2020 AMC 10A Problems/Problem 25

The following problem is from both the 2020 AMC 12A #23 and 2020 AMC 10A #25, so both problems redirect to this page.

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?

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

Solution 1

Consider the probability that rolling two dice gives a sum of $s$ , where $s \leq 7$ . There are $s - 1$ pairs that satisfy this, namely $(1, s - 1), (2, s - 2), ..., (s - 1, 1)$ , out of $6^2 = 36$ possible pairs. The probability is $\frac{s - 1}{36}$ .

Therefore, if one die has a value of $a$ and Jason rerolls the other two dice, then the probability of winning is $\frac{7 - a - 1}{36} = \frac{6 - a}{36}$ .

In order to maximize the probability of winning, $a$ must be minimized. This means that if Jason rerolls two dice, he must choose the two dice with the maximum values.

Thus, we can let $a \leq b \leq c$ be the values of the three dice, which we will call $A$ , $B$ , and $C$ respectively. Consider the case when $a + b < 7$ . If $a + b + c = 7$ , then we do not need to reroll any dice. Otherwise, if we reroll one die, we can roll dice $C$ in the hope that we get the value that makes the sum of the three dice $7$ . This happens with probability $\frac16$ . If we reroll two dice, we will roll $B$ and $C$ , and the probability of winning is $\frac{6 - a}{36}$ , as stated above.

However, $\frac16 > \frac{6 - a}{36}$ , so rolling one die is always better than rolling two dice if $a + b < 7$ .

Now consider the case where $a + b \geq 7$ . Rerolling one die will not help us win since the sum of the three dice will always be greater than $7$ . If we reroll two dice, the probability of winning is, once again, $\frac{6 - a}{36}$ . To find the probability of winning if we reroll all three dice, we can let each dice have $1$ dot and find the number of ways to distribute the remaining $4$ dots. By stars and bars, there are ${6\choose2} = 15$ ways to do this, making the probability of winning $\frac{15}{6^3} = \frac5{72}$ .

In order for rolling two dice to be more favorable than rolling three dice, $\frac{6 - a}{36} > \frac5{72} \rightarrow a \leq 3$ .

Thus, rerolling two dice is optimal if and only if $a \leq 3$ and $a + b \geq 7$ . The possible triplets $(a, b, c)$ that satisfy these conditions, and the number of ways they can be permuted, are

$(3, 4, 4) \rightarrow 3$ ways.

$(3, 4, 5) \rightarrow 6$ ways.

$(3, 4, 6) \rightarrow 6$ ways.

$(3, 5, 5) \rightarrow 3$ ways.

$(3, 5, 6) \rightarrow 6$ ways.

$(3, 6, 6) \rightarrow 3$ ways.

$(2, 5, 5) \rightarrow 3$ ways.

$(2, 5, 6) \rightarrow 6$ ways.

$(2, 6, 6) \rightarrow 3$ ways.

$(1, 6, 6) \rightarrow 3$ ways.

There are $3 + 6 + 6 + 3 + 6 + 3 + 3 + 6 + 3 + 3 = 42$ ways in which rerolling two dice is optimal, out of $6^3 = 216$ possibilities, Therefore, the probability that Jason will reroll two dice is $\frac{42}{216} = \boxed{\textbf{(A) }\frac7{36}}$

Solution 2

We count the numerator. Jason will pick up no dice if he already has a 7 as a sum. We need to assume he does not have a 7 to begin with. If Jason decides to pick up all the dice to re-roll, by the stars and bars rule ways to distribute, ${n+k-1 \choose k-1}$ , there will be 2 bars and 4 stars(3 of them need to be guaranteed because a roll is at least 1) for a probability of $\frac{15}{216}=\frac{2.5}{36}$ . If Jason picks up 2 dice and leaves a die showing $k$ , he will need the other two to sum to $7-k$ . This happens with probability $\frac{6-k}{36}$ for integers $1 \leq k \leq 6$ . If the roll is not 7, Jason will pick up exactly one die to re-roll if there can remain two other dice with sum less than 7, since this will give him a $\frac{1}{6}$ chance which is a larger probability than all the cases unless he has a 7 to begin with. We have $\frac{1}{6} > \underline{\frac{5,4,3}{36}} > \frac{2.5}{36} > \frac{2,1,0}{36}.$ We count the underlined part's frequency for the numerator without upsetting the probability greater than it. Let $a$ be the roll we keep. We know $a$ is at most 3 since 4 would cause Jason to pick up all the dice. When $a=1$ , there are 3 choices for whether it is rolled 1st, 2nd, or 3rd, and in this case the other two rolls have to be at least 6(or he would have only picked up 1). This give $3 \cdot 1^{2} =3$ ways. Similarly, $a=2$ gives $3 \cdot 2^{2} =12$ because the 2 can be rolled in 3 places and the other two rolls are at least 5. $a=3$ gives $3 \cdot 3^{2} =27$ . Summing together gives the numerator of 42. The denominator is $6^3=216$ , so we have $\frac{42}{216}=\boxed{\textbf{(A) } \frac{7}{36}}$

