Thread "Extreme Math Challenge... Squared!" in forum "MAFIA... and other forum games"

Posted at 10:04PM, Apr. 1, 2008

=) Good call

*when Bacon realizes he did something smart by accident and plays it off as if it were intentional*

XD

Posted at 11:04PM, Apr. 1, 2008

Okay, time for a word problem:

You are in a room with three doors and a guard. The guard tells you that behind one door is an exit and behind the other two are rabid tigers. He also, says that he knows which door the exit is behind. You choose a random door, but before you open it, the guard opens a different door which contains a tiger and then immediately closes it. He, then, gives you the option of choosing the other unopened door.

The question: Which door would you choose?

Let me clarify:

- You chose door #1, a tiger is behind door #2, you have the option of choosing door #3

- This ISN'T a lateral thinking puzzle in the sense that you can't stay in the room until the tigers die or make the tigers attack the guards while you escape or do some abstract stunt that "magically" gets you out.

- This question DOES involve math (mainly probability), so to get full credit, SHOW YOUR WORK (AKA Don't guess, think)!!!

Posted at 2:04AM, Apr. 6, 2008

so the chances of a door not having a tiger are 1/3
but he eliminates one of the doors which does have a tiger so the chances are 1/2

but the chances of a door having a tiger are 2/3
with the revealed tiger that makes it… 1/3?

so the chances of your chosen door having a tiger are 1/3 still… i think…

I'd stay with my chosen door

Posted at 9:04PM, Apr. 14, 2008

I hate showing work… but I'd have to say door number three.

The Psycho… Guard I mean… shows the middle door, so, the two tigers would be in adjacent rooms. By doing this, I'd feel MUCH safer with the next door… on the chance that he showed me the SECOND tiger.

X*X=Y
Y+X=Z
Z/X=B
B-X=1

This is a little problem I figured out by accident in math once: All of the above problems leads to the next. Solve for X.

Posted at 10:04PM, Apr. 15, 2008

Okay, time for a word problem:

You are in a room with three doors and a guard. The guard tells you that behind one door is an exit and behind the other two are rabid tigers. He also, says that he knows which door the exit is behind. You choose a random door, but before you open it, the guard opens a different door which contains a tiger and then immediately closes it. He, then, gives you the option of choosing the other unopened door.

The question: Which door would you choose?

Well, if you initially pick the door with the exit, and the guard shows you a door with a tiger, then you have the choice between your exit and a tiger door, that's 1/2 chance each.

But, if you choose a door with a tiger initially, then, since the guard reveals the other door with a tiger, by switching, you'd end up at the exit.

Since there are two doors with tigers, this switching to the exit can occur twice, while switching to a tiger can only occur once. Thus, the probability of switching doors to the exit is 2/3, and switching to a door is 1/3.

And since there are 2 doors with a tiger, and one with the exit, by keeping the door you originally chose, that's 2/3 chance of failure, and 1/3 chance of success.

Thus it is best to switch doors.

X*X=Y
Y+X=Z
Z/X=B
B-X=1

This is a little problem I figured out by accident in math once: All of the above problems leads to the next. Solve for X.

B-X=1
B=1+X

Z/X=B
Z/X=1+X
Z=X+X^2

Y+X=Z
Y+X=X+X^2
Y=X^2

X*X=y
X*X=X^2
X^2=X^2

Err, yeah there is no real "solving" for x here.

———–

Here's a cute and simple one:

How do you make 1000 by the addition of numbers whose digits are only 8's and where the sum of the number of digits of all numbers used is 8?

(incase you don't understand that second part of the question, I mean say you add 34 and 233, then the sum of the number of their digits is 5)

Posted at 12:04PM, Apr. 24, 2008

^Bump.