Oops I broke the laws of math.

[Flaming_Dingleberry]

MrRubix, here's a problem that was given to all six-seven periods of a class I was in. I was the only one to solve it within the week and I want you to solve it. This is probably ****ing easy as hell, but still, it's the hardest math problem I retained and know the answer to lmfao.

There are 12 bags of gold, one bag has fake gold in it, you don't know whether the fake gold is lighter or heavier than real gold, you have a scale, you can only use it three times, what variations of weighing the gold will tell you which bag is fake gold?

Everyone else is welcome to try to solve my crappy algebra problem.

[leonid]

I've already solved that question in the past, so I won't spoil it..

[Flaming_Dingleberry]

FFRers are smarter than my high school xD

Good thing I was destined to be an FFRer.

That's why I was able to solve it of course.

[MrRubix]

Originally Posted by Flaming_Dingleberry

MrRubix, here's a problem that was given to all six-seven periods of a class I was in. I was the only one to solve it within the week and I want you to solve it. This is probably ****ing easy as hell, but still, it's the hardest math problem I retained and know the answer to lmfao.

There are 12 bags of gold, one bag has fake gold in it, you don't know whether the fake gold is lighter or heavier than real gold, you have a scale, you can only use it three times, what variations of weighing the gold will tell you which bag is fake gold?

Everyone else is welcome to try to solve my crappy algebra problem.

Split the 12 bags into three groups of four where each sub-group has a single bag and a set of three bags, which we'll denote as X1 and X3, Y1 and Y3, and Z1 and Z3.

(Weighing 1) Compare: X1+X3 to Y1+Y3 (If equal, group Z has irregular bag)
If this happens: (Weighing 2) Compare: Y3 to Z3
If Z3 heavier/lighter, (Weighing 3) compare two bags within Z3 to deduce the irregular.
If Y3=Z3 though, then (Weighing 3) compare Y1 to Z1 to see whether Z1 is heavier/lighter.
But if Weighing 1 gave an imbalance, then compare (Weighing 2) X1+Y3 to Y1+Z3.
If they are equal, then the irregular bag is in X3, so (Weighing 3) comparing any two there will single the irregular bag.
But, otherwise, (Weighing 3) comparing X1 to Z1 will finish things off if X1+Y3 was heavier, or (Weighing 3) comparing two bags within Y3 will single the irregular bag out if X1+Y3 was heavier.

[MrRubix]

Originally Posted by leonid

@Rubix 9/19

I'm having trouble proving whether a certain rule is optimal. Would be good if I knew the optimal expected gain beforehand

Yes, 9/19 is correct -- good job!

[Flaming_Dingleberry]

Well, of course, that's right Rubix, dude how does your brain work? How did you decide to group the bags the way you did?

[leonid]

Solving a LOT of puzzles would work i guess

[MrRubix]

Originally Posted by Flaming_Dingleberry

Well, of course, that's right Rubix, dude how does your brain work? How did you decide to group the bags the way you did?

I've done problems like that countless times before -- it's kinda second nature by now. In my opinion, one of the best ways to learn how to problemsolve is to just solve problems. Over the years, you do so many of them that it becomes easier to figure out when to apply certain logic to certain problems. For instance, in this case it's a good idea to work with triplet groups, since triplets allow for the easiest way to operate on a "if these two groups don't equate, then the residual group has the odd one out" style of solving.

[Flaming_Dingleberry]

Are you good at keybeat Rubix?

And what does a Rubik's cube look like when you do your siggy?

[MrRubix]

Nope


My ring fingers are retarded

[leonid]

Originally Posted by Flaming_Dingleberry

And what does a Rubik's cube look like when you do your siggy?

LOL

[MrRubix]

Originally Posted by Flaming_Dingleberry

And what does a Rubik's cube look like when you do your siggy?

Each algorithm is on its own line -- one algorithm per cube.

[emerald000]

Expected gain for your 52 cards draw game is 3.7316053312...$

[MrRubix]

Under what strategy?

The optimal solution has many possible (weighted) payoffs.

[korny]

I only learned the math that was necessary for every day living.

[Divinefury]

Ugh rubix you could have said that the algorithms were seperate i tried doing the whole thing on one cube

[emerald000]

Originally Posted by MrRubix

Under what strategy?

The strategy is weird, but it is composed as followed. Format: (Number of blacks drawn---stop if you have drawn that much reds)

0---8
1---9
2---10
3---11
4---12
5---13
6---14
7---15
8---16
9---16
10---17
11---18
12---19
13---20
14---21
15---22
16---23
17---24
18---25
19---26
20---26
21---27
22---28
23---29
24---30
25---31
26---32
27---33
28---34
29---34
30---35
31---36
32---37
33---38
34---39
35---40
36---40
37---41
38---42
39---43
40---44
41---45
42---45
43---46
44---47
45---48
46---49
47---49
48---50
49---51
50---51
51---52
52---52

[leonid]

Are you assuming a deck has 52 blacks and 52 reds?

[emerald000]

*Goes hide in a corner...*

Let me modify that a bit.

Expected outcome: 2.62447554899

0---6
1---7
2---8
3---8
4---9
5---10
6---11
7---12
8---13
9---14
10---14
11---15
12---16
13---17
14---18
15---19
16---19
17---20
18---21
19---22
20---23
21---23
22---24
23---25
24---25
25---26
26---26

Should be better.

A big lol at me for actually figuring out how to do the problem but not how to count how much blacks/reds there are in a deck of cards.

[stargroup]

Originally Posted by MrRubix

Come on man, sack up and at least give it a try. It is hard (it took me literally all day to solve before I was certain of my answer), but it isn't impossible.

dude, you don't understand

I have one week to finish college apps, annotations for a a book and a paper to go along with it

stop making me look like an ass do you really think I would pass up a math problem