Dice roll probability

John wrote:
> Has anyone ever sat down and worked out a formula or
> method to determine the probability that rolling N dice
> will produce a score above or below X?
>
> I've tried to come up with one, as it sounds rather
> useful, but I keep having to refer to things I haven't
> looked at for ten years. So before I reinvented the wheel
> I thought I'd ask if anyone has already done the work for
> me.

Perhaps you may be interested in the mathematics describing
such problems. Keywords that will help you explore beyond
what is given below are "convolution" and "z-transform".

When you roll 1d6 if the principle of indifference applies
then each of the possible numbers are equally likely. We can
express this knowledge as a uniform probability distribution
on the integers as below (use a fixed-width font):

p(k) = { 0 : k < 1
1/6 : 1 <= k <= 6
0 : k > 6 }

|
1/6 -| * * * * * *
|
|
0 -*--------------------*--*
| | | | | | | | |
0 1 2 3 4 5 6 7 8

When you roll 2d6 and total the numbers the sum can range
between 2 and 12. However, as you know, some totals are more
likely than others. Out of the the 36 possible rolls only
one (double 1's or "snake eyes") totals 2 while six of the
36 total 7. Now the probability distribution on the possible
sums is the "convolution" of the individual distributions.

You can find lot's of information, explanations, examples,
and methods of computing convolutions online. However, for
this particular case (discrete domain) there is a relatively
simple tool called the "z-transform" that you can apply.

The z-transform maps a function in one space (here k-space
usually used in situations where k would be the discrete
"time" domain) to a function in an alternate z-space.

Z[f(k)] = Sum_k_0_inf [ f(k) * z^(-k) ]

The usefulness is that some operations in k-space become
more simple in z-space. Specifically, Convolution of two
functions in k-space C(f(k,g(k)) becomes multiplication in
z-space Z[C(f(k),g(k))] = Z[f(k) * Z[g(k)] = f(z)*g(z).
This is called the z-transform "convolution theorem".

Let me show you how this is useful here. First z-transform
1d6

Z[1d6] = 1/6 * (z^-1 + z^-2 + z^-3 + z^-4 + z^-5 + z^-6) .

Now the convolution of two d6's Z[2d6] is simply

Z[2d6] = Z[1d6] * Z[1d6]
Z[2d6] = 1/6 * (z^-1 + z^-2 + z^-3 + z^-4 + z^-5 + z^-6) *
1/6 * (z^-1 + z^-2 + z^-3 + z^-4 + z^-5 + z^-6) .

In this simple case the power of z simply gives the value of
the sum and the multiplier (above simply 1/6) of each term
is the probability of that sum. Since the polynomials are in
a nice simple form we can multiply them easily using some
shorthand that simply tracks the coefficients and using
position to represent power of z^-k.

2 3 4 5 6 7 8 9 10 11 12
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
--------------------------------
1 2 3 4 5 6 5 4 3 2 1 .

Thus the probability of the sum s for 2d6 is

s : 2 3 4 5 6 7 8 9 10 11 12
p * 36 : 1 2 3 4 5 6 5 4 3 2 1 .

note that is p * 36 so you need to divide the numbers by 36
to get normalized probabilities. Now to add another 1d6

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
-----------------------------------------------
1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1 .

Thus the probability of the sum s for 3d6 is

s : 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
p * 216 : 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1

The hand computations above are in fact the same as those
you would perform by doing a naive convolution. We just
arrived at them by multiplying polynomials in z-space as
opposed to directly applying the definition of convolution
to properly zero-padded representation of the k-space
function.

For arbitrary dice it's simple to setup an excel sheet to
perform computations like the above. Or you can use the
built-in convolution functions in other software such as
Igor Pro, Matlab, Mathematica, etc. Or roll your own code.
And of course Eric B. Smith already provided a link to d6
tables up to 20d6.

