Problem:
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
Answer Choices:
A. 61β
B. 7213β
C. 367β
D. 245β
E. 92β
Solution:
Each roll of the three dice can be recorded as an ordered triple (a,b,c) of the three values appearing on the dice. There are 63 equally likely triples possible. For the sum of two of the values in the triple to equal the third value, the triple must be a permutation of one of the triples (1,1,2), (1,2,3),(1,3,4),(1,4,5),(1,5,6),(2,2,4),(2,3,5),(2,4,6), or (3,3,6). There are 3!=6 permutations of the values (a,b,c) when a,b, and c are distinct, and 3 permutations of the values when two of the values are equal. Thus there are 6β
6+3β
3=45 triples where the sum of two of the values equals the third. The requested probability is 6345β=(D)245ββ.
OR
There are 36 outcomes when a pair of dice are rolled, and the probability of rolling a total of 2,3,4,5, or 6 is 361β,362β,363β,364β, and 365β, respectively. The probability that another die matches this total is 61β, and there are 3 ways to choose the die that matches the total of the other two. Thus the requested probability is 3(361ββ
61β+362ββ
61β+363ββ
61β+364ββ
61β+365ββ
61β)=3β
3615ββ
61β=(D)245ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions