2020 AMC 12A Problem 23

Question

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?

\dfrac{7}{36}

\dfrac{5}{24}

\dfrac{2}{9}

\dfrac{17}{72}

\dfrac{1}{4}

Answer: A

Concepts:dice (probability)optimization casework

Difficulty rating: 2270

Small hint:

Compare the win chance from rerolling one die (keep two summing to at most

6

) with rerolling two (keep one small die)

Big hint:

Rerolling exactly two is best when no two dice sum to

6

or less, yet the smallest die is small enough to beat rerolling all three

Solution:

Rerolling one die, keeping two dice that sum to

s,

wins with probability

\tfrac16

when

s \le 6

and

0

otherwise. Rerolling two dice, keeping a die of value

v,

wins with probability equal to the number of ways two dice sum to

7 - v,

over

36;

this is largest when

v

is smallest. For

v=1,2,3,

these probabilities are

\tfrac5{36},\tfrac4{36},\tfrac3{36},

all greater than the reroll-all probability

\tfrac{15}{216}=\tfrac{5}{72};

for

v\ge4,

rerolling all is better.

Rerolling exactly two dice is strictly best precisely when the two smallest dice sum to at least

7

(so rerolling one cannot reach

7

) while the smallest die is

1, 2,

or

3

(so keeping it beats rerolling all three).

Sort the roll as

u\le v\le w.

If

u=1,

the only possibility is

(1,6,6),

with

3

orderings. If

u=2,

the possibilities

(2,5,5),(2,5,6),(2,6,6)

have

3+6+3=12

orderings. If

u=3,

choose

v,w

with repetition from

\{4,5,6\};

the six resulting triples have

3+6+6+3+6+3=27

orderings. Thus there are

3+12+27=42

qualifying ordered rolls out of

216,

a probability of

\dfrac{42}{216} = \dfrac{7}{36}.

Thus, A is the correct answer.