a*****h
发帖数: 484 | 1 上周Credit Suisse的笔试题，谢了先!
First person pick a number from 1 to 100,then the second make a guess, and
being rewarded with the amount equal to the number (say the number is 5,
then he gets $5). Now the question is how much the second person will be
willing to pay for the game?
对了有人收到回音了吗? |
M********t
发帖数: 163 | |
J**********g
发帖数: 213 | 3 sorry i don't understand your question...
【在 a*****h 的大作中提到】
: 上周Credit Suisse的笔试题，谢了先!
: First person pick a number from 1 to 100,then the second make a guess, and
: being rewarded with the amount equal to the number (say the number is 5,
: then he gets $5). Now the question is how much the second person will be
: willing to pay for the game?
: 对了有人收到回音了吗?
|
f*******s
发帖数: 57 | 4 Consider the strategy of picking number i with probability proportional to 1
/i, regardless of the counterparty's strategy, the expected payoff (for the
buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
that makes sure there is no-arbitrage, hence the fair value of the game.
【在 a*****h 的大作中提到】
: 上周Credit Suisse的笔试题，谢了先!
: First person pick a number from 1 to 100,then the second make a guess, and
: being rewarded with the amount equal to the number (say the number is 5,
: then he gets $5). Now the question is how much the second person will be
: willing to pay for the game?
: 对了有人收到回音了吗?
|
p******5
发帖数: 138 | 5 If the first person picks a number randomly from 1 to 100, and the second
person can only guess once, of course he should guess 100 (100 has the same
probability as other numbers to appear and he will get more reward), and the
expected reward is 1/100 * 100 = 1$.
So the second one would like to pay at most 1$ to play this game. |
p******5
发帖数: 138 | 6 Excellent!
to 1
the
price
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
a***n
发帖数: 423 | 7 Assume that the first person pick a number i with a probability of 1/100,
then the second person have a probability of 1/100 to guess right and 99/100
to guess wrong. for each correct guess, the second person can get $i.
So expectation of the second person's pay off is 1/100*sum of ($1/100 to $
100/100)=$0.505
How do you think about it?
【在 a*****h 的大作中提到】
: 上周Credit Suisse的笔试题，谢了先!
: First person pick a number from 1 to 100,then the second make a guess, and
: being rewarded with the amount equal to the number (say the number is 5,
: then he gets $5). Now the question is how much the second person will be
: willing to pay for the game?
: 对了有人收到回音了吗?
|
c****o
发帖数: 1280 | 8 Is it something about counter-party risk? Is we did it in the following way,
we are subject to "conterparty risk"? just some naive thought.....
100
【在 a***n 的大作中提到】
: Assume that the first person pick a number i with a probability of 1/100,
: then the second person have a probability of 1/100 to guess right and 99/100
: to guess wrong. for each correct guess, the second person can get $i.
: So expectation of the second person's pay off is 1/100*sum of ($1/100 to $
: 100/100)=$0.505
: How do you think about it?
|
a*****h
发帖数: 484 | 9 This looks cool to me~~~Where did you learn the stuffs from?
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
a*****h
发帖数: 484 | 10 That was the first one could come up with, but I think fiverings gives a
much cooler answer:)
100
【在 a***n 的大作中提到】
: Assume that the first person pick a number i with a probability of 1/100,
: then the second person have a probability of 1/100 to guess right and 99/100
: to guess wrong. for each correct guess, the second person can get $i.
: So expectation of the second person's pay off is 1/100*sum of ($1/100 to $
: 100/100)=$0.505
: How do you think about it?
|
|
|
a***n
发帖数: 423 | 11 One question: in this strategy, the probability to pick number 1 is 1. does
it mean the first person will always pick up number 1? does it require sum
of probability to pick number i (i = 1 to 100) is 100 percent?
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
a*****h
发帖数: 484 | 12 Nope, he picks i with prob. (1/i)/(1+1/2+...), so they sum up to 1.
does
sum
【在 a***n 的大作中提到】
: One question: in this strategy, the probability to pick number 1 is 1. does
: it mean the first person will always pick up number 1? does it require sum
: of probability to pick number i (i = 1 to 100) is 100 percent?
:
: 1
: the
|
b******e
发帖数: 118 | 13 My answer is same as yours.
