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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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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