Solution 3 (Illustration of Solution 2)

We have the following observations:

Jason rerolls exactly zero dice if and only if the initial sum is $7,$ in which the probability of winning is always $1.$

If rerolling exactly one die is possible to win, then the probability of winning is always $\frac16.$
The initial sum of the two other dice must be $2,3,4,5,$ or $6.$ In all cases, exactly $1$ of the $6$ possible outcomes of the die rerolled gives a win.

If Jason (re)rolls all three dice, then the probability of winning is always $\frac{\binom62}{6^3}=\frac{15}{216}=\frac{5}{72}.$

For the denominator, rolling three dice gives a total of $6^3=216$ possible outcomes.

For the numerator, this is the same as counting the ordered triples of positive integers $(a,b,c)$ for which $a+b+c=7.$ Suppose that $7$ balls are lined up in a row. There are $6$ gaps in between the balls, and placing dividers in $2$ of the gaps separates the balls into $3$ piles. From left to right, the numbers of balls in the piles match with $a,b,$ and $c,$ respectively. There are $\binom62=15$ ways to place the dividers. Note that the divider placements and the ordered triples have one-to-one correspondence, and all such ordered triples satisfy that $1\leq a,b,c\leq6.$

If rerolling exactly two dice is possible to win, then we can apply casework to the outcome of the remaining die: $\begin{array}{c|c|c} & & \\ [-2.5ex] \textbf{Remaining Die} & \textbf{Sum Needed for the Two Other Dice} & \textbf{Probability of Winning} \\ [0.5ex] \hline & & \\ [-2ex] 1 & 6 & 5/36 \\ 2 & 5 & 4/36 \\ 3 & 4 & 3/36 \\ 4 & 3 & 2/36 \\ 5 & 2 & 1/36 \end{array}$ Note that if Jason rerolls exactly two dice, then the probability of winning is at most $\frac{5}{36}.$

The optimal strategy is that:

If Jason needs to reroll at least zero, one, two, or three dice to win, then he rerolls exactly zero dice.

If Jason needs to reroll at least one, two, or three dice to win, then he rerolls exactly one die.

If Jason needs to reroll at least two or three dice to win, then he rerolls exactly two dice if and only if the probability of winning is greater than $\frac{5}{72}.$ The first three cases in the table above meet this requirement. We will analyze these cases by considering the initial outcomes of the two other dice:

$\begin{array}{c|c|c} & & \\ [-2.5ex] \textbf{Remaining Die} & \textbf{Initial Outcomes of the Two Other Dice} & \textbf{\# of Ways} \\ [0.5ex] \hline & & \\ [-2ex] 1 & \{4,5,6\} & \hspace{2mm}3\cdot3^2=27 \\ 2 & \{5,6\} & \hspace{2mm}3\cdot2^2=12 \\ 3 & \{6\} & 3\cdot1^2=3 \end{array}$ Since any of the three dice can be the remaining die, we need a factor of $3$ for the counts, as indicated in the third column of the table above.

Finally, the requested probability is $\frac{27+12+3}{216}=\frac{42}{216}=\boxed{\textbf{(A) } \frac{7}{36}}.$

~MRENTHUSIASM

Video Solution

https://youtu.be/B8pt8jF04ZM - Happytwin

https://youtu.be/3W4jOpCiBx8

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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All AMC 10 Problems and Solutions

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

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