100
【在 a***n 的大作中提到】
: Assume that the first person pick a number i with a probability of 1/100,
: then the second person have a probability of 1/100 to guess right and 99/100
: to guess wrong. for each correct guess, the second person can get $i.
: So expectation of the second person's pay off is 1/100*sum of ($1/100 to $
: 100/100)=$0.505
: How do you think about it?
|
f*******s
发帖数: 57 | 14 I think you can start with the special case of two numbers (1 and 2), where
you notice there is some strategy to make sure the buyer/seller makes 2/3 on
average. The general case then follows. I didn't realize this is a non-
arbitrage problem before solving it, only retrospectively, I found it's
actually connected with finance.
【在 a*****h 的大作中提到】
: This looks cool to me~~~Where did you learn the stuffs from?
:
: 1
: the
|
a*****h
发帖数: 484 | 15 Thanks! I wonder for a non-finance, Engineering PhD to learn financial math,
which one would be recommended to start with?
where
on
【在 f*******s 的大作中提到】
: I think you can start with the special case of two numbers (1 and 2), where
: you notice there is some strategy to make sure the buyer/seller makes 2/3 on
: average. The general case then follows. I didn't realize this is a non-
: arbitrage problem before solving it, only retrospectively, I found it's
: actually connected with finance.
|
a*****h
发帖数: 484 | 16 That's more like an upperbound to me, while if the first player don't pick
the number randomly, but strategically, you're never gonna get this much ($1
).
,
make |
k*******d
发帖数: 1340 | 17 赌场一方如果就是不选100，那么啥都得不到
1,
make |
t********s
发帖数: 54 | 18 Why would you consider the strategy of picking number i with probability
proportionaly to 1/i?
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
b***k
发帖数: 2673 | 19 这个题我觉得正确理解题目的意思更重要。
第一个人是要pay out money的，所以他自然是要想让这个payout不受
第二个人的选择方式(或者说赌博方式)的影响，而且他还要尽量使这个payout
越小越好。于是第一个人选择strategy的时候就基于这两个原则展开。
所以这个题目初看像个概率题，其实是个game theory问题，是允许第一个制定一个
最优策略，使这个payout不受第二个人支配，同时也能使payout尽可能小。
而第二个人，要在“看穿”第一个人的把戏的基础上来决定how much willing to pay
to play the game。毫无疑问，这个钱就应该是第一个人最优策略下算出来的钱。
综上分析，可以看出楼上fiverings的解答是正确的，而那个值约等于$0.19，可见这
个游戏payoff很低，在最优策略下，如果需要高于0.19的入场费来玩这个游戏，对第
二个人来说都是不合算的。
【在 a*****h 的大作中提到】
: 上周Credit Suisse的笔试题，谢了先!
: First person pick a number from 1 to 100,then the second make a guess, and
: being rewarded with the amount equal to the number (say the number is 5,
: then he gets $5). Now the question is how much the second person will be
: willing to pay for the game?
: 对了有人收到回音了吗?
|
Y***e
发帖数: 1030 | 20 其实这题我一直没读明白，我觉得讲的不清楚，面试的时候还可以问问来clarify, 考
试的时候啥
也不能干。。啧啧
【在 a*****h 的大作中提到】
: 上周Credit Suisse的笔试题，谢了先!
: First person pick a number from 1 to 100,then the second make a guess, and
: being rewarded with the amount equal to the number (say the number is 5,
: then he gets $5). Now the question is how much the second person will be
: willing to pay for the game?
: 对了有人收到回音了吗?
|
|
|
x*a
发帖数: 48 | 21 So this is minimax strategy. What you describe makes good sense except that the 2nd really doesn't and can't care what the 1st's strategy is.
pay
【在 b***k 的大作中提到】
: 这个题我觉得正确理解题目的意思更重要。
: 第一个人是要pay out money的，所以他自然是要想让这个payout不受
: 第二个人的选择方式(或者说赌博方式)的影响，而且他还要尽量使这个payout
: 越小越好。于是第一个人选择strategy的时候就基于这两个原则展开。
: 所以这个题目初看像个概率题，其实是个game theory问题，是允许第一个制定一个
: 最优策略，使这个payout不受第二个人支配，同时也能使payout尽可能小。
: 而第二个人，要在“看穿”第一个人的把戏的基础上来决定how much willing to pay
: to play the game。毫无疑问，这个钱就应该是第一个人最优策略下算出来的钱。
: 综上分析，可以看出楼上fiverings的解答是正确的，而那个值约等于$0.19，可见这
: 个游戏payoff很低，在最优策略下，如果需要高于0.19的入场费来玩这个游戏，对第
|
a*****h
发帖数: 484 | 22 Yeah. Well, Blook gave a nice explanation already--it is more likely to be a
game theory problem than a probability one. Sorry for the misleading title.
