s*****i
发帖数: 93 | 1 3/4 or 1/2？ i think is 3/4. right? thanks... | a****y
发帖数: 268 | | j*****4
发帖数: 292 | 3 3/4. The general solution is n/2^(n-1)
【在 s*****i 的大作中提到】
: 3/4 or 1/2？ i think is 3/4. right? thanks...
| s*****i
发帖数: 93 | 4 我说3/4，面试的人说是1/2，真郁闷。。。不知道我错还是他错了
【在 j*****4 的大作中提到】
: 3/4. The general solution is n/2^(n-1)
| a****y
发帖数: 268 | 5 说说咋算的3/4。。
【在 s*****i 的大作中提到】
: 我说3/4，面试的人说是1/2，真郁闷。。。不知道我错还是他错了
| v**m
发帖数: 706 | 6 It is 1/2.
1 *(1/2*1/2 + 1/2*1/2)
Point a, b, c.
Pa = 1; Set a as the start point.
if b in [0, PI], then c should be [0,PI], it is 1/2*1/2
vice versa,
if b in [0, -PI], then c should be [0,-PI], it is 1/2*1/2 | v*******y
发帖数: 1586 | 7 The first one and the second you can put whereever you want
Then do integration for the third one? | j*****4
发帖数: 292 | 8 why not tell him how you get the result?
【在 s*****i 的大作中提到】
: 我说3/4，面试的人说是1/2，真郁闷。。。不知道我错还是他错了
| v*******y
发帖数: 1586 | 9 这个方法太彪悍了
第一二个怎么放都是半圆
【在 v**m 的大作中提到】
: It is 1/2.
: 1 *(1/2*1/2 + 1/2*1/2)
: Point a, b, c.
: Pa = 1; Set a as the start point.
: if b in [0, PI], then c should be [0,PI], it is 1/2*1/2
: vice versa,
: if b in [0, -PI], then c should be [0,-PI], it is 1/2*1/2
| j*****4
发帖数: 292 | 10 if you choose a as starting point,then just [0,pi] is enough.
and b,c could be the starting point as well. so it's 3*(1/2)^2
【在 v**m 的大作中提到】
: It is 1/2.
: 1 *(1/2*1/2 + 1/2*1/2)
: Point a, b, c.
: Pa = 1; Set a as the start point.
: if b in [0, PI], then c should be [0,PI], it is 1/2*1/2
: vice versa,
: if b in [0, -PI], then c should be [0,-PI], it is 1/2*1/2
| | | b******r
发帖数: 548 | 11 suppose r=1, the minimum distance between a,b is R
then R is uniform(0,pi). Draw two diameters through a and b,
you will find that, if c lies in 2*pi-R, a,b,c will be in half of
the circle. so the answer is E((2*pi-R)/2*pi) = 3/4.
【在 s*****i 的大作中提到】
: 3/4 or 1/2？ i think is 3/4. right? thanks...
| v**m
发帖数: 706 | 12 you are right, it was my mistake.
there is another method using integral.
/PI
| 1 x 3
1-2 * | --- * --- dx = --
| 2PI 2PI 4
/0 | v**m
发帖数: 706 | 13 Exactly, this is the perfect answer.
【在 b******r 的大作中提到】
: suppose r=1, the minimum distance between a,b is R
: then R is uniform(0,pi). Draw two diameters through a and b,
: you will find that, if c lies in 2*pi-R, a,b,c will be in half of
: the circle. so the answer is E((2*pi-R)/2*pi) = 3/4.
| c*********n
发帖数: 86 | 14 设x，y，360-x－y为3各点之间的夹角，
所以 x>180, y>180 or x+y<180为三个满足的条件，所以在x+y<=360的整个区域内，这
三个条件的区域正好占据3/4.
【在 s*****i 的大作中提到】
: 3/4 or 1/2？ i think is 3/4. right? thanks...
| c******r
发帖数: 300 | 15 Here is one solution without calculus:
Let A, B, C be the three points, WLOG, we can fixed one point, say point A,
consider the diameter that has A as one end-point, let the center of the
circle be O and the other end point be A', the only case where the three
points doesn't lie on some semi-circle is when BC intersects with OA', whose
probability is
1/2 * 1/2
the first 1/2 is the probability that BC intersects the diameter AA', the
second one is the conditional probability that BC intersects OA
【在 s*****i 的大作中提到】
: 3/4 or 1/2？ i think is 3/4. right? thanks...
| w*****e
发帖数: 197 | 16 This is a very good answer, just using symmetry twice.
