Solve this Riddle if you can

Johny's mother had three children.

The oldest child's name was April. The middle child's name was May.

What was the youngest child's name?

Morgan

Two strong chessplayers have been enjoying a casual game on an open-air board in a public park. When there is nothing left on the board except two kings and a black rook, White offers a draw. Black (to move) grins and says "Agreed - there's no win here". And they go off for an ice cream.

Why?

The rook is a crow, not a chesspiece.

Bah. Almost took me longer to write the riddle than it did for you to answer it.

That grins gave me the answer. I would smile too when a rook would land on the chess board.

Ok, this one is for you:

You think that it is specially for you, but I think it is not for you at all. Anyone thinks that. Except someone in Life of Brian.

Individual.

Close

Here's a very nice riddle I was recently asked. I was able to solve if with a bit of help.

There's a 36 story building, and you have two identical balls made of glass.

If you throw a ball from any story, it may or may not break. But you do not know if it would break or not. So it is possible that the ball would break from the first floor, and it is also possible that it will not break from the 36th floor.

Your goal is to find out precisely at which floor the ball would break (or not break at all). To do so, you may throw the balls from any floor you like, and as long as it doesn't break, you may throw it again. It does not matter if at the end both balls break, as long as you find out the exact floor where the ball breaks (or not at all). However - you must do so using the least amount of throws possible.

For example, if you find out that a ball breaks from the 11th floor, but does not break from the 9th floor, you still cannot tell exactly where it breaks. Because it could break from both the 11th or the 10th floors.

Good luck!

Does the part about having 2 balls have any significance? At least I can't see hoe it does. I would say just first try from the 18th floor, then if it doesn't break try from the 9th, etc... I think you get the idea.

It makes for all the difference.

If you have just one ball, there is only one way to do it. You start on the first floor. If it doesn't break - you go up one floor, and so on. This way, if you have really bad luck, it may take 36 tries to find out the exact floor where a ball breaks (or not at all). You cannot start from the second floor with just one ball, because if it breaks, you cannot tell whether the ball also breaks on the first floor or not.

If you have two balls, you have more flexibility, because you can afford to lose one and carry on with the second. In your example, if you throw a ball from the 18th floor and it breaks - what do you do? throwing the second ball from the 9th then is bad, because if it breaks, for all you know the ball may break anywhere between the first and 9th floors.

I think that you make the least throws when you break the 36 floors in three equally big parts. Hence you first walk to the twelfth floor and drop a ball from that floor. If it does not break, then will you walk to the 24th floor and repeat the exercise. If it does not break, then will you go up one by one until you or reach the 36th floor or the ball breaks.

If it breaks at the twelfth floor, then will you start with the second ball at the first floor, otherwise will you start at the thirdteenth floor and drop the ball from each consecutive floor.

Good. Going by this strategy, if you have the worst luck (which we must assume), the ball breaks on the 36th floor, or not at all. That means it would take you 19 throws to find out.

See if you can improve it!

It will take less then 19 throws:

1 on the twelfth,

2. on the 24th,

3 on the 25th etc. All in all 14 throws max.

Is the good solution the least maximum or the least average times to throw a ball? I expect the first.

Counting fail 

Yes, your strategy does it in 14 throws. And you are right to assume we should do it in the least maximum (that sounds funny lol). Either way, 14 can be improved.

That's thirten tries if we don't try it on the 36th floor if it's the last one left, since it is the only floor not tried, and will therefore break from it.

I clearly stated that it is possible that the ball doesn't even break from the 36th floor.

That's right, you cannot tell.

you drop it every 6th floor for the least maximum. But wouldn't the ball wear out and eventually break with all the drops :p

Edit: every 5th and 7th and 8th floor seems to work too? hmm

Well you have to be more specific. For instance, if you try every 6th floor, let's assume it does not break on the 6th, nor the 12th, nor 18th, 24th, and 30th. Which floor do you go next to?

And no, the ball does not wear out (it's not about realism, mind you ).