Difference between revisions of "1993 AIME Problems/Problem 11"
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==Solution 2 (Kind of Bashy but very nice) == | ==Solution 2 (Kind of Bashy but very nice) == | ||
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− | + | In order to begin this problem, we need to calculate the probability that Alfred will win on the first round. | |
− | < | + | Because he goes first, Alfred has a <math>\frac{1}{2}</math> chance of winning (getting heads) on his first flip. |
− | + | Then, Bonnie, who goes second, has a <math>\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}</math>, chance of winning on her first coin toss. | |
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+ | Therefore Alfred’s chance of winning on his second flip is <math>\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}</math> | ||
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+ | |||
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+ | From this, we can see that Alfred’s (who goes first) chance of winning the first round is: <math>\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots = \frac{2}{3}</math> | ||
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+ | Bonnie’s (who goes second) chance of winning the first round is then <math>1 - \frac{2}{3} = \frac{1}{3}</math>. | ||
This means that the person who goes first has a <math>\frac{2}{3}</math> chance of winning the round, while the person who goes second has a <math>\frac{1}{3}</math> chance of winning. | This means that the person who goes first has a <math>\frac{2}{3}</math> chance of winning the round, while the person who goes second has a <math>\frac{1}{3}</math> chance of winning. | ||
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+ | |||
Now, through casework, we can calculate Alfred’s chance of winning the second round. | Now, through casework, we can calculate Alfred’s chance of winning the second round. | ||
− | Case 1: Alfred wins twice; <math>\frac{2}{3}</math> | + | Case 1: Alfred wins twice; <math>\frac{2}{3} \times \frac{1}{3}</math> (Bonnie goes first this round) <math>=\frac{2}{9}</math>. |
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+ | Case 2: Alfred loses the first round, but wins the second; <math>\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}</math>. | ||
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+ | Adding up the cases, we get <math>\frac{2}{9} + \frac{2}{9} = \frac{4}{9}</math>. | ||
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+ | Alfred, therefore, has a <math>\frac{4}{9}</math> of winnning the second round, and Bonnie has a <math>1-\frac{4}{9} = \frac{5}{9}</math> chance of winning this round. | ||
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+ | |||
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+ | From here, it is not difficult to see that the probabilities alternate in a pattern. | ||
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+ | Make <math>A</math> the probability that Alfred wins a round. | ||
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+ | The chances of Alfred and Bonnie, respectively, winning the first round are <math>A</math> and <math>A - \frac{1}{3}</math>, which can be written as <math>2A - \frac{1}{3} = 1</math> | ||
− | + | The second round’s for chances are <math>A</math> and <math>A + \frac{1}{9}</math>, which can also be written as <math>2A + \frac{1}{9}</math> | |
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− | From | + | From this, we can conclude that for the <math>n</math>th even round, the probability that Alfred (<math>A</math>) wins can be calculated through the equation <math>2A + \frac{1}{3^n} = 1</math>. |
− | + | Solving the equation for <math>n = 6</math>, we get <math>A = \frac{364}{729}</math>. | |
− | + | <math>364 + 729 = 1093</math>. | |
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− | + | So our answer is <math>\boxed{093}</math>. | |
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=10|num-a=12}} | {{AIME box|year=1993|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:25, 23 August 2020
Problem
Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is , where and are relatively prime positive integers. What are the last three digits of ?
Solution
The probability that the th flip in each game occurs and is a head is . The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is , and the probability of the second person winning is .
Let be the probability that Alfred wins the th game, and let be the probability that Bonnie wins the th game.
If Alfred wins the th game, then the probability that Alfred wins the th game is . If Bonnie wins the th game, then the probability that Alfred wins the th game is .
Thus, .
Similarly, .
Since Alfred goes first in the st game, .
Using these recursive equations:
Since , .
Solution 2 (Kind of Bashy but very nice)
In order to begin this problem, we need to calculate the probability that Alfred will win on the first round.
Because he goes first, Alfred has a chance of winning (getting heads) on his first flip.
Then, Bonnie, who goes second, has a , chance of winning on her first coin toss.
Therefore Alfred’s chance of winning on his second flip is
From this, we can see that Alfred’s (who goes first) chance of winning the first round is:
Bonnie’s (who goes second) chance of winning the first round is then .
This means that the person who goes first has a chance of winning the round, while the person who goes second has a chance of winning.
Now, through casework, we can calculate Alfred’s chance of winning the second round.
Case 1: Alfred wins twice; (Bonnie goes first this round) .
Case 2: Alfred loses the first round, but wins the second; .
Adding up the cases, we get .
Alfred, therefore, has a of winnning the second round, and Bonnie has a chance of winning this round.
From here, it is not difficult to see that the probabilities alternate in a pattern.
Make the probability that Alfred wins a round.
The chances of Alfred and Bonnie, respectively, winning the first round are and , which can be written as
The second round’s for chances are and , which can also be written as
From this, we can conclude that for the th even round, the probability that Alfred () wins can be calculated through the equation .
Solving the equation for , we get .
.
So our answer is .
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.