Difference between revisions of "2021 Fall AMC 10B Problems/Problem 20"
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Solution <math>1</math> is incorrect, it only considered the first round and didn't consider there might be second round or more. | Solution <math>1</math> is incorrect, it only considered the first round and didn't consider there might be second round or more. | ||
− | + | Solutions <math>2</math> and <math>3</math> uses [https://en.wikipedia.org/wiki/Bayes%27_theorem Bayes' theorem]. | |
<math>P \left( A | B \right) = \frac{P \left( B | A \right) P \left( A \right) }{P \left( B \right)} </math> | <math>P \left( A | B \right) = \frac{P \left( B | A \right) P \left( A \right) }{P \left( B \right)} </math> |
Revision as of 01:30, 13 June 2022
Contents
[hide]Problem 20
In a particular game, each of players rolls a standard -sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a given that he won the game?
Solution 1
Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a is just the probability that the highest roll in the first round is .
Let indicate the probability that event occurs. We find that ,
so
~kingofpineapplz
Solution 2 (Conditional Probability)
The conditional probability formula states that , where means A given B and means A and B. Therefore the probability that Hugo rolls a five given he won is , where A is the probability that he rolls a five and B is the probability that he wins. In written form,
The probability that Hugo wins is by symmetry since there are four people playing and there is no bias for any one player. The probability that he gets a 5 and wins is more difficult; we will have to consider cases on how many players tie with Hugo...
No Players Tie
In this case, all other players must have numbers from 1 through four.
There is a chance of this happening.
One Player Ties
In this case, there are ways to choose which other player ties with Hugo, and the probability that this happens is . The probability that Hugo wins on his next round is then because there are now two players rolling die.
Therefore the total probability in this case is .
Two Players Tie
In this case, there are ways to choose which other players tie with Hugo, and the probability that this happens is . The probability that Hugo wins on his next round is then because there are now three players rolling the die.
Therefore the total probability in this case is .
All Three Players Tie
In this case, the probability that all three players tie with Hugo is . The probability that Hugo wins on the next round is , so the total probability is .
Finally, Hugo has a probability of rolling a five himself, so the total probability is
Finally, the total probability is this probability divided by which is this probability times four; the final answer is
~KingRavi
~Edits by BakedPotato66
Solution 3
We use to refer to Hugo. We use to denote the outcome of Hugo's th toss. We denote by , , the other three players. We denote by the number of players among , , whose first tosses are 5. We use to denote the winner.
We have
Now, we compute .
We have The first equality follows from the law of total probability. The second equality follows from the property that Hugo's outcome is independent from other players' outcomes.
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Remark (Bayes' Theorem)
Solution is incorrect, it only considered the first round and didn't consider there might be second round or more.
Solutions and uses Bayes' theorem.
In this problem
Video Solution
Video Solution by Interstigation
~Interstigation
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.