Difference between revisions of "2006 AIME II Problems/Problem 10"
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Thus, <math>319 + 512 = \boxed{831} </math>. | Thus, <math>319 + 512 = \boxed{831} </math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | After the first game, there are <math>10</math> games we care about-- those involving <math>A</math> or <math>B</math>. There are <math>3</math> cases of these <math>10</math> games: <math>A</math> wins more than <math>B</math>, <math>B</math> wins more than <math>A</math>, or <math>A</math> and <math>B</math> win the same number of games. Also, there are <math>2^{10} = 1024</math> total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs <math>\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252</math> times, by a special case of Vandermonde's Identity (see [[Combinatorial identity]]). There are therefore <math>\frac{1024-252}{2} = 386</math> possibilities for each of the other two cases. | ||
+ | |||
+ | If <math>B</math> has more wins than <math>A</math> in its <math>5</math> remaining games, then <math>A</math> cannot beat <math>B</math> overall. However, if <math>A</math> has more wins or if <math>A</math> and <math>B</math> are tied, <math>A</math> will beat <math>B</math> overall. Therefore, out of the <math>1024</math> possibilites, <math>386+252 = 638</math> ways where <math>A</math> wins, so the desired probability is <math>\frac{638}{1024} = \frac{319}{512}</math>, and <math>m+n=\boxed{831}</math>. | ||
== See also == | == See also == |
Revision as of 22:09, 21 March 2018
Problem
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team
beats team
The probability that team
finishes with more points than team
is
where
and
are relatively prime positive integers. Find
Contents
Solution
Solution 1
The results of the five remaining games are independent of the first game, so by symmetry, the probability that scores higher than
in these five games is equal to the probability that
scores higher than
. We let this probability be
; then the probability that
and
end with the same score in these five games is
.
Of these three cases (), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ways to
to have
victories, and
ways for
to have
victories. Summing for all values of
,
![$1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.$](http://latex.artofproblemsolving.com/3/3/9/3396c0524cf771796a5435ee8399fa5234e5e39a.png)
Thus . The desired probability is the sum of the cases when
, so the answer is
, and
.
Solution 2
You can break this into cases based on how many rounds wins out of the remaining
games.
- If
wins 0 games, then
must win 0 games and the probability of this is
.
- If
wins 1 games, then
must win 1 or less games and the probability of this is
.
- If
wins 2 games, then
must win 2 or less games and the probability of this is
.
- If
wins 3 games, then
must win 3 or less games and the probability of this is
.
- If
wins 4 games, then
must win 4 or less games and the probability of this is
.
- If
wins 5 games, then
must win 5 or less games and the probability of this is
.
Summing these 6 cases, we get , which simplifies to
, so our answer is
.
Solution 3
We can apply the concept of generating functions here.
The generating function for is
for the first game where
is winning n games. Since
lost the first game, the coefficient for
is 0. The generating function for the next 5 games is
. Thus, the total generating function for number of games he wins is
.
The generating function for is the same except that it is multiplied by
instead of
.
Thus, the generating function for
is
.
The probability that wins 0 games is
. Since the coefficients for all
where
sums to 32, the probability that
wins more games is
.
Thus, the probability that has more wins than
is
.
Thus, .
Solution 4
After the first game, there are games we care about-- those involving
or
. There are
cases of these
games:
wins more than
,
wins more than
, or
and
win the same number of games. Also, there are
total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs
times, by a special case of Vandermonde's Identity (see Combinatorial identity). There are therefore
possibilities for each of the other two cases.
If has more wins than
in its
remaining games, then
cannot beat
overall. However, if
has more wins or if
and
are tied,
will beat
overall. Therefore, out of the
possibilites,
ways where
wins, so the desired probability is
, and
.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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