The Math Thread

[NanTheDark]

Did I fail it?

[*Emelia K. Fletcher]

P(G) [1/3, no greens left]
P(B/R) [1, returns a green]
P(G) [1/5, no greens left]
1/15

P(B) [1/3, two greens left]
P(G) [2/4, one green left]
P(G) [1/5, no greens left]
1/30

P(R) [1/3, two greens left]
P(G) [2/4, one green left]
P(G) [1/5, no greens left]
1/30

2/15 overall



i missed that during copy-pasting lmao

[lordpat]

Yaaaay! That's correct.

[Kimonio]

So this happened last night, and I forgot to get the logs.

Let's imagine that an atom of Hydrogen started the birth of the universe. This atom is a sphere.

How would you guys calculate the estimated size of the universe, if the universe is a sphere?

[Oranjui]

i literally just spent the past half hour sketching a diagram of every possible drawing

no i'm not really sure what i'm doing with my life

P(no green remaining after 3 draws) = 2/15





more symbolically or whatever you'd have something like

case 1: draw #1 is green [aka {G}]
P(G) = 1/3; 2r,2b (0 greens after draw #1)
P(R or B) = 4/4 = 1; 1r,1g,3b or 3r,1g,1b (1 green after draw #2, no matter what the colour was)
P(G) = 1/5; (0 greens after draw #3)
==> 1/3 * 1 * 1/5 = 1/15

case 2: draw #1 is not green [aka {G^C}={R,B}]
P(R or B) = 2/3; 2r,2g or 2g,2b (2 greens after draw #1, no matter what the colour was)
P(G) = 2/4 = 1/2; 3r,1g,1b or 1r,1g,3b (1 green after draw #2, no matter what the first draw was)
P(G) = 1/5; (0 greens after draw #3)
==> 2/3 * 1/2 * 1/5 = 2/30 = 1/15

P(case 1 or case 2) = 1/15 + 1/15 = 2/15

[lordpat]

Wow, that is dedication, now I know my problem is correct.

[NanTheDark]

I fail so hard xD

[BrawlerEX]

Quite a bit of reading: show

A face-down stack of 8 playing cards consisted of 4 Aces (A's) and 4 Kings (K's). After I revealed and then removed the top card, I moved the new top card to the bottom of the stack without revealing the card. I repeated this procedure until the stack was left with only 1 card . I repeated this procedure until the stack was left with only 1 card, which I then revealed. The cards revealed were AKAKAKAK, in that order. If my original stack of 8 cards had simply been revealed one card at a time, from top to bottom (without ever moving cards to the bottom of the stack), in what order would they have been revealed?



I got this problem from the CAML (California Mathematics League) competition in good ol' bone dry California. It's part of a six problem test. All the schools compete. They give 'bout 5-6 of these tests every school year... Pretty fun stuff

[Oranjui]

Lel they had that exact same question on one here in Wisconsin a few months ago. I never did figure it out. Mostly because we only had 20 minutes to complete it and I had like 30 seconds left by the point I got to that question and then never thought about it again until now. I'll give it a stab, though:

Spoiler: show

Beginning order: 12345678
--> 1, 3456782
--> 13, 567824
--> 135, 78246
--> 1357, 2468
--> 13572, 684
--> 135726, 48
--> 1357264, 8
--> 13572648 is the final order of cards revealed.

13572648 = AKAKAKAK means that
1=A
2=A
3=K
4=A
5=A
6=K
7=K
8=K
which means the original order was AAKAAKKK?

Except your question was "in what order would they have been revealed," so technically the answer is AKAKAKAK :p

[NanTheDark]

I'll just ignore this topic until I can get all of my lost math knowledge back. I suck right now.

*goes to the open fields to punch out math monsters*