Difference between revisions of "2020 AMC 10A Problems/Problem 25"
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==See Also== | ==See Also== | ||
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Revision as of 20:05, 1 February 2020
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 
Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?
Solution 1
Consider the probability that rolling two dice gives a sum of 
, where 
. There are 
pairs that satisfy this, namely 
, out of 
possible pairs. The probability is 
.
Therefore, if one dice has a value of 
and Jason rerolls the other two dice, then the probability of winning is 
.
In order to maximize the probability of winning, 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 
be the values of the three dice, which we will call 
, 
, and 
respectively. Consider the case when 
. If 
, then we do not need to reroll any dice. Otherwise,
if we reroll one dice, we can roll dice in the hope that we get the value that makes the sum of the three dice 
. This happens with probability 
. If we reroll two dice, we will roll and , and the probability of winning is 
, as stated above.
However, 
, so rolling one dice is always better than rolling two dice if .
Now consider the case where 
. Rerolling one dice will not help us win since the sum of the three dice will always be greater than . If we reroll two dice, the probability of winning is, once again, . To find the probability of winning if we reroll all three dice, we can let each dice have 
dot and find the number of ways to distribute the remaining 
dots. By Stars and Bars, there are 
ways to do this, making the probability of winning 
.
In order for rolling two dice to be more favorable than rolling three dice, 
.
Thus, rerolling two dice is optimal if and only if 
and . The possible triplets 
that satisfy these conditions, and the number of ways they can be permuted, are
ways.
ways.
ways.
ways.
ways.
ways.
ways.
ways.
ways.
ways.
There are 
ways in which rerolling two dice is optimal, out of 
possibilities, Therefore, the probability that Jason will reroll two dice is 
Another way to finish
As the denominator is very obviously 
, we consider the numerator. When 
, 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. This give 
ways.
Similarly, 
gives 
because the 2 can be rolled in 3 places and the other two rolls are at least 5.
gives 
.
Summing together gives the numerator of 42.
Video Solution
https://youtu.be/3W4jOpCiBx8 - 
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