..
【在 Y***e 的大作中提到】
: 其实这题我一直没读明白，我觉得讲的不清楚，面试的时候还可以问问来clarify, 考
: 试的时候啥
: 也不能干。。啧啧
|
x******a
发帖数: 6336 | 23 how to prove this is the best strategy for the first person?
thanks
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
w****i
发帖数: 143 | 24 if the proportional is p=1/(1+1/2+...+1/100),should it be 100p?
So payoff is 100/(1+1/2+...+1/100).
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
s*****e
发帖数: 115 | 25 If this problem is viewed as a game, why would the 1st person even consider
picking any number other than 1?
If it's game, the best strategy for the first person is picking 1 every time
, isn't it?
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
s*****e
发帖数: 115 | 26 Sorry, please ignore my previous post. I think I agree with your answer now.
Here's how I got it:
Person 1 choose a number from 1 to 100 according to the following the
probabilities:
p1, p2, ..., p100
Person 2 guess a number from 1 to 100 according to the following the
probabilities:
q1, q2, ..., q100
The expected payoff to Person 2 is:
E=p1*q1*1+p2*q2*2+p3*q3*3+...+p100*q100*100
Person 1 wants to minimize E while Person 2 tries to maximize E, both
subject to the constraints:
p1+p2+...+p100=1
q1+q2+...+q100=1
The equilibrium can be found using the Lagrange multipliers "a" and "b":
Let F = E + a*(p1+p2+...+p100) + b*(q1+q2+...+q100)
and the equilibrium occurs at:
DF/Dpi=0, for i=1,...,100 ===> qi*i+a=0 ===> qi = -a/i, for i=1,...,100
DF/Dqi=0, for i=1,...,100 ===> pi*i+b=0 ===> pi = -b/i, for i=1,...,100
Thus, the optimal strategy for Person 1 is to choose number i with a
probability proportional to 1/i
and the optimal strategy for Person 2 is to guess number i also with a
probability proportional to 1/i.
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
x*a
发帖数: 48 | 27 schiele, very elegant math and leads to the same results. Kudos.
However, I believe Lagrange does not apply here since
1. It's not a joint optimization. For 2, all you can control is {q_i},
likewise {p_i} for 1.
2. Hypothetically consider a joint optimization, what does this p_i~1/i mean
really? It's probably just a saddle point (鞍点), not one of its local
extrema (it's not quadratic). In fact, the maximum is when p100=q100=1 and
minimum is whenever all (p_i,q_i) pairs have at least one zero value.
now.
【在 s*****e 的大作中提到】
: Sorry, please ignore my previous post. I think I agree with your answer now.
: Here's how I got it:
: Person 1 choose a number from 1 to 100 according to the following the
: probabilities:
: p1, p2, ..., p100
: Person 2 guess a number from 1 to 100 according to the following the
: probabilities:
: q1, q2, ..., q100
: The expected payoff to Person 2 is:
: E=p1*q1*1+p2*q2*2+p3*q3*3+...+p100*q100*100
|
s*****e
发帖数: 115 | 28 This is exactly a MinMax problem for the objective function
E=p1*q1*1+...+p100*q100*100
subject to the constraints
p1+p2+...+p100=1
q1+q2+...+q100=1
The solution achieved here using the Lagrange multipliers is precisely a
saddle point, which is in fact THE type of solution you want for MinMax
problem. This is completely different from "maximzing E subject to the
constraints" or "minimzing E subject to the constraints" -- to both case,
the solutions would occur on the boundary of the domain for this problem,
just as you have pointed out.
The method of Lagrange multipliers doesn't discriminate the cases where you'
re looking for a local min, a local max, or a saddle. To all three cases,
the "necessary" condition is that: the gradient of the objective function is
a linear combination of the normal vectors of the contraint surfaces. The
algebraic formulation of this geometrical idea gives the set of formulas for
the Lagrange multiplier method.
mean
【在 x*a 的大作中提到】
: schiele, very elegant math and leads to the same results. Kudos.
