Hard probability problem

Reply #4

db101 — Sat, 04 Feb 2023 02:23:21 GMT

Great explanation. I thought that was pretty nifty when my kid showed me the solution. Who'd expect pi to show up in such a problem

Reply #3

cottoneyedjoe — Fri, 03 Feb 2023 20:08:25 GMT

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

1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... = pi/4

has the rearrangement

(1 + 1/5 - 1/3) + (1/9 + 1/13 - 17) + (1/17 + 1/21 - 1/11) + (1/25 + 1/29 - 1/15) + ... = pi/4 + ln(2)/4

That's a good one that mixes two kinds of transcendental numbers. Another curious one is

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + ... = ln(2)

vs... [ More ]

Reply #2

Wavepack — Fri, 03 Feb 2023 18:46:21 GMT

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... [ More ]

Reply #1

cottoneyedjoe — Fri, 03 Feb 2023 18:06:13 GMT

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... [ More ]

Hard probability problem

db101 — Tue, 31 Jan 2023 15:58:54 GMT

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... [ More ]