l*********t
发帖数: 89 | 1 Teams play game with each other. In each round, if one team wins, it gets 1
point; if draw, it gets 0.5 point; if lose, it gets 0. After each team
played with every other team, now consider the 10 teams with the lowest
points, we know each of the lowest 10 teams got exactly half of their
individual total points by playing with every other team within the 10
lowest-ranked teams. Question: how many teams are there in total? | u****g
发帖数: 402 | 2 我怎么觉得好像只有lower bound？
假设有n个team满足上面的要求，再加一个team，它赢得所有对那10个team的比赛，这
样就不会改变那10个team的分数。
the lower bound is 19. | a****y
发帖数: 99 | 3
1
~~~~~~~~~~~~~~~~~~~~~~
i think it should be every team including the lowest 10 .
but i could not figure out.
my equation
C_n^2 /2 = C_{10}^2 + (n-10)^10
but it doenst work.
maybe i misunderstood the meaning of the problem.
【在 l*********t 的大作中提到】
: Teams play game with each other. In each round, if one team wins, it gets 1
: point; if draw, it gets 0.5 point; if lose, it gets 0. After each team
: played with every other team, now consider the 10 teams with the lowest
: points, we know each of the lowest 10 teams got exactly half of their
: individual total points by playing with every other team within the 10
: lowest-ranked teams. Question: how many teams are there in total?
| k****o
发帖数: 11 | 4 19, easy to get the answer, assume that all the games are draw.
To prove it is 19, assume that n is the number of the rest, then:
C(10,2)+10n+C(n,2)=C(10+n,2)
solve for n=9,
total=10+9=19 | w*****e
发帖数: 197 | 5 Are you joking or something?
C(10,2) + 10n + C(n,2) = C(10+n, 2)
is tautology. Any n >=0 will satisfy this!
【在 k****o 的大作中提到】
: 19, easy to get the answer, assume that all the games are draw.
: To prove it is 19, assume that n is the number of the rest, then:
: C(10,2)+10n+C(n,2)=C(10+n,2)
: solve for n=9,
: total=10+9=19
| i***y
发帖数: 285 | 6 设有10+X 个队。
c(x,2) + 10x- C（10，2） >= 9x
x>=9
c(x,2)是前X支队的两两比赛得分。
10X-C（10，2）是前X支队和后10对比赛前X支队得分，因为后10两两得分C（10，2）＝
前X支队和后１０对比赛后１０支队得分
９是后１０支队的平均分。
1
【在 l*********t 的大作中提到】
: Teams play game with each other. In each round, if one team wins, it gets 1
: point; if draw, it gets 0.5 point; if lose, it gets 0. After each team
: played with every other team, now consider the 10 teams with the lowest
: points, we know each of the lowest 10 teams got exactly half of their
: individual total points by playing with every other team within the 10
: lowest-ranked teams. Question: how many teams are there in total?
| w*****e
发帖数: 197 | 7 the average of the bottom 10 is 9/2 not 9.
your result is correct though.
i have yet to find a formal proof.
【在 i***y 的大作中提到】
: 设有10+X 个队。
: c(x,2) + 10x- C（10，2） >= 9x
: x>=9
: c(x,2)是前X支队的两两比赛得分。
: 10X-C（10，2）是前X支队和后10对比赛前X支队得分，因为后10两两得分C（10，2）＝
: 前X支队和后１０对比赛后１０支队得分
: ９是后１０支队的平均分。
:
: 1
| z****0
发帖数: 1 | 8 看看这样对不对：
假设第i个人得分是 p_i, 其中和第j个人比赛的得分是p_i^j，则 p_i = \sum_{j \neq
i} p_i^j.
每个人得分的总和为
C(N,2) = \sum_i \sum_{j \neq i} p_i^j
= 2\sum_i \sum{ j \in S, j \neq i} p_i^j
= 2\sum_{i \in S} \sum{ j \in S, j \neq i} p_i^j + 2\sum_{i \notin S} \sum{
j \in S, j \neq i} p_i^j
= 2\sum_{i \in S} \sum{ j \in S, j \neq i} p_i^j + 2\sum_{i \notin S} \sum{
j \in S, j \neq i} ( 1 - p_j^i )
= 2\sum_{i \in S} \sum{ j \in S, j \neq i} p_i^j + 2*|S|* |S^c| - 2\sum_{j \
in S} \sum_{i \notin S} p_j^i
= 2\sum_{i \in S} \sum{ j \in S, j \neq i} p_i^j + 2*|S|* |S^c| - 2\sum_{j \
in S} \sum_{i \in S, i \neq j} p_j^i
= 2|S|*|S^c|
所以 N(N-1)/2 = 2*10*(N-10)
解得 N = 16 或 N = 25. 排除 N = 16 因为 S 里面是得分最低的10人。 | k****i
发帖数: 162 | 9 The lowest 10 teams has a total points of 2*C(10,2)=90, and an average point
of 9.
Apparently, we will only get a lower bound of n, achieved when all the game
draw.
The lower bound is (n-1)*0.5=9 or n=19.
iLucy's solution is fine. | w*****e
发帖数: 197 | 10 Sorry I was not thinking clearly earlier.
Yes, the solution is exactly the correct one.
Overall this problem is not that interesting at all.
point
game
【在 k****i 的大作中提到】
: The lowest 10 teams has a total points of 2*C(10,2)=90, and an average point
: of 9.
: Apparently, we will only get a lower bound of n, achieved when all the game
: draw.
: The lower bound is (n-1)*0.5=9 or n=19.
: iLucy's solution is fine.
| | | l*********t
发帖数: 89 | 11 Thanks to all of you for your inputs. | b******e
发帖数: 118 | 12 The lowest 10 teams has a total points of 2*C(10,2)=90, and an average point
of 9.
请问这里为什么要乘以2?
point
game
【在 k****i 的大作中提到】
: The lowest 10 teams has a total points of 2*C(10,2)=90, and an average point
: of 9.
: Apparently, we will only get a lower bound of n, achieved when all the game
: draw.
: The lower bound is (n-1)*0.5=9 or n=19.
: iLucy's solution is fine.
| m*********g
发帖数: 646 | 13 C(10,2)="total points by playing with every other team within the 10
lowest-ranked teams" which is only a half of their total points. | b******e
发帖数: 118 | 14 (n-1)*0.5=9
这里其实是C(n,2)/n = 9 为什么这里就不用乘以2了呢？
【在 m*********g 的大作中提到】
: C(10,2)="total points by playing with every other team within the 10
: lowest-ranked teams" which is only a half of their total points.
| m*********g
发帖数: 646 | 15 ....
here it means: in the scenario that every team draw, every team should score
(n-1)*0.5. because there will be totally n-1 games for each team.
then make this equal to 9.
You may want to spend more time to read the original post...
【在 b******e 的大作中提到】
: (n-1)*0.5=9
: 这里其实是C(n,2)/n = 9 为什么这里就不用乘以2了呢？
| b******e
发帖数: 118 | 16 Got it. Thanks a lot!
score
【在 m*********g 的大作中提到】
: ....
: here it means: in the scenario that every team draw, every team should score
: (n-1)*0.5. because there will be totally n-1 games for each team.
: then make this equal to 9.
: You may want to spend more time to read the original post...
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