Solve this Riddle if you can

ok every 6th floor gets you 11 tries, but I think thats wrong. You try every 6th floor, so 6th, 12th, 18th, 24etc.. if it breaks on 18th, you start throwing the other ball from 13th. So worst case being 35th, that is 11 tries.

But you can throw 9th floor, then 17th floor, then 24th, then 30th, then 35th (if it breaks on any of them, try from previously tested floor + 1), which gives you 9 least maximum tries.

throwig 8th, 15th, 21st, 26th, 30th, 33rd then 35th (if it breaks on any of them, try from previously tested floor + 1), which gives you 8 least maximum tries.

Ding ding ding, we have a winner. This is the most efficient way to do this. Well done!

Wow that's cool! By the way, I got the night watchman one before I read the answer!

The horse's name was Bobtail...wait, wrong puzzle...that's Jingle Bells.

i think that anything glass dropped for 12 to 15 feet will break....I only need one drop to prove it and still have a ball left.....

I want to revitalize this thread, so I will start with a variation of my first riddle. Here it is:

FakeName6 wrote:

Here is a probability problem of my own invention:

A man has a drawer with 150 socks: 50 red, 40 orange, 30 yellow, 20 green, and 10 blue. He takes six socks out. What is the chance that at least two of them are the same color?

My new one, unlike this earlier one (the answer was 100%, for those who missed it), is not a trick question. It involves real statistics. I am not totally sure that I got it right, so you guys might get me on it. Here it is:

The situation is the same as the first one. Except now you just pull two socks.

Simple, right? Or is it....

The chance of drawing a red sock followed by another red is (50/150) x (49/149). Add to that the chance of an orange followed by another orange (40/150) x (39/149). Add to that the chances of other-coloured pairs worked out similarly. The numerical answer is left as an exercise for the student.

they were playing (if you win you lose) or 960 were thos moves aren't possible

Fakename6's question wasn't a trick question, it was just an application of the pidgeonhole principle.

Your riddle starts with 'a man is playing a chess game'.

He is playing without an opponent, hence the two possible solutions are:

1. there is no opponent,

2. the man who is losing is also the man who is winning.

They are playing FischerRandom.

The would-be loser circled the wrong result on the scoresheet and the would-be winner signed it without paying attention.

His opponent is his little son, who started to cry after the announcement of his father that he would be checkmated. His father regretted his behaviour, apologized to his son, retracted the move and told the boy that he had not lost. It was all ok and play would continu with 4. d3 Nf6.

The opponent had given the man Queen-odds, so he had a flight square for his king.

They were playing chess 960.

The opponent was white.

Ringed piece odds, the ringed piece wasn't the queen

The announcement of checkmate was forbidden, because it was not allowed to speak during the game. Hence was the man disqualified by the referee and was the point given to his opponent.