3 probability math puzzles

Reply #18

db101 — Tue, 26 Jul 2022 23:56:51 GMT

That matches what I got doing random samples.

Reply #17

cottoneyedjoe — Mon, 25 Jul 2022 16:35:33 GMT

Average distance between two random points on the boundary of a 1x1 square is

(3 + sqrt(2) + 5*LN(sqrt(2) + 1)) / 12 0.7351

Average area of a triangle whose vertices are random points on the boundary of a circle with radius 1 is

3 / (2*pi) 0.4775

Reply #16

Orange71 — Sun, 24 Jul 2022 15:06:08 GMT

Agree. Relative to the problem statement, I had erroneously used polar coordinates assuming uniform r on the interval [0,R] and random theta on the interval [0,2*Pi]. The probability per unit area as a function of the random variable r here is 1/(Pi*R*r), so as we approach the origin the probability per unit area becomes asymptotically infinite.

I will also think about the correct integral for true random uniform x/y Cartesian probabilities on/inside the circle.

Reply #15

cottoneyedjoe — Sun, 24 Jul 2022 03:55:58 GMT

If you set the coordinates of your random points as (r*cos(theta), r*sin(theta)) and (p*cos(phi), p*sin(phi)), where r and p are uniformly distributed over (0, 1) and theta and phi are informly distributed over (0, 2*pi) you will get (2/3)R^2 as your answer. However, there is a problem with that approach. The random point (r*cos(theta), r*sin(theta)) (where r and theta are uniformly random) is not uniformly distributed over the circle. The points are concentrated in the center.

So if you

Reply #14

Orange71 — Sat, 23 Jul 2022 22:41:29 GMT

Here's another puzzle similar to the original #3.

Suppose we have a circle of radius R. Two points are chosen at random on or inside the radius of the circle. All locations on or inside the circle have equal (uniform) probability. A line segment L is drawn between the two points. If a square is then drawn with side length L, what is the average area of the square?

Spoiler: I get (2/3)R^2. Can anyone confirm he gets the same answer

Reply #13

db101 — Sat, 23 Jul 2022 07:38:32 GMT

Problem 3 is good because it can spawn a lot of related questions. Here are a couple I came up with.

Given a 1x1 square and 2 random points on the boundary (not inside), what is the average distance between them? I worked out that you have to take the weighted average of 3 cases: both points on the same side (1/4), points on adjacent sides l1/2), and points on opposite sides (1/4).

Given a circle of radius R and 2 random points on the circumference, the average distance between them

Reply #12

Orange71 — Fri, 22 Jul 2022 21:18:24 GMT

Thanks. Very useful. I'll use that going forward.

Another interesting problem:

We have a line segment of length 1, so if x is the variable representing distance from the origin, x is defined on the interval [0,1]. Two points (x_1 and x_2) on the [0,1] line segment are chosen at random according to the Uniform Distribution. What is the average distance between the two points (in terms of absolute value, regardless whether x_1 or x_2 is greater)? It turns out to be 1/3, our familiar friend.... [ More ]

Reply #11

cottoneyedjoe — Fri, 22 Jul 2022 19:32:49 GMT

After you create the image, upload it to a third-party image hosting site like imgur. Then the grab the url of the image where its hosted on imgur. Then when you type a post, click on the little image icon and paste the imgur url in the pop up window. (some people use other sites besides imgur, but I find that one the easiest to use since it doesn't require sign in.

Reply #10

Orange71 — Fri, 22 Jul 2022 18:53:01 GMT

Yes, I didn't include the 1/(b-a) constant term, to the 4th power, in my post since it's 1.

I found the same with the line segment problem. Much harder since you have to deal with integrating the square root. Isn't it interesting that the solution is not just the square root of the solution for the problem you posted.

