*humorman*

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.

*BloodTh*

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)