I***e
发帖数: 1136 | 1 设X为一随机变量,
P(X=k)=2^(-k), k=1,2,...
1. 取两个信封, 一个之中放3^X元, 另一个中放3^(X+1)元.
2. 从两个信封中随机(等概率)取出一个.
现在, 如果你看到是3^N元在信封中, 你的推理是: X只能是(N-1)或N
P(X=N|信封中有3^N)=P(X=N, 信封中有3^N元)/[P(X=N-1,信封中有3^N元)
+P(X=N, 信封中有3^N元)]=2^(-N)/(2^(-N+1)+2^(-N))=1/3.
所以, E(另一个信封中的钱数|看到3^N)=1/3*3^(N+1)+2/3*3^(N-1)=3^N+2*3(N-2)
>看到的3^N.
所以, 你应该换取另外的一个信封.
可是, 这个结论对任何看到的数都成立, 包含如果你看到3元的情形. 于是你得出结论,
不论给你的信封中装着多少钱你都应该要另一个. 所以, 你甚至没有必要打开信封就
应该决定换到另一个.
从另外角度, 这两个信封应该是等价的, 哪里逻辑出现问题了? | x******g
发帖数: 318 | 2 这个问题dntx曾经在水母的iqdoor版出过，当时的讨论真是激烈^_^
其实我也在《蚁迹寻踪》上见过这个题目
原因在于这里面根本就没有什么矛盾，换句话说矛盾只是一种幻觉.
应邀撤出一个矛盾来那就是：设一个信封里期望为k元，我们大概可以得到k=7k/9
之类的怪式子，但这并不是矛盾，因为k是无穷大
【在 I***e 的大作中提到】
: 设X为一随机变量,
: P(X=k)=2^(-k), k=1,2,...
: 1. 取两个信封, 一个之中放3^X元, 另一个中放3^(X+1)元.
: 2. 从两个信封中随机(等概率)取出一个.
: 现在, 如果你看到是3^N元在信封中, 你的推理是: X只能是(N-1)或N
: P(X=N|信封中有3^N)=P(X=N, 信封中有3^N元)/[P(X=N-1,信封中有3^N元)
: +P(X=N, 信封中有3^N元)]=2^(-N)/(2^(-N+1)+2^(-N))=1/3.
: 所以, E(另一个信封中的钱数|看到3^N)=1/3*3^(N+1)+2/3*3^(N-1)=3^N+2*3(N-2)
: >看到的3^N.
: 所以, 你应该换取另外的一个信封.
| r****y
发帖数: 1437 | 3 here when you put 3^(x) and 3^(x+1) into two envelopes,
x is a R.V., but x+1 is not R.V.. Because once your x is selected,
x+1 is fixed. So P(x+1 for this case) = P(x).
then P(x=n|3^N in envelope) = 1/2, and then equal probablity
【在 I***e 的大作中提到】
: 设X为一随机变量,
: P(X=k)=2^(-k), k=1,2,...
: 1. 取两个信封, 一个之中放3^X元, 另一个中放3^(X+1)元.
: 2. 从两个信封中随机(等概率)取出一个.
: 现在, 如果你看到是3^N元在信封中, 你的推理是: X只能是(N-1)或N
: P(X=N|信封中有3^N)=P(X=N, 信封中有3^N元)/[P(X=N-1,信封中有3^N元)
: +P(X=N, 信封中有3^N元)]=2^(-N)/(2^(-N+1)+2^(-N))=1/3.
: 所以, E(另一个信封中的钱数|看到3^N)=1/3*3^(N+1)+2/3*3^(N-1)=3^N+2*3(N-2)
: >看到的3^N.
: 所以, 你应该换取另外的一个信封.
| x******g
发帖数: 318 | 4 对了，土豆兄
看到我在science的帖子了吗?你那个证明是错误的.
【在 I***e 的大作中提到】
: 设X为一随机变量,
: P(X=k)=2^(-k), k=1,2,...
: 1. 取两个信封, 一个之中放3^X元, 另一个中放3^(X+1)元.
: 2. 从两个信封中随机(等概率)取出一个.
: 现在, 如果你看到是3^N元在信封中, 你的推理是: X只能是(N-1)或N
: P(X=N|信封中有3^N)=P(X=N, 信封中有3^N元)/[P(X=N-1,信封中有3^N元)
: +P(X=N, 信封中有3^N元)]=2^(-N)/(2^(-N+1)+2^(-N))=1/3.
: 所以, E(另一个信封中的钱数|看到3^N)=1/3*3^(N+1)+2/3*3^(N-1)=3^N+2*3(N-2)
: >看到的3^N.
: 所以, 你应该换取另外的一个信封.
| J**i
发帖数: 166 | 5 ......
【在 r****y 的大作中提到】
: here when you put 3^(x) and 3^(x+1) into two envelopes,
: x is a R.V., but x+1 is not R.V.. Because once your x is selected,
: x+1 is fixed. So P(x+1 for this case) = P(x).
: then P(x=n|3^N in envelope) = 1/2, and then equal probablity
| I***e
发帖数: 1136 | 6 我知道它是错误的--发贴后不久就发现反例了. 你贴的方法很精妙...
