Randomly choose $n+1$ points on a $S^n-1$, probability of $n$-simplex containing center Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to simplify complex integration?probabilty of random points on perimeter containing centerSimpson's rule error rate for N-dimensionProbability that one of a set of four points lies within the triangle formed by the other threeInformation matrix for a Student's T distributionProbability of center enclosed by polygon of random pointsProbability that convex hull of $2n$ random points contains the $n-$sphere's centerProbability that a random triangle with vertices on a circle contains an arbitrary point inside said circleProbability that a stick randomly broken in three places can form a triangleProbability that the distance between two random points inside a circle is less than some value
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Randomly choose $n+1$ points on a $S^n-1$, probability of $n$-simplex containing center
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to simplify complex integration?probabilty of random points on perimeter containing centerSimpson's rule error rate for N-dimensionProbability that one of a set of four points lies within the triangle formed by the other threeInformation matrix for a Student's T distributionProbability of center enclosed by polygon of random pointsProbability that convex hull of $2n$ random points contains the $n-$sphere's centerProbability that a random triangle with vertices on a circle contains an arbitrary point inside said circleProbability that a stick randomly broken in three places can form a triangleProbability that the distance between two random points inside a circle is less than some value
$begingroup$
Randomly choose $n+1$ points on a $S^n-1$(surface of ball in $n$-dim space). What's the probability that the $n$-simplex formed by these $n+1$ points contain the center of the sphere?
I conjecture that the result is $frac12^n$.
For $n=2$, it's easy:
$$int_0^2pi frac12pi fracpi-theta2pidtheta = frac14$$
For $n=3$, I compute and the result is $1/8$ :
By the same method, but need to use spherical triangle area formula and spherical triangle cosine rule
The detailed computation is following:
Very complicated function integration, I can hardly imagine that such a numerical integration has analytical solution $1/8$. And this complicated method is hard to be generalized to higher dimension. There should be some simple explanation and can be generalized to higher dimension case.
calculus probability-theory geometric-probability
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
$endgroup$
|
show 2 more comments
$begingroup$
Randomly choose $n+1$ points on a $S^n-1$(surface of ball in $n$-dim space). What's the probability that the $n$-simplex formed by these $n+1$ points contain the center of the sphere?
I conjecture that the result is $frac12^n$.
For $n=2$, it's easy:
$$int_0^2pi frac12pi fracpi-theta2pidtheta = frac14$$
For $n=3$, I compute and the result is $1/8$ :
By the same method, but need to use spherical triangle area formula and spherical triangle cosine rule
The detailed computation is following:
Very complicated function integration, I can hardly imagine that such a numerical integration has analytical solution $1/8$. And this complicated method is hard to be generalized to higher dimension. There should be some simple explanation and can be generalized to higher dimension case.
calculus probability-theory geometric-probability
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
$endgroup$
$begingroup$
Any reason you have that conjecture?
$endgroup$
– Arctic Char
Apr 8 at 19:49
1
$begingroup$
This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily.
$endgroup$
– Arthur
Apr 8 at 19:53
$begingroup$
@ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$
$endgroup$
– maplemaple
Apr 8 at 19:53
$begingroup$
@Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case.
$endgroup$
– maplemaple
Apr 8 at 19:58
1
$begingroup$
Check out Wendel's Theorem for a generalization.
$endgroup$
– Mike Earnest
Apr 9 at 0:20
|
show 2 more comments
$begingroup$
Randomly choose $n+1$ points on a $S^n-1$(surface of ball in $n$-dim space). What's the probability that the $n$-simplex formed by these $n+1$ points contain the center of the sphere?
I conjecture that the result is $frac12^n$.
