AMC 10 · 2020 · #25

Grade 7 probability

probability-basicexpected-valuecaseworksystematic-enumerationconditional-probability caseworkidentify-subproblemscomplementary-counting ↑ Prerequisites: probability-basicexpected-value

📏 Long solution 💡 4 insights

Problem

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?

$\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}$

Pick an answer.

(A)

$\frac{7}{36}$

(B)

$\frac{5}{24}$

(C)

$\frac{2}{9}$

(D)

$\frac{17}{72}$

(E)

$\frac{1}{4}$

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Toolkit + CCSS Solution

Understand

Restated: Jason rolls three fair six-sided dice, sees the result, then chooses which subset of the three to reroll so as to maximize his probability of ending with sum exactly $7$. What is the probability — over the random initial roll — that the optimal subset to reroll contains exactly two dice?

Givens: Three fair standard six-sided dice; $6^3 = 216$ equally likely initial rolls; After seeing the rolls, Jason picks a subset (size $0, 1, 2,$ or $3$) to reroll; He wins iff the final sum is exactly $7$; He plays optimally — picks the subset that maximizes win probability; Answer choices: (A) $\tfrac{7}{36}$, (B) $\tfrac{5}{24}$, (C) $\tfrac{2}{9}$, (D) $\tfrac{17}{72}$, (E) $\tfrac{1}{4}$

Unknowns: Probability that the optimal action is to reroll exactly $2$ dice

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #7 (Subproblems): four strategies (reroll $0, 1, 2,$ or $3$) — compute each one's win probability as a separate subproblem, then compare. Tool #9 (Easier Problem): the count of two-dice sums equaling $s$ is a tiny lookup ($s-1$ for $s \le 7$). Tool #2 (Systematic List): enumerate sorted triples $(a, b, c)$ with $a \le b \le c$ satisfying the derived conditions, multiplying by permutation counts. Tool #3 (Eliminate Possibilities): inequality reroll-$2$ vs. reroll-$1$ rules out a whole family of states; another inequality vs. reroll-$3$ restricts the kept die to $a \in \{1,2,3\}$.

Execute — Answer: A

#7 Identify Subproblems 7.SP.C.8 Step 1

Compute each strategy's win probability.
Let the initial roll be $(d_1, d_2, d_3)$, sorted as $a \le b \le c$.
Reroll $0$ (keep all): win iff $a+b+c = 7$ exactly, otherwise probability $0$.
Reroll $3$ (keep none): count ordered triples summing to $7$.
Unordered partitions of $7$ into three dice values: $\{1,1,5\}\;(3 \text{ perms}), \{1,2,4\}\;(6), \{1,3,3\}\;(3), \{2,2,3\}\;(3)$, total $15$.
Probability $= 15/216 = 5/72$.

$$P(\text{reroll 3}) = \dfrac{15}{216} = \dfrac{5}{72}$$

💡 List the four unordered triples that sum to $7$ and count permutations.

#7 Identify Subproblems 7.SP.C.7 Step 2

Reroll $1$ die.
Suppose Jason keeps two dice with values summing to $s$.
The rerolled die must show $7 - s$, which is possible iff $1 \le 7-s \le 6$, i.e.
$1 \le s \le 6$.
When possible, the win probability is $1/6$.
Best pair to keep: the one with smallest sum, i.e.
the two smallest dice $(a, b)$.
So reroll-$1$ is viable iff $a + b \le 6$, with win probability $1/6$.
(If $a + b \ge 7$, no pair has sum $\le 6$, so reroll-$1$ wins with probability $0$.)

$$P(\text{reroll 1}) = \begin{cases} 1/6 & \text{if } a+b \le 6 \\ 0 & \text{if } a+b \ge 7 \end{cases}$$

💡 Keep the smallest two; the rerolled die has one winning face out of six.

