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转一道有趣的题！(quant).

转一道有趣的题！(quant).# BrainTeaser - 大脑工作室

s*12008-01-04 08:01

1 楼

刚看到，不知道有人给出答案没有。
A poll was taken between two candidates A and B. A gets X votes, and B gets
Y votes, with X>Y. If the votes were counted one by one, what is the
probability that there was a tie at any time during the counting?

o*n2008-01-04 08:01

2 楼

2Y/(X+Y)
solution已经有人贴在quant版上了，合集里头的最后一篇说得很清楚。
还可以问个相关的问题：what is the probability that A was always ahead as the
votes were counted?
答案是(X-Y)/(X+Y).

gets

【在 s******1 的大作中提到】

: 刚看到，不知道有人给出答案没有。
: A poll was taken between two candidates A and B. A gets X votes, and B gets
: Y votes, with X>Y. If the votes were counted one by one, what is the
: probability that there was a tie at any time during the counting?

o*n2008-01-04 08:01

3 楼

a harder question is:
what is the probability that B, the final loser, was ever leading as the
votes were counted?
it is certainly solvable with a specific pair of X and Y using computers,
but I don't know how to get the general answer by hand.

the

【在 o***n 的大作中提到】

: 2Y/(X+Y)
: solution已经有人贴在quant版上了，合集里头的最后一篇说得很清楚。
: 还可以问个相关的问题：what is the probability that A was always ahead as the
: votes were counted?
: 答案是(X-Y)/(X+Y).
:
: gets

s*12008-01-04 08:01

4 楼

现在才看清题，我还以为计算在总共X＋Y次唱票的时候，出现Tie的概率，或者说出现
Tie次数的期望。

c*s2008-01-04 08:01

5 楼

the answer is (X+Y,Y-1)/(X+Y,Y),
here (X+Y,Y) means the number of choosing Y from X+Y.
the reduction of the ratio is Y/(X+1).

【在 o***n 的大作中提到】

: a harder question is:
: what is the probability that B, the final loser, was ever leading as the
: votes were counted?
: it is certainly solvable with a specific pair of X and Y using computers,
: but I don't know how to get the general answer by hand.
:
: the

o*n2008-01-04 08:01

6 楼

不是很明白分子，能不能给个hint？

【在 c******s 的大作中提到】

: the answer is (X+Y,Y-1)/(X+Y,Y),
: here (X+Y,Y) means the number of choosing Y from X+Y.
: the reduction of the ratio is Y/(X+1).

o*n2008-01-04 08:01

7 楼

给一个反例：X=3, Y=2
一共10个可能的paths，其中有5个B一度领先（四个以B开头，另外一个以ABB开头），
即1/2的概率，但是Y/(X+1) = 3/4
另外，当X～Y>>1时，Y/(X+1)->1，这个好像不make sense。
当然，你这个解在Y=1的时候是正确的。

【在 c******s 的大作中提到】

: the answer is (X+Y,Y-1)/(X+Y,Y),
: here (X+Y,Y) means the number of choosing Y from X+Y.
: the reduction of the ratio is Y/(X+1).

c*s2008-01-04 08:01

8 楼

Y=2, Y/(X+1)=1/2, 不是反例。
如果X和Y接近而且很大，就好比一场篮球比赛的比分，
你应该很少看到失败方一直落后的情况吧，所以概率趋近1也合理。

【在 o***n 的大作中提到】

: 给一个反例：X=3, Y=2
: 一共10个可能的paths，其中有5个B一度领先（四个以B开头，另外一个以ABB开头），
: 即1/2的概率，但是Y/(X+1) = 3/4
: 另外，当X～Y>>1时，Y/(X+1)->1，这个好像不make sense。
: 当然，你这个解在Y=1的时候是正确的。

o*n2008-01-04 08:01

9 楼

ftft,sorry啦
那请给个hint吧, :)

【在 c******s 的大作中提到】

: Y=2, Y/(X+1)=1/2, 不是反例。
: 如果X和Y接近而且很大，就好比一场篮球比赛的比分，
: 你应该很少看到失败方一直落后的情况吧，所以概率趋近1也合理。

c*s2008-01-04 08:01

10 楼

也是用reflection，
不过是在你第一次跨过对角线的那个点之后，
那个点的坐标始终是Y比X大一个。
reflect之后， Y减少了一个， X增加了一个，
所以是(X+Y,X+1)

【在 o***n 的大作中提到】

: ftft,sorry啦
: 那请给个hint吧, :)