Sven's Official BrainTeaser Thread !! and Sander.

[Jono]

you take over the goatio, then you come back, then you take over the lionio and bring the goatio back, then you take over the cornio, then come back, then take over the goatio ^_^

you got it  :thumbsup:

im guessin you get +1 karma for that  :cheers:

[Mike]

no you cant.

because taking the top and bottom rows, they have 7 spaces left, so obviously you can fill it horizontally (and thereby fill all the rest).

therefore you would have to have 3 horizontal and one verticle in each... but (as you can test in something like excel, or just common sense) that is impossible. so its impossible.

No you don't.

You may have 7 verticle if you'd like.

[Sander]

i'd like to thank the people honouring me and i thank sven for making this tread partly mine  :cheers: please make this tread the coolest  :biggrin:

and mike, in a minute i'm going to try yours in real

[sputnik]

no you cant.

because taking the top and bottom rows, they have 7 spaces left, so obviously you can fill it horizontally (and thereby fill all the rest).

therefore you would have to have 3 horizontal and one verticle in each... but (as you can test in something like excel, or just common sense) that is impossible. so its impossible.

No you don't.

You may have 7 verticle if you'd like.

yes but then if you have 7 verticle, if you look at the extreme left&right rows, the must also be verticle, and there is only 7 spaces there so you will have a space left.

[TW89]

Yeah man Sander some genius.

And at this point i'd like you Sven for starting this thread  :biggrin:  :thumbsup:

You like me? Aww..

Muthafucka I loooove you  lol    This thread cool though

[Mike]

no you cant.

because taking the top and bottom rows, they have 7 spaces left, so obviously you can fill it horizontally (and thereby fill all the rest).

therefore you would have to have 3 horizontal and one verticle in each... but (as you can test in something like excel, or just common sense) that is impossible. so its impossible.

No you don't.

You may have 7 verticle if you'd like.

yes but then if you have 7 verticle, if you look at the extreme left&right rows, the must also be verticle, and there is only 7 spaces there so you will have a space left.

They must not be verticle?

You are trying to find certain combinations which are not possible. What you do need is a proof for which none combination works. There is one, and it's quite simple.

Think of it as there are millions of possible ways to try this one. Why are all those million ways impossible to complete?

[Mike]

Didn't I post the answer to this one?

Strange, I thought I did. Well, here is goes.

What you achieve by taking the two bricks off is that you take away to bricks with the same color. Then you have 30 bricks of one color, and 32 of the other.

And as stated, two pieces with the same color is never next to eachother and one domino brick covers up exactly two pieces, one of each color. So to cover the board, you'll somehow be forced to cover two pieces of one color, without covering the other one, and therefore it's impossible.

New riddle
This one's pretty good, though it's hard.

A terrorist plans to highjack an aeroplane. The plane has 100 seats and after he buys his ticket it's all booked up. The highjacker is the first man to enter the plane, but he's so nervous that he forgot which seat he took so he takes on a random seat. The next person who enters the plane goes to his seat, but if the highjacker sits there, the person chooses another seat randomized. The next person tries to find his seat but if there's someone sitting there, he takes a random.

The procedure continues untill the last person gets aboard. Funnily enough, this person's a CIA-agent, and if someone's sitting on his seat, he'll get suspicious and will therefore cancel the flight.

What's the propability of the flight getting cancelled?

[oLi]

There are 100 seats and this is 1 person so to me the only logical answer is 1%

[sputnik]

the way i see it, we are finding the probability that the cia seat will be taken.

if 5 ppl are on plane, p1 (terrorist) has a 4/5 chance of not taking the cia seat, keeping the plane uncancelled
p2 has a 3/4 chance of not taking his seat
p3 has a 2/3 chance of not taking his seat
p4 has a 1/2 chance of not taking his seat
obv p5 is cia agent.

so 4/5 x 3/4 x 2/3 x 1/2 = 1/5 chance no one will take his seat

so a 4/5 chance that his seat will be taken and flight cancelled.

so in this riddle, should be a 99/100 chance that the flight gets cancelled :3 i think

but as oli said, that seems too simple

[Mike]

The people don't choose their seats at random, unless their seat's taken.

[Sander]

there's 1% chance that the highjacker will take his own place.
so 99% that someone elses place is seated by him.

that person will take another seat (random) and he has 98/99 chance that he takes a seat of someone that has not entered the plane yet (the other one is that he takes the highjackers seat and that will cause that all others seat their own seat)

so far there's a (99/100)*(98/99) chance that the flight will be cancelled

well the person that noticed that his seat will take another random place and he has 97/98 chance that he doesn't take the highjackers place (if he does everything will become normal again)

well you understand that this goes on until person 99 has the choice to pick the seat of the highjacker or the ciaman. this is 1/2.
now if you mulitply all these chances: (99/100)*(98/99)*(97/98)......(2/3)*(1/2)
persons that know maths know that you can multiply all numerators and all denominators. there for the final answer is: (1*2*3....*99)/(2*3*4.....*100)*100. you can take away all number that are both above and below the dash so what remains is: 1/100. there's therefore exactly 1% chance that the flight will get cancelled. oli and sputnik were right

[Mike]

What you're calculating is that if you let 99 people choose a random spot in a plane with 100 seats, what's the propability that the last person gets the seat he booked.

What if the highjacker takes a random seat which is not the seat of the person number 2, so that number 2 just can get seated on his own seat?

The second person does only choose a random seat, if the highjacker stole his seat, which is a 1/100 chance.

Besides, you calculated what the propability of the flight not getting cancelled is, (if everyone were to choose at random), the propability of it getting cancelled would then be 99%, but that's not the correct answer, since they do not choose their seats at random unless their seat is taken.