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Quant版 - One interview question, tough
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1 (共1页)
s*********k
发帖数: 1989
1
Here is 1000(m) poker player tournament. It is a knock-out tournament.
Say buy-in is 100$/10$. 100$ goes the final prize tool. 10$ is the knock out
bonus. If you win the showdown hand and the opponent has no chip left, then
you
knock him out and collect the bonus(10$).
Player X finshed 100th(n).
Quest, what is the expectation (on how many knock-out bonus) that player X
won/collected? Formulate the result with m/n
s*u
发帖数: 2240
2
看不懂,原题就是这么描述的?

out
then

【在 s*********k 的大作中提到】
: Here is 1000(m) poker player tournament. It is a knock-out tournament.
: Say buy-in is 100$/10$. 100$ goes the final prize tool. 10$ is the knock out
: bonus. If you win the showdown hand and the opponent has no chip left, then
: you
: knock him out and collect the bonus(10$).
: Player X finshed 100th(n).
: Quest, what is the expectation (on how many knock-out bonus) that player X
: won/collected? Formulate the result with m/n

k**x
发帖数: 2611
3
如果假设每次出局的人有平均的概率是输给剩下的任何人。
第一个出局的人的10块钱平均去了n-1个人那里,平均每人10*(1/n-1)
第二个出局的人的10块钱平均去了n-2个人那里,平均每人10*(1/n-2)
...
所以第m名平均可以得到10*(1/(n-1)+1/(n-2)+...+1/m)~10*(log(n-1)-log(m))
s*********k
发帖数: 1989
4
NIU, That is the solution almost. Just the sanity check at n does not match.
If the guy finish at n (first one to be knocked out), he gets zero.
Your formular gives a negative number.

【在 k**x 的大作中提到】
: 如果假设每次出局的人有平均的概率是输给剩下的任何人。
: 第一个出局的人的10块钱平均去了n-1个人那里,平均每人10*(1/n-1)
: 第二个出局的人的10块钱平均去了n-2个人那里,平均每人10*(1/n-2)
: ...
: 所以第m名平均可以得到10*(1/(n-1)+1/(n-2)+...+1/m)~10*(log(n-1)-log(m))

d****m
发帖数: 119
5
Assume the sittings are rearranged randomly after each knock-out.
F(m-k,m) = F(m-k,m-k+1)+F(m-k+1,m)
where F(n,m)*10 is the solution we are looking for.
F(a-1,a)=1/(a-1)
F(n,m) is the sum of 1/(m-1), ... 1/n

out
then

【在 s*********k 的大作中提到】
: Here is 1000(m) poker player tournament. It is a knock-out tournament.
: Say buy-in is 100$/10$. 100$ goes the final prize tool. 10$ is the knock out
: bonus. If you win the showdown hand and the opponent has no chip left, then
: you
: knock him out and collect the bonus(10$).
: Player X finshed 100th(n).
: Quest, what is the expectation (on how many knock-out bonus) that player X
: won/collected? Formulate the result with m/n

1 (共1页)
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