I'd love to know how you posted LaTex equations into the editor for this site. I have had no luck uploading any image file. Their tutorial is not helpful.... [ More ]

Reply #9

cottoneyedjoe — Fri, 22 Jul 2022 18:02:03 GMT

Yeah that's how I think most would solve it. The area integral is not too bad because it's just polynomial, and as you discovered you can split it into two easier integrals of two variables. If you let one endpoint be (x, y) and the other be (u, v), it's

The factor in front is because in these types of problems you technically need to divide by the length of each interval to find the average, but conveniently in this problem they all = 1.

On the other hand, the average length... [ More ]

Reply #8

Orange71 — Fri, 22 Jul 2022 17:33:09 GMT

It is a quadruple integral, at least that's one way to solve it. Suppose we have two points A and B. Then, for two orthogonal axes, x and y, we have 4 variables x_A, x_B, y_A, and y_B, where _ means sub . The side/length of the solution square is therefore sqrt((x_A - x_B)^2 + (y_A - y_B)^2). So if we square that, for the area of the solution square, it's just f(x_A,x_B,y_A,y_B) = (x_A - x_B)^2 + (y_A - y_B)^2. Here, x_A, x_B, y_A, and y_B random uniform variables in the range [0,1].

If you q... [ More ]

Reply #7

Orange71 — Fri, 22 Jul 2022 16:32:01 GMT

Say what

Reply #6

Mayte — Fri, 22 Jul 2022 12:56:40 GMT

I believe that mathematical puzzles have to do with the lottery, for example those puzzles, like what they say, think of a number, add such an amount, subtract such an amount, the result is this, so in the lottery it is the same, since in the lottery the results are predetermined, and in computing there are only 2 types of basic operations that is addition and subtraction that's all, regardsssss... .

Reply #5

db101 — Fri, 22 Jul 2022 08:04:19 GMT

when I took samples to find the average length it was significantly less than the root of 1/3. This makes sense to me because if you have a 9x9 square and 1x1 square the avg area is 41 but the avg side length is 5. I think the right mathematical approach is to look at areas rather than lengths. Seeing as there are 2+2=4 variables at play I think it must be a quadruple integral. 2 for the xy coordinates of one endpoint and 2 for the xy coordinates of the other.

Reply #4

Orange71 — Fri, 22 Jul 2022 02:08:11 GMT

Wow. I'm missing something on the last problem. If we take one dimension, a line segment, between 0 at 1, the average distance between two points chosen at random is 1/3. You can show that as result of a double definite integral of |y-x| from 0 to 1 on dy dx. Thus, in the perpendicular direction (say y, instead of x), it would also be 1/3. If you take the hypotenuse, the length would be sqrt(2)/3. Square that for the area, it would be 2/3. So I'm off by a factor of 2. Any help on the solution ex... [ More ]

Reply #3

cottoneyedjoe — Thu, 21 Jul 2022 23:16:00 GMT

It is 1/3. You can do that problem with calc, or doing evenly spaced random samples in a square and taking the limit as the space between them decreases, but there may also be a clever solution that requires less computation.

Good job on the answers.

Reply #2

db101 — Thu, 21 Jul 2022 20:33:10 GMT

Okay I did the third one with random samples and the answer was very close to 1/3. Didnt expect that.

Reply #1

db101 — Tue, 19 Jul 2022 19:50:56 GMT

The first one is a good puzzler. The answer is 2/3.

The second one is okay. I think the answer is 12/70

Dont know how to do the third but I could estimate it with random samples. Looks like a calculus problem.

3 probability math puzzles

cottoneyedjoe — Fri, 15 Jul 2022 21:06:49 GMT

I don't remember the sources for these, but they combine probability and geometry in a nice way. Not applicable to the lottery in any way, shape or form.

1) (Easier)

Let C be a circle with a diameter of 2. Let P and Q be two points picked at random along the circumference of C (assume a uniform random distribution). What is the probability that the length of the chord connecting P and Q is at least 1?

2) (Easier)

Let Q be a cube. Let A, B, C, and D be four distinct vertices of Q pick... [ More ]