For $n=2$, it's easy:
$$int_0^2pi frac12pi fracpi-theta2pidtheta = frac14$$
For $n=3$, I compute and the result is $1/8$ :
By the same method, but need to use spherical triangle area formula and spherical triangle cosine rule
The detailed computation is following:
Very complicated function integration, I can hardly imagine that such a numerical integration has analytical solution $1/8$. And this complicated method is hard to be generalized to higher dimension. There should be some simple explanation and can be generalized to higher dimension case.
calculus probability-theory geometric-probability
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
$endgroup$
Randomly choose $n+1$ points on a $S^n-1$(surface of ball in $n$-dim space). What's the probability that the $n$-simplex formed by these $n+1$ points contain the center of the sphere?
I conjecture that the result is $frac12^n$.
For $n=2$, it's easy:
$$int_0^2pi frac12pi fracpi-theta2pidtheta = frac14$$
For $n=3$, I compute and the result is $1/8$ :
By the same method, but need to use spherical triangle area formula and spherical triangle cosine rule
The detailed computation is following:
Very complicated function integration, I can hardly imagine that such a numerical integration has analytical solution $1/8$. And this complicated method is hard to be generalized to higher dimension. There should be some simple explanation and can be generalized to higher dimension case.
calculus probability-theory geometric-probability
calculus probability-theory geometric-probability
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
edited Apr 8 at 20:06
maplemaple
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
asked Apr 8 at 19:46
maplemaplemaplemaple
35419
35419
$begingroup$
Any reason you have that conjecture?
$endgroup$
– Arctic Char
Apr 8 at 19:49
1
$begingroup$
This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily.
$endgroup$
– Arthur
Apr 8 at 19:53
$begingroup$
@ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$
$endgroup$
– maplemaple
Apr 8 at 19:53
$begingroup$
@Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case.
$endgroup$
– maplemaple
Apr 8 at 19:58
1
$begingroup$
Check out Wendel's Theorem for a generalization.
$endgroup$
– Mike Earnest
Apr 9 at 0:20
|
show 2 more comments
$begingroup$
Any reason you have that conjecture?
$endgroup$
– Arctic Char
Apr 8 at 19:49
1
$begingroup$
This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily.
$endgroup$
– Arthur
Apr 8 at 19:53
$begingroup$
@ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$
$endgroup$
– maplemaple
Apr 8 at 19:53
$begingroup$
@Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case.
$endgroup$
– maplemaple
Apr 8 at 19:58
1
$begingroup$
Check out Wendel's Theorem for a generalization.
$endgroup$
– Mike Earnest
Apr 9 at 0:20
$begingroup$
Any reason you have that conjecture?
$endgroup$
– Arctic Char
Apr 8 at 19:49
$begingroup$
Any reason you have that conjecture?
$endgroup$
– Arctic Char
Apr 8 at 19:49
1
1
$begingroup$
This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily.
$endgroup$
– Arthur
Apr 8 at 19:53
$begingroup$
This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily.
$endgroup$
– Arthur
Apr 8 at 19:53
$begingroup$
@ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$
$endgroup$
– maplemaple
Apr 8 at 19:53
$begingroup$
@ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$
$endgroup$
– maplemaple
Apr 8 at 19:53
$begingroup$
@Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case.
$endgroup$
– maplemaple
Apr 8 at 19:58
$begingroup$
@Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case.
$endgroup$
– maplemaple
Apr 8 at 19:58
1
1
$begingroup$
Check out Wendel's Theorem for a generalization.
$endgroup$
– Mike Earnest
Apr 9 at 0:20
$begingroup$
Check out Wendel's Theorem for a generalization.
$endgroup$
– Mike Earnest
Apr 9 at 0:20
|
show 2 more comments
1 Answer
1
active
oldest
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$begingroup$
In your picture, you've first chosen two points on the circle, then drawn the diagonals, then found the probability that the third point is on the correct cord. This will get difficult to generalize.
Instead, first choose the two diagonals, then choose the third point, then find the probability that the first two points are on the "correct ends" of the two diagonals. The probability is immediately seen to be $frac14$.