,
whose
that
【在 c******r 的大作中提到】
: Here is one solution without calculus:
: Let A, B, C be the three points, WLOG, we can fixed one point, say point A,
: consider the diameter that has A as one end-point, let the center of the
: circle be O and the other end point be A', the only case where the three
: points doesn't lie on some semi-circle is when BC intersects with OA', whose
: probability is
: 1/2 * 1/2
: the first 1/2 is the probability that BC intersects the diameter AA', the
: second one is the conditional probability that BC intersects OA
| d*j
发帖数: 13780 | 17 what if N points?
【在 w*****e 的大作中提到】
: This is a very good answer, just using symmetry twice.
:
: ,
: whose
: that
| s***e
发帖数: 267 | 18 For N points A_1,..., A_n, suppose that they are on half circle. If you
start from any point and collect points in clockwise order, then you get a
list of n points. Obviously, one and only one such list is within half
circle (with probability 1).
WLOG, assume such list start from A_1. Then for the points A_2, .., A_n to
be within the half circle starting from A_1, the probability is 0.5*0.5*...=
0.5^(n-1). Consider symmetry, the result is
0.5^(n-1) * n
In the case of 3 points the result is 3/4.
【在 d*j 的大作中提到】
: what if N points?
| c******r
发帖数: 300 | 19 Good point, I think there is also a way to get the answer n/2^{n-1} with
some similar idea: let A1, ... , An be the n points chosen, consider again
the corresponding end-points A1',..., An' such that Ak and Ak' forms a
diameter. remember that to choose the n points uniformly, we can first
choose the diameter (the angle) first and then choose one of the two end-
points randomly. Therefore, here Ak and Ak' are exchangeable. For n points
to lie on some semi-circle, there must exist some Ai such tha
【在 d*j 的大作中提到】
: what if N points?
| p*****k
发帖数: 318 | 20 probably i have posted this before, the general case with
N points on a D-dimensional sphere results the prob of
Pr(N,D) = sum{from m=0 to D} C(N-1,m) / 2^(N-1)
where C(n,m)=n!/[m!(n-m)!] is the binomial coefficient.
the essential idea was outlined by shede and chopinor.
[OP is asking for Pr(3,1)]
more details can be found at:
J.Wendel, 'A problem in geometric probability'
Math. Scand., 11 (1962), 109-111.
or
http://www.mathpages.com/home/kmath327/kmath327.htm | | | f***a
发帖数: 329 | 21 any point has the same probability to be the "first" point in the chain of
all n points. once the "first" point is determined, each of the res (n-1)
points has 1/2 chance to be in the half-circle. so,
n*(1/2)^(n-1) | c******r
发帖数: 300 | 22 Nice, I was actually trying to solve the 3-dimensional case since I remember
it's a classic problem. Thanks for the reference.
【在 p*****k 的大作中提到】
: probably i have posted this before, the general case with
: N points on a D-dimensional sphere results the prob of
: Pr(N,D) = sum{from m=0 to D} C(N-1,m) / 2^(N-1)
: where C(n,m)=n!/[m!(n-m)!] is the binomial coefficient.
: the essential idea was outlined by shede and chopinor.
: [OP is asking for Pr(3,1)]
: more details can be found at:
: J.Wendel, 'A problem in geometric probability'
: Math. Scand., 11 (1962), 109-111.
: or
| i*****e
发帖数: 159 | 23 once the first point is determined, how is there 1/2 chance that the second
one will be in the half-circle? what is the "half-circle" in terms of the
first point?
【在 f***a 的大作中提到】
: any point has the same probability to be the "first" point in the chain of
: all n points. once the "first" point is determined, each of the res (n-1)
: points has 1/2 chance to be in the half-circle. so,
: n*(1/2)^(n-1)
| j*****g
发帖数: 454 | 24 i think 1/2
Consider 3 points as a triangle in an circle. it is the probability that
the main angle is bigger than 90 degree
【在 s*****i 的大作中提到】
: 我说3/4，面试的人说是1/2，真郁闷。。。不知道我错还是他错了
| b******r
发帖数: 548 | 25 still 3/4
suppose A, B, C are three angles. find P(max(A,B,C)>90)
max(A,B,C) is uniform(60,180).
so it 90/120 = 3/4
【在 j*****g 的大作中提到】
: i think 1/2
: Consider 3 points as a triangle in an circle. it is the probability that
: the main angle is bigger than 90 degree
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