Advanced book on Mathematics Olympiad

6.3 Probability 311

Example.Show that the probability of getting a six when a die is rolled four times is
greater than the probability of getting a double six when two dice are rolled 24 times.

Here is a brief history of the problem. Chevalier de Méré, a gambler of the seventeenth
century, observed while gambling that the odds of getting a six when rolling a die four
times seem to be greater than^12 , while the odds of getting a double six when rolling two
dice 24 times seem to be less than^12. De Méré thought that this contradicted mathematics
because^46 =^2436. He posed this question to B. Pascal and P. Fermat. They answered the
question...and probability theory was born. Let us see the solution.

Solution.The probability that a six does not occur when rolling a die four times is(^56 )^4 ,
and so the probability that a six occurs is 1−(^56 )^4 ≈ 0 .5177. The probability that a double
six does not occur when rolling two dice 24 times is(^3536 )^24 , whence the probability that
a double six occurs is 1−(^3536 )^24 ≈ 0 .4914. The second number is smaller. 

Example.Considernindistinguishable balls randomly distributed inmboxes. What is
the probability that exactlykboxes remain empty?

Solution.Number the boxes 1, 2 ,...,mand letxibe the number of balls in theith box.
The number of ways one can distributenballs inmboxes is equal to the number of
nonnegative integer solutions to the equation

x 1 +x 2 +···+xm=n.

These solutions were counted in problem 887 from Section 6.2.3 and were found to be(
m+n− 1
m− 1

)

. This is the total number of cases.
If we fixkboxes and distribute the balls in the remainingn−kboxes such that each
box receives at least one ball, then the number of ways to do this is equal to the number
of positive integer solutions to the equation

x 1 +x 2 +···+xm−k=n.

This was also computed in one of the examples from Section 6.2.3 and was shown to be(
n− 1
m−k− 1

)

. Thekboxes can be chosen in

(m

k

)

ways. We find the number of favorable cases

to be

(m

k

)( n− 1

m−k− 1

)

. The required probability is therefore
(m
k

)(n− 1

m−k− 1

)

(m+n− 1

m− 1

). 

If you grabnballs and place them one at a time randomly in boxes, you will find that
they do not seem to fit the probabilities just calculated. This is because they are not really
indistinguishable balls: the order of placement and the fact that they are macroscopic
balls makes them distinguishable. However, the example above does correspond to a real