[pittsy]

Try this:

http://video.yahoo.com/network/10114...9466&l=5218106

Wondering how it works.

[stovesax08]

Worked for me. I wonder how much of that is based on pure statistics?

[fdsasdasdf]

The entire thing is based upon mathematics. I'd need to do it a couple of times to notice the pattern, but what it looks like is basically the way he does it makes it so that you will end up in that corner no matter what you do.

After doing it a couple of times...
Basically from what I have seen, it is mathematically impossible to fall upon a square that he will chose to be "gone." If he chose even numbers, then it would be possible to choose a square to be gone that would be gone if it was odd, but because he uses only odd numbers, you absolutely cannot chose a square that will be gone no matter what you start off with. You basically have a lower probability of choosing specific squares each time a square is taken away (AKA without replacement). Because it is only odd, you can only choose a certain number of combinations. He has figured out what those will be, and found which will be impossible to choose. Those that are impossible to choose are taken away; therefore, once he gets that specific combination and specific number of squares taken away, you absolutely must land on that one specific square.

This is a little beyond my current comprehension of mathematics as I have not taken a proofs class yet, but I can almost guarantee you that it has something to do with odd numbers and not being able to choose what you first selected due to odd number properties and shrinking of the space where you can move.

This is a combination of conditional probability and mathematical properties. Gotta love math

[radix]

This is pretty easy. Take the first count to seven for example. There is no way to get to the square that he removes at the end of the count in seven steps (the house) without moving diagonally. Note that squares immediately adjacent the current square (the one your finger is on), can only be reached in an odd number of jumps. Squares not immediately adjacent (positioned diagonally) can only be reached in an even number of jumps. Both starting points are always an odd even number of jumps from the house icon, so there is no way you could have landed on it in an odd number of jumps. Seven is an odd number, therefore he can remove the square, because there is no way you could have followed the instructions and landed on it. It's also worth noting that two starting points are diagonally situated, and that after the first count to seven there are only four possible squares you could land on regardless of starting point (the squares with odd jump counts from the starting points). After the second count to three, there are only three possible squares that you could jump to in a three count (Q, :-|, 0). Repeat recursively.

[radix]

Originally Posted by Zell

This is a combination of conditional probability and mathematical properties. Gotta love math

It's easy to figure out using just basic arithmetic combined with simple observation. You're over-analyzing this.

[fdsasdasdf]

Originally Posted by radix

You're over-analyzing this.

I get that a lot. Simple math is the hardest for me yet hard math is the easiest for me...

[radix]

Originally Posted by Zell

I get that a lot. Simple math is the hardest for me yet hard math is the easiest for me...

I used to get that a lot as well, then I found this:

http://en.wikipedia.org/wiki/Occam's_razor

haven't had much of a problem since.