The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Rolling Dice and Probability

Date: 11/07/2007 at 07:26:09
From: Harri
Subject: Probability, rolling dice

Having thrown a die 10 times, what is the probability of rolling all 
six numbers?  Order does not matter.

Date: 11/07/2007 at 10:03:55
From: Doctor Tom
Subject: Re: Probability of rolling numbers 1-6 in ten throws.

Hello Harri,

This is a tough problem, and you can find details if you look up
"coupon collector's problem".  (Waiting to get all 6 numbers is like a
coupon collector who is trying to get a complete set where each new
coupon is equally likely to be any of them.)

There are easy tricks to find the expected (average) time to get all
six numbers and that turns out to be about 14.7 rolls.  In other
words, if you repeat the experiment, rolling until you get all six,
and write down the number of rolls it takes you to do so each time,
you'll find that the average of those numbers is about 14.7.

To solve your problem exactly for small numbers of rolls, like 10, we
can proceed as follows.  Let p0, p1, p2, p3, p4, p5 and p6 be the
probabilities that at a certain stage in the process you've seen 0, 1,
2, ..., 6 different faces.  Before you start, p0=1 and p1=p2=...=p6=0.

After one roll, p0=0 (and will be zero from now on), p1 = 1,

In general, there is a nice formula for figuring out the new p0, ...,
p6 from the old ones, one roll later.  Suppose they are p0, ..., p6 at
one stage, and you want to compute p0', p1', p2', ..., p6' one roll
later.  Here are the formulas:

  p0' = 0
  p1' = p1/6
  p2' = 5 p1/6 + 2 p2/6
  p3' = 4 p2/6 + 3 p3/6
  p4' = 3 p3/6 + 4 p4/6
  p5' = 2 p4/6 + 5 p5/6
  p6' =   p5/6 + p6

Let's look at one of these carefully; the others can be argued in the
same way.  Let's look at how we obtain p3'.  Well, either we had three
faces before and didn't get a new one, and this will happen 3 times
out of 6, which accounts for the term 3 p3/6.  OR, we had two 
faces and got a new one.  If we had two, then 4 of the six rolls from
there would yield a new face, which is where the term 4 p2/6 comes
from.  There are no other ways to arrive at three faces.

Now just do the arithmetic.  Here are the first few steps.  You'll
have to continue, step by step, to stage 10 if you want the answer for
10 rolls.  The numbers in parentheses represent (p0, p1, p2,  ...,
p6).  The "stage i" means the situation after i rolls.

  stage 0:  (1, 0, 0, 0, 0, 0, 0)
  stage 1:  (0, 1, 0, 0, 0, 0, 0)
  stage 2:  (0, 1/6, 5/6, 0, 0, 0, 0)
  stage 3:  (0, 1/36, 15/36, 20/36, 0, 0, 0)
  stage 4:  (0, 1/216, 35/216, 120/216, 60/216, 0, 0)

For practical purposes, you don't need exact fractions; you can do
each stage with a calculator, and keep the decimal expansion.

This technique of keeping track of the probability of being in various
"states" as time progresses is called a "Markov chain", which you can
also look up.  There is a very easy way to work out the numbers from
my formulas by repeated matrix multiplication.  Look up "Markov chain"
for more information.

As time goes on, the probabilities will shift to the right, toward p6,
but there is always some small probability that smaller numbers will
be present.  Thus at every stage after 5 rolls, the odds of having
completed the set of faces increases since the probability of being in
stage 5 is always positive, but after a certain point it will get
smaller and smaller.

- Doctor Tom, The Math Forum 

Date: 01/29/2008 at 19:00:08
From: Doctor Anthony
Subject: Re: Probability of rolling numbers 1-6 in ten throws.

Hi Harri -

I would add a technique that can be used in this type of problem.  
The number of ways that 10 numbered balls could be distributed in 6 
numbered boxes such that no box is empty is a suitable model.  The 
number of ways is given by the coefficient of x^10/10! in the 
expansion of [e^x - 1]^6 and is denoted by T(10,6).  The number of 
unrestricted ways of distributing the balls is 6^10 so that the 
required probability is


We have

  [e^x - 1]^6 = e^(6x) - C(6,1)e^(5x) + C(6,2)e^(4x) - C(6,3)e^(3x)
                + C(6,4)e^(2x) - C(6,5)e(x) + 1

We now pick out the terms in x^10/10!

  = (x^10/10!)[6^10 - 6(5^10) + 15(4^10) - 20(3^10) + 15(2^10) - 6]

  = (x^10/10!)[16435440]

and so the coefficient of x^10/10! is 16435440.

If you throw a die 10 times the total number of possible outcomes, 
with 6 possibilities at each throw, is 6^10.

So the required probability is 16435440/6^10  =  0.271812128

- Doctor Anthony, The Math Forum 
Associated Topics:
College Probability

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.