2020 AMC 10A Problem 25

P21-25~12 minPrint

Basic Probability Basic Counting Games & Processes Organized listing / direct enumeration Split into exhaustive cases and handle each

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7.$ Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?

Show solution without answering

Key insight

Keeping one die and rerolling two is best only when every pair of dice adds to 7 or more and the smallest die is 1, 2 or 3.

Solution

This question has two layers. First decide what Jason should do for any roll. Then count the rolls where his best move is to reroll exactly two dice. Write every chance out of $216 = 6^3$ so they are easy to compare.

Part 1: how good is each choice?

Reroll nothing. He wins only if the sum is already $7$ .
Reroll one die. He keeps two dice with sum $s$ . The new die must show $7 - s$ . That is possible when $s$ is $6$ or less, and then the chance is $\frac16 = \frac{36}{216}$ .
Reroll two dice. He keeps one die showing $a$ . The two new dice must add to $7 - a$ . For example, if $a = 1$ they must add to $6$ , which happens in $5$ ways ( $1{+}5, 2{+}4, 3{+}3, 4{+}2, 5{+}1$ ) out of $36$ . In general there are $6 - a$ ways, so the chance is $\frac{6 - a}{36} = \frac{6(6 - a)}{216}$ . The best is $\frac{30}{216}$ , when $a = 1$ .
Reroll all three. Three dice add to $7$ in $15$ ways (such as $1{+}1{+}5$ in $3$ orders, $1{+}2{+}4$ in $6$ orders, and so on). The chance is $\frac{15}{216}$ .

Part 2: what is the best move for a given roll?

5. If some two of his dice add to $6$ or less, keeping them and rerolling one gives $\frac{36}{216}$ , which beats every two-dice reroll (at most $\frac{30}{216}$ ). If the sum is already $7$ , he simply stops. Either way he does not reroll two dice.
6. Otherwise every pair adds to $7$ or more. Then rerolling one die is useless. He compares keeping his smallest die $a$ against rerolling all three:
- $a = 1$ : $\frac{30}{216}$ ; $a = 2$ : $\frac{24}{216}$ ; $a = 3$ : $\frac{18}{216}$ . These all beat $\frac{15}{216}$ , so he rerolls two.
- $a = 4$ : $\frac{12}{216}$ ; $a = 5$ : $\frac{6}{216}$ . These lose to $\frac{15}{216}$ , so he rerolls all three.
7. So he rerolls exactly two dice when the smallest die is $1$ , $2$ or $3$ and the two smallest dice add to $7$ or more.

Part 3: count those rolls. Sort the dice as $a \le b \le c$ . A set with all three numbers different can appear in $6$ orders; one with two equal numbers in $3$ orders.

$a = 1$ : need $b \ge 6$ . Only $(1, 6, 6)$ : $3$ rolls.
$a = 2$ : need $b \ge 5$ . $(2,5,5)$ , $(2,5,6)$ , $(2,6,6)$ : $3 + 6 + 3 = 12$ rolls.
$a = 3$ : need $b \ge 4$ . $(3,4,4)$ , $(3,4,5)$ , $(3,4,6)$ , $(3,5,5)$ , $(3,5,6)$ , $(3,6,6)$ : $3 + 6 + 6 + 3 + 6 + 3 = 27$ rolls.
Total: $3 + 12 + 27 = 42$ rolls out of $216$ . The chance is $\frac{42}{216} = \frac{7}{36}$ .

The answer is $\boxed{\textbf{(A)}\ \frac{7}{36}}$ .

Why this works

"Plays to win" problems have two steps: work out the best move for every possible situation, then count the situations that lead to the move asked about. Putting all the chances over the same bottom ( $216$ ) makes the comparisons easy. When counting dice rolls, list sorted groups and then multiply by the number of orders.

Concise solution

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

First evaluate each option's chance of winning.

Reroll none: wins only if the sum is already $7$ .
Reroll one die, keeping two dice with sum $s$ : the rerolled die must show $7 - s$ , possible exactly when $s \le 6$ , with probability $\frac16 = \frac{36}{216}$ .
Reroll two dice, keeping one die showing $a$ : the two new dice must sum to $7 - a$ , which has $6 - a$ outcomes, probability $\frac{6-a}{36}$ ; at best $\frac{5}{36} = \frac{30}{216}$ when $a = 1$ .
Reroll all three: the number of ways three dice sum to $7$ is $\binom{6}{2} = 15$ , probability $\frac{15}{216}$ .

So if some pair of dice sums to at most $6$ (which includes every roll with total $7$ ), Jason keeps that pair and rerolls one die or none; he never rerolls two. If every pair sums to at least $7$ , he compares keeping his smallest die $a$ , worth $\frac{6(6-a)}{216}$ , against $\frac{15}{216}$ : keeping wins when $6 - a \ge 3$ , i.e. $a \le 3$ , and rerolling all wins when $a \ge 4$ .

Hence he rerolls exactly two dice when the smallest die $a \le 3$ and the two smaller dice sum to at least $7$ . With dice sorted $a \le b \le c$ :

$a = 1$ : need $b \ge 6$ : $(1,6,6)$ , $3$ orderings.
$a = 2$ : need $b \ge 5$ : $(2,5,5)$ , $(2,5,6)$ , $(2,6,6)$ : $3 + 6 + 3 = 12$ .
$a = 3$ : need $b \ge 4$ : $(3,4,4)$ , $(3,4,5)$ , $(3,4,6)$ , $(3,5,5)$ , $(3,5,6)$ , $(3,6,6)$ : $3 + 6 + 6 + 3 + 6 + 3 = 27$ .

Total $3 + 12 + 27 = 42$ of $216$ equally likely rolls: $\frac{42}{216} = \frac{7}{36}$ .

The answer is $\boxed{\textbf{(A)}\ \frac{7}{36}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Intermediate Counting & Probability, Chapter 13: Events With States (§13.4)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)