#9 Solve an Easier Related Problem 7.SP.C.8 Step 3

Reroll $2$ dice.
Keep one die of value $k$; the rerolled pair must sum to $7-k$.
The number of ordered pairs from two dice summing to $s$ is $s-1$ for $s \le 7$ (a tiny easier subproblem).
For $k \ge 1$, $7-k \le 6$, so the count is $(7-k)-1 = 6-k$ out of $36$.
To maximize, keep the smallest die $a$: win probability $= (6-a)/36$, for $a \in \{1, 2, 3, 4, 5\}$ (probability $0$ if $a = 6$).

$$P(\text{reroll 2}) = \dfrac{6 - a}{36}$$

💡 Keep the smallest die so the pair has the widest target sum.

#3 Eliminate Possibilities 7.RP.A.3 Step 4

Filter 1: reroll $2$ must beat reroll $1$.
$\dfrac{6-a}{36}$ at best is $\dfrac{5}{36}$ (when $a=1$), which is strictly less than $\dfrac{6}{36} = \dfrac{1}{6}$.
So whenever reroll-$1$ is viable (i.e.
$a+b \le 6$), Jason prefers it.
Hence reroll-$2$ is optimal only when reroll-$1$ is NOT viable, i.e.
$a + b \ge 7$.

$$\dfrac{6-a}{36} < \dfrac{6}{36} = \dfrac{1}{6} \;\Rightarrow\; \text{need } a+b \ge 7$$

💡 If reroll-$1$ is on the table, it always wins — so we need to kill it first.

#3 Eliminate Possibilities 7.RP.A.3 Step 5

Filter 2: reroll $2$ must beat reroll $3$.
$\dfrac{6-a}{36} > \dfrac{5}{72}$ becomes $\dfrac{2(6-a)}{72} > \dfrac{5}{72}$, i.e.
$12 - 2a > 5$, i.e.
$a < 3.5$.
So $a \in \{1, 2, 3\}$.

$$a \in \{1, 2, 3\}$$

💡 Bigger kept die hurts — too small a target sum gives too few rerolled pairs.

#3 Eliminate Possibilities 7.SP.C.8 Step 6

Filter 3: reroll $2$ must beat reroll $0$.
If $a + b + c = 7$ initially, reroll-$0$ wins outright.
But if $a+b \ge 7$ (from Filter 1) and $a \le b \le c$, then $a+b+c \ge 7+c \ge 7+b \ge 7+a \ge 8 > 7$.
So Filter 1 automatically rules out initial sum $= 7$.
No extra work needed.

$$a + b \ge 7 \;\Rightarrow\; a+b+c \ge 8 \;\Rightarrow\; \text{initial sum} \ne 7$$

💡 Once $a+b \ge 7$, the total is at least $8$, so reroll-$0$ wins with probability $0$.

#2 Make a Systematic List 7.SP.C.8 Step 7

Enumerate sorted $(a, b, c)$ with $a \le b \le c$ satisfying $a \in \{1,2,3\}$ AND $a+b \ge 7$.
Case $a=1$: need $b \ge 6$, so $b = c = 6$.
Triple $(1,6,6)$ — $3$ orderings.
Case $a=2$: need $b \ge 5$.
Sub-case $b=5$: $c \in \{5, 6\}$ — $(2,5,5)$ has $3$ orderings, $(2,5,6)$ has $3! = 6$.
Sub-case $b=6$: $c=6$, $(2,6,6)$ has $3$ orderings.
Sub-total $a=2$: $3+6+3 = 12$.
Case $a=3$: need $b \ge 4$.
Sub-case $b=4$: $c \in \{4,5,6\}$ — $(3,4,4): 3, (3,4,5): 6, (3,4,6): 6$.
Sub-case $b=5$: $c \in \{5,6\}$ — $(3,5,5): 3, (3,5,6): 6$.
Sub-case $b=6$: $c=6$, $(3,6,6): 3$.
Sub-total $a=3$: $3+6+6+3+6+3 = 27$.

$$|a=1|=3, \;\; |a=2|=12, \;\; |a=3|=27$$

💡 Systematic by $a$, then $b$, then $c$ — each gives a small number of multisets.

