Three Fair Dice

Date: 05/16/99 at 08:01:51
From: Laura
Subject: Probability dealing with combinations of dice

Dear Dr. Math,

I have been trying to solve this problem for quite a while but I have 
not been able to come up with an answer.  

a) Find the probability of a total of 9 and the probability of a 
   total of 10 when three fair dice are tossed once. Compare the 
   probabilities.

b) Find the probability of at least one six in 4 tosses of a fair die 
   and the probability of at least one double-six in 24 tosses of two  
   fair dice. Compare the probabilities.

c) Find the probability of at least one six when 6 dice are rolled and 
   the probability of at least two sixes when 12 dice are rolled.   
   Compare the probabilities.

For a), I found that there are 342 possible total outcomes when 
rolling 3 dice, but I do not think this is correct. I got 342 by 
writing out all the possible outcomes - I counted the triple digits 
(i.e. getting a 3, a 3, and a 3) as six ways instead of one. Also, is 
it correct to take the number of ways to get a nine and divide by the 
total possible outcomes, to find the probability of getting a nine?

For b), I again was not sure how to solve this. Do I take 1/6, the 
probability of getting a 6, and multiply it by 4 - the number of times 
tossed? And for the second part, I have no idea.

For c), is it 1/6 times 6 or 1/6 ^6, or.....?

Finally, how do I compare? Would it be reasonable to take the two 
outcomes and state their difference? What, do you think, should I 
include in the statement?

If you could answer even part of this question, I would greatly 
appreciate it.

Thank you very much.

Laura Elisibeth Mazzeo
11th grade IB student

Date: 05/16/99 at 16:24:04
From: Doctor Anthony
Subject: Re: Probability dealing with combinations of dice

a)

The generating function is (x + x^2 + x^3 + x^4 + x^5 + x^6)^3

and you require the coefficient of x^9 and x^10

The term in brackets is  

           [x(1-x^6)]^3
           -------------  = x^3(1 - 3x^6 + 3x^12 - x^18)(1-x)^(-3)
              (1-x)^3

=  (x^3 - 3x^9 + 3x^15 - x^21)(1 + C(3,1)x + C(4,2)x^2 + C(5,3)x^3 +..

We must pick out the terms in x^9

    x^3.C(8,6)x^6 - 3x^9    Required coefficient = C(8,6) - 3

                                                 = 28 - 3 = 25

And the coefficient of x^10

    x^3.C(9,7)x^7 - 3x^9.C(3,1)x  Required coefficient = C(9,7) - 9

                                                       = 36 - 9 = 27

The number of outcomes with 3 dice is 6^3 = 216, so we get

Probability of a total of 9 = 25/216

Probability of a total of 10 = 27/216 =  1/8
            

b)

In the first situation we need the probability of AT LEAST one six in 
4 throws.

That is   1 - P(0)  where P(0) = prob. no six in 4 throws

    P(0) = (5/6)^4 = 0.48225

So the probability of at least 1 six = 0.51775  

For the second situation the probability of a double six is 1/36

Probability of no double six = 35/36

In 24 throws   P(0) = (35/36)^24  = 0.508596

and therefore the probability of at least one double six = 0.4914

The probabilities are 0.51775 for at least one six in 4 throws and a 
probability of 0.4914 for at least one double six in 24 throws.


c)

Probability of no six with 6 dice = (5/6)^6 = 0.334898
So the probability of at least one six = 0.6651

Probability of no six with 12 dice = (5/6)^12 = 0.1121566
Probability of 1 six with 12 dice = C(12,1)(1/6)(5/6)^11 = 0.26917597

So the probability of 0 or 1 six with 12 dice = 0.38133257

Probability of at least 2 sixes with 12 dice = 0.618667

Therefore the probability is 0.6651 for at least one six with 6 dice 
and the probability is 0.618667 for at least 2 sixes with 12 dice.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/