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 a perfect clear if 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:
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 explain why this is indeterminant, hopefully you can agree that it is indeterminant!
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.
