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. 61​
B. 7213​
C. 367​
D. 245​
E. 92​
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 6⋅6+3⋅3=45 triples where the sum of two of the values equals the third. The requested probability is 6345​=(D)245​​.
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 361​,362​,363​,364​, and 365​, respectively. The probability that another die matches this total is 61​, and there are 3 ways to choose the die that matches the total of the other two. Thus the requested probability is 3(361​⋅61​+362​⋅61​+363​⋅61​+364​⋅61​+365​⋅61​)=3⋅3615​⋅61​=(D)245​​.
The problems on this page are the property of the MAA's American Mathematics Competitions