两个信封问题还真是令人惊叹: 直觉是多么不严格.
【在 x******g 的大作中提到】
: 对了，土豆兄
: 看到我在science的帖子了吗?你那个证明是错误的.
| I***e
发帖数: 1136 | 7 你的推理并不正确.
X is a R.V., X+1 is also an R.V. Conditional on you observed 3^N dollars you d
o not
know for sure what X is right? And there is a perfect probability distribution
.
To further illustrate, let's say you do the following many many times:
1. Pick a X
2. 2 envelopes one with 3^X, one with 3^(X+1)
3. Pick one at random, give it to you and you open it.
Now let's say that you have observed $27 in it: there must be a posterior dist
ribution
of X right? I am telling you that P(X=3|observed 27)
【在 r****y 的大作中提到】
: here when you put 3^(x) and 3^(x+1) into two envelopes,
: x is a R.V., but x+1 is not R.V.. Because once your x is selected,
: x+1 is fixed. So P(x+1 for this case) = P(x).
: then P(x=n|3^N in envelope) = 1/2, and then equal probablity
| r****y
发帖数: 1437 | 8 Yeah, you are right. No matter what number you got, you always has more ch
ance to hold it as X rather than (X+1), so you should exchange for another one
, which is more likely a X+1.
contribute a small matlab script to simulate this problem
LEN = 1e7;
x = rand(LEN, 1);
y = 1 - x;
x = round(log2(y));
x = x + 1;
base = input('the base ');
total_money_exchange = 0;;
total_money = 0;
for i = 1:LEN
m = base^(-x(i));
n = base^(-x(i)-1);
% >0.5 for m and <0.5 for n
tmp = rand;
i
【在 I***e 的大作中提到】
: 你的推理并不正确.
: X is a R.V., X+1 is also an R.V. Conditional on you observed 3^N dollars you d
: o not
: know for sure what X is right? And there is a perfect probability distribution
: .
: To further illustrate, let's say you do the following many many times:
: 1. Pick a X
: 2. 2 envelopes one with 3^X, one with 3^(X+1)
: 3. Pick one at random, give it to you and you open it.
: Now let's say that you have observed $27 in it: there must be a posterior dist
| J**i
发帖数: 166 | 9 嘿嘿，你这个程序其实根本得不到正确的结论的，多运行几次就知道了
信封里的钱的期望值是无穷大，所以是没办法蒙特卡罗的
【在 r****y 的大作中提到】
: Yeah, you are right. No matter what number you got, you always has more ch
: ance to hold it as X rather than (X+1), so you should exchange for another one
: , which is more likely a X+1.
: contribute a small matlab script to simulate this problem
: LEN = 1e7;
: x = rand(LEN, 1);
: y = 1 - x;
: x = round(log2(y));
: x = x + 1;
: base = input('the base ');
| q*s
发帖数: 26 | 10 Is the sigmma-field of the probability properly defined?
R.V is defined on sigma-field
【在 I***e 的大作中提到】
: 你的推理并不正确.
: X is a R.V., X+1 is also an R.V. Conditional on you observed 3^N dollars you d
: o not
: know for sure what X is right? And there is a perfect probability distribution
: .
: To further illustrate, let's say you do the following many many times:
: 1. Pick a X
: 2. 2 envelopes one with 3^X, one with 3^(X+1)
: 3. Pick one at random, give it to you and you open it.
: Now let's say that you have observed $27 in it: there must be a posterior dist
| r***w
发帖数: 35 | 11 For rigorous proof, I do not support the computer programming as IMS has done
before regarding the Four Color Problem. I think the most important problem is
what the dual above declared, when X goes to infinity, it will blow up; so if
you wanna discuss this, the X must be finite number
【在 I***e 的大作中提到】
: 你的推理并不正确.
: X is a R.V., X+1 is also an R.V. Conditional on you observed 3^N dollars you d
: o not
: know for sure what X is right? And there is a perfect probability distribution
: .
: To further illustrate, let's say you do the following many many times:
: 1. Pick a X
: 2. 2 envelopes one with 3^X, one with 3^(X+1)
: 3. Pick one at random, give it to you and you open it.
: Now let's say that you have observed $27 in it: there must be a posterior dist
| I***e
发帖数: 1136 | 12 The sigma-algebra is well defined.
The traditional 2-envelope problem has an undefined probability space.
But this enhanced version has a well defined probability space yet still
gives seemingly paradoxical conclusion.
【在 q*s 的大作中提到】
: Is the sigmma-field of the probability properly defined?
: R.V is defined on sigma-field
| p***o
发帖数: 836 | 13 Somebody has a correct solution or explanation ?
【在 I***e 的大作中提到】
: The sigma-algebra is well defined.
: The traditional 2-envelope problem has an undefined probability space.
: But this enhanced version has a well defined probability space yet still
: gives seemingly paradoxical conclusion.
| k***e
发帖数: 8 | 14 Search exchange paradox. I think the explanation is similar to
the simple version.
http://www.pitt.edu/~jdnorton/papers/Exchange_paradox.pdf
【在 p***o 的大作中提到】
: Somebody has a correct solution or explanation ?
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