = 1 we get

    B[1][3] += 1
    B[2][4] += 1
    B[2][5] += 1
    B[2][6] += 1

    etc.

    At the end, we'll set A = B and start over.

    If you need clarification, please let me know. It's a bit hard to illustrate even for 4-sided dice because of the amount of typing and such, but at least it's better than listing every possible combination.
    Last edited by undefined; June 18th 2010 at 09:24 PM.
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