"Get Out of The Hell !"

You’re new to hell, and you’re given a choice: You can go directly to the fourth circle, or you can play simultaneous chess games against Alexander Alekhine and Aron Nimzowitsch. Alekhine always plays black and smokes a pipe of brimstone. Nimzowitsch plays white and wears cuff links made of human teeth. Neither has ever lost.

If you can manage even a draw against either player, you’ll be set free. But if they both beat you, you’ll go to the eighth circle for eternity.

What should you do?

This trick will save your life! 

Freedom From The Hell !

How you are trapped in hell?

Alekhine always plays black and Nimzowitsch plays always white. Obviously, you will be forced to choose white against Alekhine and black against Nimzowitsch. 

Wait for Nimzowitsch's first move and then play the same move on Alekhine’s board. Note how Alekhine responds to move & copy that move on Nimzowitsch's board.

This way, effectively Nimzowitsch is playing against Alekhine and you are just transferring moves between to 2 masters.

Since, they never lost a single chess game, there are high chances that the game between two ends in draw.

Even if Alekhine wins with his black then you are winning against Nimzowitsch as you are copying Alekhine's move against Nimzowitsch's white using your black. Same is case if Nimzowitsch wins.

In short, you will end up with either draw with both or win against at least one (against both is impossible). You will be free in any case.

And in fact, you need not to have knowledge of how to play chess to get freedom from the hell.

Tricky Probability Puzzle of 4 Balls

I place four balls in a hat: a blue one, a white one, and two red ones. Now I draw two balls, look at them, and announce that at least one of them is red. What is the chance that the other is red?

Well, it's not 1/3!

Tricky Probability Puzzle of 4 Balls : Solution

What was the puzzle?

It's not 1/3. It would have been 1/3 if I had taken first ball out, announced it as red and then taken second ball out. But I have taken pair of ball out. So, there are 6 possible combinations.

red 1 - red 2
red 1 - white
red 2 - white
red 1 - blue
red 2 - blue
white - blue 

Out of those 6, last is invalid as I already announced the first ball is red. That leaves only 5 valid combinations.

And out of 5 possible combinations only first has desired outcome i.e. both are red balls.
Hence, there is 1/5 the chance that the other is red