數學知識交流---幾何概率

I would say the question is not well-defined.

(Triangle refers to the triangle enclosed by the first three point, if any)

You can actually get two solutions: 0 and 1/2, based on two views:
1) Area ratio, then it's 0 since 0/infinity = 0

2) Number of points, and in fact points inside and outside the point can be related by a bijective function f. Then we would say points inside the triangle is as many as points outside the triangle. Since point ratio = 1:1, prob. = 1/2.

Based on two valid and different answer, we would say this is not well-defined.

(If you can't understand how the points can be related by a bijective function, you may consider this analogy:
"Find the prob. that 3 random points A,B,C on a plane forms an acute triangle."

1) Rotate, enlarge and shift the triangle so that A is on (0,0) and B is on (1,0). C is acute if x-value of C is between 0 and 1, so prob. = 1/infinity = 0
2) (0,1) can be related to R/(0,1) by a bijective function so that points between (0,1) is as many as R/(0,1), therefore prob. = 1/2.

but by the method of chung, area of B,D,F shouldnt =A,C,E...

when it is infinitive, there are still different between them![limit]

i think we have to find a way to calculate the area of them within a limited size

2011-08-29 14:43:52 補充：
ah yes!they are same! (vert opp angles)

我的數學不算太好,但想到少少頭緒,但不知是否正確,若有錯誤,還望指正
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,我先假設平面上是一點都沒有,
我是一點一點放下去,
最先放的三點是影響不到能否構成凹四邊形,
其實要構成一個凹四邊形,第四點才是最重要。
若把最先放的三點連起來,並延長線段,(如下圖),
會發現凡是把第四放到A,C,E區任何一個位置,出來的必定是凸四邊形
而凡是把第四放到B,D,F區任何一個位置,出來的必定是凹四邊形
若該平面的大小是無限大,A,B,C,D,E,F的面積也是無限大
所以概率為3/6=1/2





圖片參考：http://imgcld.yimg.com/8/n/HA00039848/o/701108270069013873449740.jpg

☂雨後陽光☀ ( 知識長 ) ,

有頭緒便可發表，但現在我不能告訴你答案。

2011-08-31 10:32:53 補充：
chung ,

留意中間的三角形。

1/2

Right ?