Difference between revisions of "2018 AMC 10B Problems/Problem 9"
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It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s. | It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s. | ||
− | So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible | + | So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible roll. The pairs with the same probability are as follows: |
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)... | (7, 42), (8, 41), (9, 40), (10, 39), (11, 38)... |
Revision as of 14:57, 24 November 2020
Contents
Problem
The faces of each of standard dice are labeled with the integers from to . Let be the probabilities that when all dice are rolled, the sum of the numbers on the top faces is . What other sum occurs with the same probability as ?
Solution 1
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.
So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible roll. The pairs with the same probability are as follows:
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...
However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is , and we are done.
Written By: Archimedes15
Solution 2
Let's call the unknown value . By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and . So,
and our answer is By: Soccer_JAMS
Solution 3
For the sums to have equal probability, the average sum of both sets of dies has to be . Since having is similar to not having , you just subtract 10 from the expected total sum. so the answer is
By: epicmonster
Solution 4
The expected value of the sums of the die rolls is , and since the probabilities should be distributed symmetrically on both sides of , the answer is , which is .
By: dajeff
Solution 5
Another faster and easier way of doing this, without using almost any math at all, is realizing that the possible sums are . By symmetry, (and doing a few similar problems in the past), you can realize that the probability of obtaining is the same as the probability of obtaining , and on and on and on. This means that , and thus the correct answer is .
By: fhdsaukfaioifk
Note
Calculating the probability of getting a sum of is also easy. There are cases:
Case :
cases
Case :
cases
Case :
cases
The probability is .
Calculating :
Therefore, the probability is
~Zeric Hang (Main writer) and fhdsaukfaioifk (Editor)
Related Problems
There is a similar to problem 11 of the AMC 10A in the same year, which is almost an replica of the problem mentioned by Zeric Hang in the Note section: https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11
Video Solution
~savannahsolver
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.