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Roll Two Dice and Use the Sum You Stay and Keep Both or You Roll Both Again and Try to Do Better

Written By Saturday, May 14, 2022 Add Comment Edit
The following problem is from both the 2020 AMC 12A #23 and 2020 AMC 10A #25, so both problems redirect to this page.

Contents

  • ane Problem
  • two Solution 1
  • 3 Solution ii (Illustration)
  • four Solution 3 (Similar to Solution 2)
  • 5 Solution 4
  • half-dozen Solution 5
  • vii Video Solutions
    • 7.1 Video Solution 1
    • 7.2 Video Solution 2 (Richard Rusczyk)
  • eight See Besides

Problem

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (mayhap empty, possibly all three dice) to reroll. Later rerolling, he wins if and simply 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 2 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 one

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

Therefore, if i die has a value of $a$ and Jason rerolls the other ii dice, so 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 2 dice with the maximum values.

Thus, we can let $a \leq b \leq c$ be the values of the 3 die. Consider the case when $a + b < 7$. If $a + b + c = 7$, and then we do not need to reroll any dice. Otherwise, if we reroll one die, we can reroll any 1 die 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 curlicue the two maximum dice, and the probability of winning is $\frac{6 - a}{36}$, equally stated to a higher place. Still, $\frac16 > \frac{6 - a}{36}$, then rolling one die is always amend than rolling 2 dice if $a + b < 7$.

At present consider the case where $a + b \geq 7$. Rerolling one die will not help us win since the sum of the iii dice will e'er be greater than $7$. If we reroll two dice, the probability of winning is, in one case over again, $\frac{6 - a}{36}$. To discover the probability of winning if we reroll all iii dice, we tin can let each die have $1$ dot and notice the number of ways to distribute the remaining $4$ dots. By stars and bars, in that location are ${6\choose2} = 15$ means to do this, making the probability of winning $\frac{15}{6^3} = \frac5{72}$.

In order for rolling two dice to exist more favorable than rolling 3 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 atmospheric condition, and the number of means they can be permuted, are

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

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

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

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

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

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

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

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

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

$(1, 6, 6) \rightarrow 3$ 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 volition reroll two die is $\frac{42}{216} = \boxed{\textbf{(A) } \frac{7}{36}}$.

~edits by eagleye

Solution 2 (Analogy)

We conclude all of the following after the initial roll:

  1. Jason rerolls exactly zero dice if and simply if the sum of the iii dice is $7,$ in which the probability of winning is ever $1.$
  2. If Jason rerolls exactly ane die, then the sum of the 2 other die must exist $2,3,4,5,$ or $6.$ The probability of winning is always $\frac16,$ as exactly $1$ of the $6$ possible outcomes of the die rerolled results in a win.
  3. If Jason rerolls exactly two dice, and then the event of the remaining die must exist $1,2,3,4,$ or $5.$ Applying casework to the remaining die produces the following table: \[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Remaining Die} & \textbf{Sum Needed for the Two Dice Rerolled} & \textbf{Probability of Winning} \\ [0.5ex] \hline & & \\ [-2ex] 1 & 6 & 5/36 \\ 2 & 5 & 4/36 \\ 3 & 4 & 3/36 \\ 4 & 3 & 2/36 \\ 5 & 2 & 1/36 \end{array}\] The probability of winning is at most $\frac{5}{36}.$
  4. If Jason (re)rolls all 3 dice, then the probability of winning is always $\frac{\binom62}{6^3}=\frac{15}{216}=\frac{5}{72}.$

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

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

The optimal strategy is that:

  • If Jason needs to reroll at least nil dice to win, and so he rerolls exactly zilch dice.
  • If Jason needs to reroll at least i die to win, then he rerolls exactly one die.
  • If Jason needs to reroll at least ii dice to win, then he rerolls exactly two dice if and only if the probability of winning is greater than $\frac{5}{72},$ the probability of winning for rerolling all 3 dice.

    The first three cases in the table to a higher place satisfy this requirement. Nosotros volition analyze these cases past considering the initial outcomes of the two dice rerolled. Note that any of the three die can exist the remaining die, so nosotros demand a cistron of $3$ for all counts in the tertiary column: \[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Remaining Die} & \textbf{Initial Outcomes of the Two Dice Rerolled} & \textbf{\# of Ways} \\ [0.5ex] \hline & & \\ [-2ex] 1 & \text{Both are in }\{6\}\text{, not necessarily distinct.} & 3\cdot1^2=3 \\ 2 & \text{Both are in }\{5,6\}\text{, not necessarily distinct.} & \hspace{2mm}3\cdot2^2=12 \\ 3 & \text{Both are in }\{4,5,6\}\text{, not necessarily distinct.} & \hspace{2mm}3\cdot3^2=27 \end{array}\]

Finally, the requested probability is \[\frac{3+12+27}{6^3}=\frac{42}{216}=\boxed{\textbf{(A) } \frac{7}{36}}.\]

~MRENTHUSIASM

Solution three (Like to Solution 2)

