I am a big fan of the oldschool games and I once noticed that there is a sort parity associated to one and only one Tetris piece, theT piece. This parity is found with no other piece in the game.

Background:The Tetris playing field has width 10. Rotation is allowed, so there are then exactly 7 unique pieces, each of which is composed of 4 blocks.For convenience, we can name each piece by a letter. See this Wikipedia page for the Image (I is for the stick piece, O for the square, and S,T,Z,L,J are the others)

There are 2 sets of 2 pieces which are mirrors of each other, namely L,J and S,Z whereas the other three are symmetric I,O,T

Language:If a row is completely full, that row disappears. We call it aperfect clearif no blocks remain in the playing field. Since the blocks are size 4, and the playing field has width 10, the number of blocks for a perfect clear must always be a multiple of 5.

My Question:I noticed while playing that the T piece is particularly special. It seems that it has some sort of parity which no other piece has. Specifically:Quote:Conjecture: If we have played some number of pieces, and we have a perfect clear, then the number of T pieces used must be even. Moreover, the T piece is the only piece with this property.

What is the result of infinity minus infinity?

From a layman's perspective, imagine that I have an infinite number of hotel rooms, each numbered 1, 2, 3, 4, ...Then I give you all of them. I would have none left, so ∞−∞=0

On the other hand, if I give you all of the odd-numbered ones, then I still have an infinite number left. So ∞−∞=∞.

Now suppose that I give you all of them except for the first seven. Then ∞−∞=7.

While this doesn't explainwhythis is indeterminant, hopefully you can agree that itisindeterminant!

In Russian roulette, is it best to go first?

Now, the idea in a 2 player game is that it is best to be player 2, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player 1 (or your captor), thus winning, changing your total odds of losing to P1 - 3/6, P2 - 2/6, Captor - 1/6

And so on.

Catchfire wrote:Seems like a logical strategy when there's only one pistol between them. If both players had their own pistol, at the commencement player 2's turn six the game would already be over.

The Russian Roulette strategy above would not apply if the rules required the chamber to be spun before each shot is taken.

I lost many a round in my misspent youth. Thank God for my immortal soul. Next time, I'll just do the math.

JP Morgan bank CEO says $2 billion dollar loss is 'a rounding error'

And in the age of computers no less. It's finished.