m*******1
发帖数: 168 | 1 You have a chest of 8 drawers. With probability 1/2, you put a letter in one
of the drawers. With probability 1/2, you don't put a letter in any drawer.
I open the first 7 drawers, all are empty. What is the probability there is
a letter in the 8th drawer? | b*****7
发帖数: 631 | 2 我也觉得是1/9。
可以转化成一个物品放入16个抽屉之一。开了7个没看到。开第八个看到的可能性是1/9。 | O*2
发帖数: 178 | 3 1／2
因为每开一个空的抽屉，都增大了在最后一个抽屉里的可能
【在 b*****7 的大作中提到】
: 我也觉得是1/9。
: 可以转化成一个物品放入16个抽屉之一。开了7个没看到。开第八个看到的可能性是1/9。
| j********e
发帖数: 1192 | 4 这是个后验条件概率问题。
A表示信放进任一个抽屉，D=i表示信放进了第i给抽屉。
P(A) = 0.5, P(D=i|A)=1/8, P(D=i)=1/16
现在要求的是P(D=8|D!=1,2,...,7).
P(D=8|D!=1,2,...7) * P(D!=1,2,...,7) = P(D=8) = 1/16.
P(D!=1,2,...,7) = P(!A) + P(A and D=8) = 1/2 + 1/16 = 9/16
所以P(D=8|D!=1,2,...,7) = 1/9.
one
drawer.
is
【在 m*******1 的大作中提到】
: You have a chest of 8 drawers. With probability 1/2, you put a letter in one
: of the drawers. With probability 1/2, you don't put a letter in any drawer.
: I open the first 7 drawers, all are empty. What is the probability there is
: a letter in the 8th drawer?
| j********e
发帖数: 1192 | 5 这个转化很有意思
/9。
【在 b*****7 的大作中提到】
: 我也觉得是1/9。
: 可以转化成一个物品放入16个抽屉之一。开了7个没看到。开第八个看到的可能性是1/9。
| x***y
发帖数: 633 | 6 I remember a similar but famous 3-gate problem:
there are 3 gates, behind one and only one of which there is a gift. You
select one gate, and of the other 2 gates, one that does not have gift is
removed. Now, there are only 2 gates left, you can switch. Should you do it?
The answer is yes, as the other door has the probablity of 2/3 to have the
gift. The conditional probability does not apply here is because they are not really independent any more; after the removal, the remaining two gates are not
really equal. | g*********e
发帖数: 14401 | 7 这个问题取决于open drawer/door的那个人事先知不知道exactly how it is assigned
. 如果那个人也不知道，碰巧open drawer/door，那就用条件概率算。如果那人知道，
则用原始分布。（一般出这种问题的人都是想考条件概率的） | g*********e
发帖数: 14401 | 8 p(8 has letter/1-7 no letter)=P(8 has letter && 1-7 no letter)/P(1-7 no
letter)=(1/16)/(1/2+1/2×1/8）=1/9 |
|