#7 Identify Subproblems 7.SP.C.7 Step 8

Total favorable ordered triples: $3 + 12 + 27 = 42$.
Probability $= \dfrac{42}{216} = \dfrac{7}{36}$, choice (A).

$$P = \dfrac{42}{216} = \dfrac{7}{36} \Rightarrow \textbf{(A)}$$

💡 Divide favorable by total — simplify the fraction.

[1] #7 7.SP.C.8 Compute each strategy's win probability. Let the initial roll be $(d_1, d_2, d_3

[2] #7 7.SP.C.7 Reroll $1$ die. Suppose Jason keeps two dice with values summing to $s$. The rer

[3] #9 7.SP.C.8 Reroll $2$ dice. Keep one die of value $k$; the rerolled pair must sum to $7-k$.

[4] #3 7.RP.A.3 Filter 1: reroll $2$ must beat reroll $1$. $\dfrac{6-a}{36}$ at best is $\dfrac{

[5] #3 7.RP.A.3 Filter 2: reroll $2$ must beat reroll $3$. $\dfrac{6-a}{36} > \dfrac{5}{72}$ bec

[6] #3 7.SP.C.8 Filter 3: reroll $2$ must beat reroll $0$. If $a + b + c = 7$ initially, reroll-

[7] #2 7.SP.C.8 Enumerate sorted $(a, b, c)$ with $a \le b \le c$ satisfying $a \in \{1,2,3\}$ A

[8] #7 7.SP.C.7 Total favorable ordered triples: $3 + 12 + 27 = 42$. Probability $= \dfrac{42}{2

Review

Reasonableness: Quick sanity. $42/216$ simplifies via $\gcd(42, 216) = 6$ to $7/36$. The cases are mutually exclusive (a sorted triple has one $a$ value), so no double-counting. Spot-check $(1,6,6)$: $a=1$, reroll-2 keeps the $1$, need pair to sum to $6$, which has $5$ ordered ways ($(1,5),(2,4),(3,3),(4,2),(5,1)$), prob $5/36 \approx 0.139$. Reroll-$1$ alternative would keep $(6,6)$ summing to $12 > 6$, infeasible — confirming reroll-$1$ ruled out. Reroll-$3$ gives $5/72 \approx 0.069$, lower than $5/36$. Reroll-$0$ has sum $13 \ne 7$, gives $0$. So reroll-$2$ wins. Spot-check $(4,5,6)$: $a=4 \notin \{1,2,3\}$ — so reroll-$2$ NOT optimal here. Reroll-$1$ feasible? $a+b = 9 > 6$, infeasible. Reroll-$3$ gives $5/72$. Reroll-$2$ keeping $4$ gives $(6-4)/36 = 2/36 = 4/72 < 5/72$ — so reroll-$3$ wins. Confirms filter $a \le 3$.

Alternative: Tool #10 (Physical Representation): roll three colored dice many times, work out the optimal strategy by hand for each outcome (or look up via a precomputed table), and tally how many of $216$ outcomes land in the reroll-$2$ bucket. A computer simulation gives the same $42/216 = 7/36$. The advantage of the analytic route is that the two filters ($a+b \ge 7$ and $a \le 3$) pin down the small region quickly.

CCSS standards used (min grade 7)

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems (Comparing win probabilities $\frac{6-a}{36}, \frac{1}{6}, \frac{5}{72}$ via a common denominator to derive the conditions $a+b \ge 7$ and $a \le 3$.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Setting up the probability model for a uniformly random die ($P = 1/6$ per face) and using it to value each strategy.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting ordered pairs of dice summing to a target (e.g. $6-a$ out of $36$) and counting ordered triples summing to $7$ ($15$ out of $216$) by enumerating multisets and multiplying by permutation counts.)

⭐ This AMC 10 problem only needs Grade 7 dice-probability and inequality comparisons you already know — rerolling $2$ dice is optimal only when (1) every pair of two dice exceeds sum $6$ ($a+b \ge 7$) and (2) the smallest die is in $\{1,2,3\}$; listing the $42$ such ordered triples out of $216$ gives $\frac{42}{216} = \frac{7}{36}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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