2014 AMC 10A Problem 17

P16-20~4 minPrint

Basic Probability Basic Counting Split into exhaustive cases and handle each Exploit symmetry to reduce work or pair up objects

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?

Show solution without answering

Key insight

Choose which die shows the total (3 ways). The other two dice must add up to at most 6, and there are 15 ordered ways to do that.

Solution

Treat the dice as three different colours, say red, white and blue. Then there are $6 \cdot 6 \cdot 6 = 216$ rolls, all equally likely. Count the rolls where one die equals the sum of the other two.

Which die is the total? It can be the red, the white or the blue one: $3$ choices.
The other two dice. They must add up to the total die, and a die shows at most $6$ , so their sum is at most $6$ . Count the ordered ways for each sum:

sum	ways (first die, second die)	count
$2$	$(1,1)$	$1$
$3$	$(1,2), (2,1)$	$2$
$4$	$(1,3), (2,2), (3,1)$	$3$
$5$	$(1,4), (2,3), (3,2), (4,1)$	$4$
$6$	$(1,5), (2,4), (3,3), (4,2), (5,1)$	$5$

That is $1 + 2 + 3 + 4 + 5 = 15$ ways. Once these two dice are chosen, the total die is fixed.
3. No roll counted twice. Could one roll have the red die as the total and also the white die as the total? Then red $=$ white $+$ blue and white $=$ red $+$ blue, which forces blue to be $0$ . A die never shows $0$ , so this cannot happen.
4. Favourable rolls: $3 \cdot 15 = 45$ .
5. Probability: $\frac{45}{216}$ . Divide top and bottom by $9$ : $\frac{5}{24}$ .

The answer is $\boxed{\textbf{(D)}\ \dfrac5{24}}$ .

Why this works

"Two of them add to the third" really says the largest die is the sum of the other two. Choosing that die first, then counting pairs with a given sum, keeps the counting clean. Keep the dice ordered (different colours) both in the total $216$ and in the good cases, so both counts are measured the same way.

Alternative approach

Count sets of values first, then count how many ways each set can be arranged on the three dice.

Two equal small dice: $(1,1,2)$ , $(2,2,4)$ , $(3,3,6)$ . Each has $3$ arrangements (choose where the odd one out goes). That gives $3 \cdot 3 = 9$ .
Three different values: $(1,2,3)$ , $(1,3,4)$ , $(1,4,5)$ , $(1,5,6)$ , $(2,3,5)$ , $(2,4,6)$ . Each has $6$ arrangements. That gives $6 \cdot 6 = 36$ .
Total $9 + 36 = 45$ , and $\frac{45}{216} = \frac{5}{24}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

There are $6^3 = 216$ equally likely ordered rolls. Since every face is positive, if two dice add to the third, the third must be the largest value, so we count ordered triples in which one die equals the sum of the other two.

Step 1: choose which of the three dice is the "total." There are $3$ choices.

Step 2: the remaining two dice, in order, must add to that total $s$ , and $s \le 6$ . The number of ordered pairs of faces adding to $s$ is $s - 1$ :

total $s$	2	3	4	5	6
ordered pairs	1	2	3	4	5

That is $1 + 2 + 3 + 4 + 5 = 15$ ordered pairs.

Step 3: no roll is counted twice. If die $1$ equals die $2$ plus die $3$ and also die $2$ equals die $1$ plus die $3$ , then die $3$ would be $0$ , impossible.

Favorable rolls: $3 \cdot 15 = 45$ . The probability is
$\frac{45}{216} = \frac{5}{24}.$

The answer is $\boxed{\textbf{(D)}\ \dfrac5{24}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)