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One of my kids showed me this problem and while I doubt anyone cottoneyedjoe can solve it, it's too cool not to share with the rest of the class. it came from a math contest. I'll paraphrase it a little for clarity.
Alice and Brad each independently and randomly select a real number over the interval (0, 1). Assume a uniform distribution function for Alice's and Brad's numbers. They then divide Alice's number by Brad's number and round the result to the nearest integer. Assume values ending in .5 are rounded up. What is the probability the rounded value is an odd integer?
Hint: The answer is of the form p +q*π where p and q are rational numbers.
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Quote: Originally posted by db101 on Jan 31, 2023
One of my kids showed me this problem and while I doubt anyone cottoneyedjoe can solve it, it's too cool not to share with the rest of the class. it came from a math contest. I'll paraphrase it a little for clarity.
Alice and Brad each independently and randomly select a real number over the interval (0, 1). Assume a uniform distribution function for Alice's and Brad's numbers. They then divide Alice's number by Brad's number and round the result to the nearest integer. Assume values ending in .5 are rounded up. What is the probability the rounded value is an odd integer?
Hint: The answer is of the form p +q*π where p and q are rational numbers.
Nice problem! If X and Y are independent and uniformly distributed over the interval (0,1), then the probability that X/Y rounds to an odd number is
(pi - 1)/4 ≈ 0.5354
Here's how I worked it out. Let f(N) be the probability X/Y rounds to a non-negative integer N. Then you get the following table of values
N
f(N)
0
1/4
1
5/12
2
1/3 - 1/5 = 2/15
3
1/5 - 1/7 = 2/35
4
1/7 - 1/9 = 2/63
5
1/9 - 1/11 = 2/99
6
1/11 - 1/13 = 2/143
7
1/13 - 1/15 = 2/295
For N >= 2, the probability that X/Y rounds to N is 1/(2N-1) - 1/(2N + 1) = 2/(4N^2 - 1). This can be derived by finding the area of the region where N - 0.5 <= X/Y < N + 0.5 inside the square region (0,1) x (0,1) in the Cartesian plane. So the probability that X/Y rounds to an odd is equal to the infinite sum
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Nice solution. You would have scored high the Putnam exam.
I was wondering what you thought of the implications of the Riemann Rearrangement Theorem and it's implications on the foundations of Mathematics built upon infinite sums. By rearranging the terms in a conditionally convergent sum, you can have it converge to any real number!!! To me, that indicates a fundamental problem in non-discrete mathematics. You usually aren't taught about this "derangement theorem" until grad school, and even then, usually only to those who specialize in real and complex analysis.
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Quote: Originally posted by Wavepack on Feb 3, 2023
Nice solution. You would have scored high the Putnam exam.
I was wondering what you thought of the implications of the Riemann Rearrangement Theorem and it's implications on the foundations of Mathematics built upon infinite sums. By rearranging the terms in a conditionally convergent sum, you can have it converge to any real number!!! To me, that indicates a fundamental problem in non-discrete mathematics. You usually aren't taught about this "derangement theorem" until grad school, and even then, usually only to those who specialize in real and complex analysis.
Yeah it's pretty cool that when you have conditionally converging series a simple rearrangement can give you a different number. For example, the basic
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Quote: Originally posted by cottoneyedjoe on Feb 3, 2023
Nice problem! If X and Y are independent and uniformly distributed over the interval (0,1), then the probability that X/Y rounds to an odd number is
(pi - 1)/4 ≈ 0.5354
Here's how I worked it out. Let f(N) be the probability X/Y rounds to a non-negative integer N. Then you get the following table of values
N
f(N)
0
1/4
1
5/12
2
1/3 - 1/5 = 2/15
3
1/5 - 1/7 = 2/35
4
1/7 - 1/9 = 2/63
5
1/9 - 1/11 = 2/99
6
1/11 - 1/13 = 2/143
7
1/13 - 1/15 = 2/295
For N >= 2, the probability that X/Y rounds to N is 1/(2N-1) - 1/(2N + 1) = 2/(4N^2 - 1). This can be derived by finding the area of the region where N - 0.5 <= X/Y < N + 0.5 inside the square region (0,1) x (0,1) in the Cartesian plane. So the probability that X/Y rounds to an odd is equal to the infinite sum