This method generalizes without much issue to higher dimensions: pick $n$ diagonals and the $(n+1)$th point at random, then place the first $n$ points on the ends of their respective diagonals. This gives you the answer $frac12^n$ that you conjectured.
answered Apr 8 at 20:03
ArthurArthur
123k7122211
$endgroup$
add a comment |
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$begingroup$
In your picture, you've first chosen two points on the circle, then drawn the diagonals, then found the probability that the third point is on the correct cord. This will get difficult to generalize.
Instead, first choose the two diagonals, then choose the third point, then find the probability that the first two points are on the "correct ends" of the two diagonals. The probability is immediately seen to be $frac14$.
This method generalizes without much issue to higher dimensions: pick $n$ diagonals and the $(n+1)$th point at random, then place the first $n$ points on the ends of their respective diagonals. This gives you the answer $frac12^n$ that you conjectured.
answered Apr 8 at 20:03
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
In your picture, you've first chosen two points on the circle, then drawn the diagonals, then found the probability that the third point is on the correct cord. This will get difficult to generalize.
Instead, first choose the two diagonals, then choose the third point, then find the probability that the first two points are on the "correct ends" of the two diagonals. The probability is immediately seen to be $frac14$.
This method generalizes without much issue to higher dimensions: pick $n$ diagonals and the $(n+1)$th point at random, then place the first $n$ points on the ends of their respective diagonals. This gives you the answer $frac12^n$ that you conjectured.
answered Apr 8 at 20:03
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
In your picture, you've first chosen two points on the circle, then drawn the diagonals, then found the probability that the third point is on the correct cord. This will get difficult to generalize.
Instead, first choose the two diagonals, then choose the third point, then find the probability that the first two points are on the "correct ends" of the two diagonals. The probability is immediately seen to be $frac14$.
This method generalizes without much issue to higher dimensions: pick $n$ diagonals and the $(n+1)$th point at random, then place the first $n$ points on the ends of their respective diagonals. This gives you the answer $frac12^n$ that you conjectured.
answered Apr 8 at 20:03
ArthurArthur
123k7122211
$endgroup$
In your picture, you've first chosen two points on the circle, then drawn the diagonals, then found the probability that the third point is on the correct cord. This will get difficult to generalize.
Instead, first choose the two diagonals, then choose the third point, then find the probability that the first two points are on the "correct ends" of the two diagonals. The probability is immediately seen to be $frac14$.
This method generalizes without much issue to higher dimensions: pick $n$ diagonals and the $(n+1)$th point at random, then place the first $n$ points on the ends of their respective diagonals. This gives you the answer $frac12^n$ that you conjectured.
answered Apr 8 at 20:03
ArthurArthur
123k7122211
answered Apr 8 at 20:03
ArthurArthur
123k7122211
answered Apr 8 at 20:03
ArthurArthur
123k7122211
answered Apr 8 at 20:03
ArthurArthur
123k7122211
123k7122211
add a comment |
add a comment |
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$begingroup$
Any reason you have that conjecture?
$endgroup$
– Arctic Char
Apr 8 at 19:49
1
$begingroup$
This is the problem considered in this 3blue1brown video. It is only solved for $n=2$ and $3$, but the method generalizes easily.
$endgroup$
– Arthur
Apr 8 at 19:53
$begingroup$
@ArcticChar I compute by brute force the case of $n=2$ the result is $1/4$ and the case $n=3$ the result is $1/8$
$endgroup$
– maplemaple
Apr 8 at 19:53
$begingroup$
@Arthur Yes, I compute above the case of $n=2, 3$. In $n=3$, by numerical integral is $0.125 = 1/8$. But I can hardly imagine this method can get analytical solution. And this method can hardly to generalize to higher dimension case.
$endgroup$
– maplemaple
Apr 8 at 19:58
1
$begingroup$
Check out Wendel's Theorem for a generalization.
$endgroup$
– Mike Earnest
Apr 9 at 0:20