-- Keith -- Fraud 6

duggar.DeleteThis@alum.mit.edu wrote:
> John wrote:
> > Has anyone ever sat down and worked out a formula or
> > method to determine the probability that rolling N dice
> > will produce a score above or below X?
> >
> > I've tried to come up with one, as it sounds rather
> > useful, but I keep having to refer to things I haven't
> > looked at for ten years. So before I reinvented the wheel
> > I thought I'd ask if anyone has already done the work for
> > me.
>
> Perhaps you may be interested in the mathematics describing
> such problems. Keywords that will help you explore beyond
> what is given below are "convolution" and "z-transform".
>
> When you roll 1d6 if the principle of indifference applies
> then each of the possible numbers are equally likely. We can
> express this knowledge as a uniform probability distribution

aaaaargh! MATH MAJORS! aaaaargh! Do a more clever google (verb) & find
some
DOWNLOADABLE SHAREWARE / FREEWARE Software! aaaaargh!
> on the integers as below (use a fixed-width font):
>
> p(k) = { 0 : k < 1
> 1/6 : 1 <= k <= 6
> 0 : k > 6 }
>
> |
> 1/6 -| * * * * * *
> |
> |
> 0 -*--------------------*--*
> | | | | | | | | |
> 0 1 2 3 4 5 6 7 8
>
> When you roll 2d6 and total the numbers the sum can range
> between 2 and 12. However, as you know, some totals are more
> likely than others. Out of the the 36 possible rolls only
> one (double 1's or "snake eyes") totals 2 while six of the
> 36 total 7. Now the probability distribution on the possible
> sums is the "convolution" of the individual distributions.
>
> You can find lot's of information, explanations, examples,
> and methods of computing convolutions online. However, for
> this particular case (discrete domain) there is a relatively
> simple tool called the "z-transform" that you can apply.
>
> The z-transform maps a function in one space (here k-space
> usually used in situations where k would be the discrete
> "time" domain) to a function in an alternate z-space.
>
> Z[f(k)] = Sum_k_0_inf [ f(k) * z^(-k) ]
>
> The usefulness is that some operations in k-space become
> more simple in z-space. Specifically, Convolution of two
> functions in k-space C(f(k,g(k)) becomes multiplication in
> z-space Z[C(f(k),g(k))] = Z[f(k) * Z[g(k)] = f(z)*g(z).
> This is called the z-transform "convolution theorem".
>
> Let me show you how this is useful here. First z-transform
> 1d6
>
> Z[1d6] = 1/6 * (z^-1 + z^-2 + z^-3 + z^-4 + z^-5 + z^-6) .
>
> Now the convolution of two d6's Z[2d6] is simply
>
> Z[2d6] = Z[1d6] * Z[1d6]
> Z[2d6] = 1/6 * (z^-1 + z^-2 + z^-3 + z^-4 + z^-5 + z^-6) *
> 1/6 * (z^-1 + z^-2 + z^-3 + z^-4 + z^-5 + z^-6) .
>
> In this simple case the power of z simply gives the value of
> the sum and the multiplier (above simply 1/6) of each term
> is the probability of that sum. Since the polynomials are in
> a nice simple form we can multiply them easily using some
> shorthand that simply tracks the coefficients and using
> position to represent power of z^-k.
>
> 2 3 4 5 6 7 8 9 10 11 12
> 1 1 1 1 1 1
> 1 1 1 1 1 1
> 1 1 1 1 1 1
> 1 1 1 1 1 1
> 1 1 1 1 1 1
> 1 1 1 1 1 1
> --------------------------------
> 1 2 3 4 5 6 5 4 3 2 1 .
>
> Thus the probability of the sum s for 2d6 is
>
> s : 2 3 4 5 6 7 8 9 10 11 12
> p * 36 : 1 2 3 4 5 6 5 4 3 2 1 .
>
> note that is p * 36 so you need to divide the numbers by 36
> to get normalized probabilities. Now to add another 1d6
>
> 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
> 1 2 3 4 5 6 5 4 3 2 1
> 1 2 3 4 5 6 5 4 3 2 1
> 1 2 3 4 5 6 5 4 3 2 1
> 1 2 3 4 5 6 5 4 3 2 1
> 1 2 3 4 5 6 5 4 3 2 1
> 1 2 3 4 5 6 5 4 3 2 1
> -----------------------------------------------
> 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1 .
>
> Thus the probability of the sum s for 3d6 is
>
> s : 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
> p * 216 : 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1
>
> The hand computations above are in fact the same as those
> you would perform by doing a naive convolution. We just
> arrived at them by multiplying polynomials in z-space as
> opposed to directly applying the definition of convolution
> to properly zero-padded representation of the k-space
> function.
>
> For arbitrary dice it's simple to setup an excel sheet to
> perform computations like the above. Or you can use the
> built-in convolution functions in other software such as
> Igor Pro, Matlab, Mathematica, etc. Or roll your own code.
> And of course Eric B. Smith already provided a link to d6
> tables up to 20d6.
>
> -- Keith -- Fraud 6