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I'm pretty sure I've figured out this week's Riddler Express. The Riddler Classic appears to be more of a challenge.

The Riddler Express: Fix a circle. Choose three points at random (uniform distribution) on the circle. (To be clear, that's on the circle itself, not in its interior.) Those three points form a triangle. What is the probability that the center of the circle lies within the triangle? My answer:

Spoiler

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The Riddler Classic: Fix a sphere. Choose four points at random (uniform distribution) on the sphere. (To be clear, that's on the sphere itself, not in its interior.) Those four points form a tetrahedron. What is the probability that the center of the sphere lies within the tetrahedron? I don't think the technique I used to figure out the Riddler Express generalizes to this case. --Bob

Bob78164 wrote:I'm pretty sure I've figured out this week's Riddler Express. The Riddler Classic appears to be more of a challenge.

The Riddler Express: Fix a circle. Choose three points at random (uniform distribution) on the circle. (To be clear, that's on the circle itself, not in its interior.) Those three points form a triangle. What is the probability that the center of the circle lies within the triangle? My answer:

Spoiler

0.25

.

The Riddler Classic: Fix a sphere. Choose four points at random (uniform distribution) on the sphere. (To be clear, that's on the sphere itself, not in its interior.) Those four points form a tetrahedron. What is the probability that the center of the sphere lies within the tetrahedron? I don't think the technique I used to figure out the Riddler Express generalizes to this case. --Bob

I don't think the uniform distribution constraint applies to the Express puzzle. Otherwise, I think the probability would be 1.00.

mrkelley23 wrote:

Bob78164 wrote:I'm pretty sure I've figured out this week's Riddler Express. The Riddler Classic appears to be more of a challenge.

The Riddler Express: Fix a circle. Choose three points at random (uniform distribution) on the circle. (To be clear, that's on the circle itself, not in its interior.) Those three points form a triangle. What is the probability that the center of the circle lies within the triangle? My answer:

Spoiler

0.25

.

The Riddler Classic: Fix a sphere. Choose four points at random (uniform distribution) on the sphere. (To be clear, that's on the sphere itself, not in its interior.) Those four points form a tetrahedron. What is the probability that the center of the sphere lies within the tetrahedron? I don't think the technique I used to figure out the Riddler Express generalizes to this case. --Bob

I don't think the uniform distribution constraint applies to the Express puzzle. Otherwise, I think the probability would be 1.00.

Definitely not. Consider any three points in the upper half of the circle, which obviously occurs with probability greater than zero. The triangle they form definitely doesn't include the center of the circle. --Bob

Bob78164 wrote:I'm pretty sure I've figured out this week's Riddler Express. The Riddler Classic appears to be more of a challenge.

The Riddler Express: Fix a circle. Choose three points at random (uniform distribution) on the circle. (To be clear, that's on the circle itself, not in its interior.) Those three points form a triangle. What is the probability that the center of the circle lies within the triangle? My answer:

Spoiler

0.25

.

The Riddler Classic: Fix a sphere. Choose four points at random (uniform distribution) on the sphere. (To be clear, that's on the sphere itself, not in its interior.) Those four points form a tetrahedron. What is the probability that the center of the sphere lies within the tetrahedron? I don't think the technique I used to figure out the Riddler Express generalizes to this case. --Bob

Express: You are correct.
Classic:

Spoiler

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Bob78164 wrote:

mrkelley23 wrote:

Bob78164 wrote:I'm pretty sure I've figured out this week's Riddler Express. The Riddler Classic appears to be more of a challenge.

The Riddler Express: Fix a circle. Choose three points at random (uniform distribution) on the circle. (To be clear, that's on the circle itself, not in its interior.) Those three points form a triangle. What is the probability that the center of the circle lies within the triangle? My answer:

Spoiler

0.25

.

The Riddler Classic: Fix a sphere. Choose four points at random (uniform distribution) on the sphere. (To be clear, that's on the sphere itself, not in its interior.) Those four points form a tetrahedron. What is the probability that the center of the sphere lies within the tetrahedron? I don't think the technique I used to figure out the Riddler Express generalizes to this case. --Bob

I don't think the uniform distribution constraint applies to the Express puzzle. Otherwise, I think the probability would be 1.00.

Definitely not. Consider any three points in the upper half of the circle, which obviously occurs with probability greater than zero. The triangle they form definitely doesn't include the center of the circle. --Bob

If all three points are in the upper half of the circle, then they are by definition not "uniformly distributed." That's why the classic uses the words "uniformly distributed" and the Express does not.

mrkelley23 wrote:

Bob78164 wrote:

mrkelley23 wrote:
I don't think the uniform distribution constraint applies to the Express puzzle. Otherwise, I think the probability would be 1.00.

Definitely not. Consider any three points in the upper half of the circle, which obviously occurs with probability greater than zero. The triangle they form definitely doesn't include the center of the circle. --Bob

If all three points are in the upper half of the circle, then they are by definition not "uniformly distributed." That's why the classic uses the words "uniformly distributed" and the Express does not.

Uniform distribution refers to the probability distribution. It's a more precise way of saying that each point is equally likely (which isn't a helpful concept for continuous probability distributions because the probability of any specific point being selected is 0). Uniform distribution (in this context) means that the probability that your randomly selected point ends up in a given arc is the ratio of the length of that arc to 2*pi.

An alternative, for instance, would be that points are twice as likely to end up in the lower semicircle than in the upper semicircle. --Bob