: However, I believe Lagrange does not apply here since
: 1. It's not a joint optimization. For 2, all you can control is {q_i},
: likewise {p_i} for 1.
: 2. Hypothetically consider a joint optimization, what does this p_i~1/i mean
: really? It's probably just a saddle point (鞍点), not one of its local
: extrema (it's not quadratic). In fact, the maximum is when p100=q100=1 and
: minimum is whenever all (p_i,q_i) pairs have at least one zero value.
:
: now.
|
t*******y
发帖数: 637 | 29 这题second makes a guess 这句话是不是去掉比较好啊
放在里面很confusing
1
the
【在 f*******s 的大作中提到】
: Consider the strategy of picking number i with probability proportional to 1
: /i, regardless of the counterparty's strategy, the expected payoff (for the
: buyer) or cost (for the seller) is 1/(1+1/2+...+1/100), which is the price
: that makes sure there is no-arbitrage, hence the fair value of the game.
|
t*******g
发帖数: 373 | 30 0.505
【在 a*****h 的大作中提到】
: 上周Credit Suisse的笔试题，谢了先!
: First person pick a number from 1 to 100,then the second make a guess, and
: being rewarded with the amount equal to the number (say the number is 5,
: then he gets $5). Now the question is how much the second person will be
: willing to pay for the game?
: 对了有人收到回音了吗?
|
|
|
k*******a
发帖数: 772 | 31 I think it would be $1
As second person, he can bet 100 all the time, and the probability he can
win is 1/100, so expected return is 100*1/100=1
if he guesses others, expected would be k/100<1
so, he should always bet on 100
The issue here is: first person is randomly pick, but second person don't
have to randomly guess |
q*n
发帖数: 134 | 32 赞，
不受影响的假设非常重要。
pay
【在 b***k 的大作中提到】
: 这个题我觉得正确理解题目的意思更重要。
: 第一个人是要pay out money的，所以他自然是要想让这个payout不受
: 第二个人的选择方式(或者说赌博方式)的影响，而且他还要尽量使这个payout
: 越小越好。于是第一个人选择strategy的时候就基于这两个原则展开。
: 所以这个题目初看像个概率题，其实是个game theory问题，是允许第一个制定一个
: 最优策略，使这个payout不受第二个人支配，同时也能使payout尽可能小。
: 而第二个人，要在“看穿”第一个人的把戏的基础上来决定how much willing to pay
: to play the game。毫无疑问，这个钱就应该是第一个人最优策略下算出来的钱。
: 综上分析，可以看出楼上fiverings的解答是正确的，而那个值约等于$0.19，可见这
: 个游戏payoff很低，在最优策略下，如果需要高于0.19的入场费来玩这个游戏，对第
|
t*******g
发帖数: 373 | 33 同意
【在 q*n 的大作中提到】
: 赞，
: 不受影响的假设非常重要。
:
: pay
|
k*******a
发帖数: 772 | 34 请问0.19这个数值怎么来的？
如果PA(k)是A选择k的概率，PB(j)是B猜j的概率
那么在B不知道A的选择策略PA(k)的情况下，应该怎么选择PB(j)?
pay
【在 b***k 的大作中提到】
: 这个题我觉得正确理解题目的意思更重要。
: 第一个人是要pay out money的，所以他自然是要想让这个payout不受
: 第二个人的选择方式(或者说赌博方式)的影响，而且他还要尽量使这个payout
: 越小越好。于是第一个人选择strategy的时候就基于这两个原则展开。
: 所以这个题目初看像个概率题，其实是个game theory问题，是允许第一个制定一个
: 最优策略，使这个payout不受第二个人支配，同时也能使payout尽可能小。
: 而第二个人，要在“看穿”第一个人的把戏的基础上来决定how much willing to pay
: to play the game。毫无疑问，这个钱就应该是第一个人最优策略下算出来的钱。
: 综上分析，可以看出楼上fiverings的解答是正确的，而那个值约等于$0.19，可见这
: 个游戏payoff很低，在最优策略下，如果需要高于0.19的入场费来玩这个游戏，对第
|