We count the numerator. Jason will option up no die if he already has a $7$ equally a sum. We need to assume he does non have a $7$ to begin with. If Jason decides to pick up all the die to re-roll, past the stars and confined dominion means to distribute, ${n+k-1 \choose k-1}$, there will exist $2$ 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 dice showing $k$, he volition 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 non $7$, Jason volition pick up exactly one die to re-roll if there tin remain ii other dice with sum less than $7$, since this will give him a $\frac{1}{6}$ take a chance which is a larger probability than all the cases unless he has a $7$ to begin with. We have \[\frac{1}{6} > \underline{\frac{5,4,3}{36}} > \frac{2.5}{36} > \frac{2,1,0}{36}.\] Nosotros count the underlined part'south frequency for the numerator without upsetting the probability greater than it. Permit $a$ exist the roll we go on. We know $a\leq3$ since $a=4$ would cause Jason to pick up all the dice. When $a=1$, at that place are $3$ choices for whether information technology is rolled $1$st, $2$nd, or $3$rd, and in this instance the other 2 rolls take to be at least $6$ (or he would take only picked up $1$). This give $3 \cdot 1^{2} =3$ means. 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 $6^3=216$, and so we have $\frac{42}{216}=\boxed{\textbf{(A) } \frac{7}{36}}$.

Solution 4

Note that Jason will curl $2$ dies if and merely if the pairwise sum of the dies is greater than $7$. Suppose that Jason doesn't whorl a $1$. This means that roll was $(1, 6, 6)$ which can exist bundled in $3$ ways. Suppose that Jason doesn't curl a $2$. This ways the roll was either $(2, 5/6, 5/6)$ which can be arranged in $12$ ways. Suppose that Jason doesn't roll a $3$. This ways that the roll was either $(3, 4/5/6, 4/5/6)$ which can be bundled in $27$ ways. Jason can't roll a $4, 5, 6$ because he can't get a pairwise sum of $7$ from those numbers. Thus he rolls two dies exactly $27 + 12 + 3 = 42$ out of the $6^3$ possible rolls. So the respond is $\frac{42}{216} = \boxed{\textbf{(A) } \frac{7}{36}}$.

~coolmath_2018

Solution five

We can quickly write out all possible rolls that result in a full of $7$, to get that at that place is a $\frac{5}{72}$ chance that rolling 3 dice will consequence in a full of $7$.

Merits i: Iff all of Jason's initial outcomes are $4$, $5$ or $6$, Jason is best off rerolling everything.

Proof of if role: If Jason rerolls ii dice, he would like for the sum of the 2 outcomes to exist $7-\text{(The initial outcome of the untouched dice)}$. Since in this case, every dice's initial outcome is $4$, $5$ or $6$, this value is $1$, $2$ or $3$. By basic die probability, the probability that they add to one is $0$; the probability that they add to two is $\frac{1}{36}$; the probability that they add to three is $\frac{1}{18}$. Each of these probabilities is less than $\frac{5}{72}$, (the probability of winning by rerolling everything) so Jason should not reroll one dice in this scenario. Rolling one die is also nonsense in this situation, since the other 2 dice volition already add up to something greater than $7$.

Proof of but-if office: If Jason rolls a $1$, $2$ or $3$, even just once, then, the probability that rerolling the other ii dice will consequence in win is $\frac{5}{36}$, $\frac{1}{9}$ and $\frac{1}{12}$, past basic die probability. Each of these is greater than $\frac{5}{72}$, so if Jason rolls a $1$, $2$ or $3$, he should not reroll everything.

Claim 2: Iff the sum of any two of the initial dice is less than $7$, Jason is best off rerolling one or zero die.

Proof of if part: If the sum of any two of the initial dice is less than $7$, when the other die is rolled, in that location is e'er a fixed $\frac{1}{6}$ hazard that this results in a win for Jason (or, he his initial coil already won, then he doesn't have to gyre anything). Rerolling that one dice is better than rerolling all three dice, since, as we have seen there is a $\frac{5}{72}$ chance that Jason wins from that. Rerolling that one die is besides meliorate than rerolling two dice: by bones die probability, if Jason wishes for the sum of two rerolled die to equal one value, the maximum probability this happens is $\frac{1}{6}$, and this is when he wishes for the sum to be $7$; however, since he wishes for the sum of all three dice to be $7$, this will never happen, so the probability that rerolling two dice results in a win is ever less than $\frac{1}{6}$.

Proof of only-if part: If the sum of each pair of the three dice is at least $7$, it would brand no sense to reroll ane die, since the final sum of all three dice will e'er be greater than $7$. As well, if the sum of each pair of the three dice is at least $7$, the total initial sum must also be greater than $7$, so Jason should non reroll zero dice.

From having shown these 2 claims, we know that Jason should reroll two dice iff at least one of them is a $1$, $2$ or $3$, and iff the sum of any two of the initial dice is at to the lowest degree $7$. We tin can write out all the possibilities of initial rolls, business relationship for their permutations, and add everything together to see that there are $42$ full outcomes, where Jason should reroll two die. Therefore, our terminal answer is $\frac{42}{216} = \boxed{\textbf{(A) } \frac{7}{36}}$.

~ ihatemath123

Video Solutions

Video Solution ane

https://youtu.exist/B8pt8jF04ZM - Happytwin

Video Solution 2 (Richard Rusczyk)

https://artofproblemsolving.com/videos/amc/2020amc10a/516

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted past the Mathematical Association of America'south American Mathematics Competitions. AMC logo.png

schaefernouse1990.blogspot.com

Source: https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_25

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