pouët.net

omg, can i have your autograph next to the probability tree, havoc?!?!?

"i won't switch because i'm an asshole"

maali, sure you can!!!1

"Hi, Cliff!"

And don't tell me that wasn't way better than the numbers thing. Just don't.

HAVOC WHY IS YOUR SIGNATURE NOT A BEE YOU AIN'T NO GENIUS

Havoc is right because:

If I pick door A, the probability that it is the right choice is 1/3.

Now let's suppose that the moderator opens another door. The question is: what's the probability that the remaining (third) door (let's call it B) is the right choice?

If A is the right one, then of course it's 0. This is the case in 1/3 of all cases.

However, if A is the wrong one, then B must be the right one. And that's in 2/3 of the cases.

BTW, to demonstrate this better:

Imagine there are 1000 doors. You choose one. The moderator opens 998 of the remaining doors. Will you switch? ...

"i won't switch because i'm an asshole"

First of all, you don't tell me what I desire. That's fascist. Also, this is very old.

havoc <3 !!!!!

Even if havoc is right in the answer, the point by Doom is much better. Doom wins.

a car behind a door? i dont get it

I have a nice problem:

It appears that even numbers can be represented by the sum of two prime numbers. For example:

2=1+1
4=2+2
6=3+3
8=5+3
10=7+3
12=7+5
14=7+7
16=13+3

etc... prove that this is true for all even numbers.

... or find a counterexample.

1 is not a prime

http://en.wikipedia.org/wiki/Goldbach's_conjecture

Well, it is unsolved yet. It looks as if a probabilistic demostration is possible...

uhh.. spoiler!1

Well, right about the 1.

The problem with posting well-known old, unsolved maths problems like that is they appear in Wikipedia. You could probably find this without knowing the name of the idiot who came up with it in the first place:

Goldbach's conjecture

Anyway, WP has this nice graphic that should satisfy most people:


Number of ways to write an even number n as the sum of two primes (4 <= n <= 1,000,000)

You should try something that has a known proof: Given any integer n>2, x^n+y^n = z^n has no integer solutions for x,y,z (apart from 0,0,0). But why? So now the trick is to search Wikipedia for "Fermat".

Oh yeah, and 1 is not prime :P

Now if somebody had actually found a proof in this thread. Pouet would have been famous!

Nah. Mensa is monitoring the BBS. They'll steal any original ideas as soon as they're posted.

I think 65481211186144781158155612318465165432165468465189412051684846546544889997878451516849844216516498745132165487949849195165321655498498479845613216469798494651321654984984651316498798798798794613213216549879812298298715149112613665498749879846124 is a counter-example, but I have had no time to test it. Can you help me? :P

65481211186144781158155612318465165432165468465189412051684846546544889997878451 516849844216516498745132165487949849195165321655498498479845613216469798494651321 654984984651316498798798798794613213216549879812298298715149112613665498749879846 121 is prime. So is 3. So it is not a counter-example. This is based on the assumption that 65481211186144781158155612318465165432165468465189412051684846546544889997878451 516849844216516498745132165487949849195165321655498498479845613216469798494651321 654984984651316498798798798794613213216549879812298298715149112613665498749879846 121 is prime. Maple is confirming that and we'll have an answer in one or two centuries.

Meanwhile, here's an actual puzzle, probably well-known but why not:

Is it possible to transform the string "A" to the string "B" using the following four rules:

1. If the string ends in "A", you may append "B" to the end of it, e.g. "BA"->"BAB"
2. You may append the string to itself (repeat it), e.g. "BAB" -> "BABBAB"
3. "AAA" anywhere in the string may be replaced by "B", e.g. "BAAAB" -> "BBB"
4. "BB" anywhere in the string may be removed, e.g. "BBBA" -> "BA"

If yes, show an example, and if no, prove it.

Starting with the string A and applying rule 2 three times and then applying rule 3 two times will give you the desired result: ABBA

Easy problem Doom. The only way of replacing/deleting "A"s is by rule number 3. So, it is clear that to get "B"s alone, you need a number of "A"s being equal or multiple of 3.
Rule 1 and rule 4 doesn't change the number of A. Rule 2 multiply by 2 the number of A.

If a number n is not divisible between 3, then n-3 is also not divisible between 3.
If a number n is not divisble between 3, then 2*n is also not divisible between 3.

Thats all.