The worlds first riddle!

KNIGHTS, KNAVES, AND CATS:
Only three of the six possible combinations are consistent. Two of these have A as the knight. I'd bet on A.

NUMBER OF 2's:
1-333

BASKETS OF BALLS:
The probability is 9/64.
red, red, non-red (black, white, or purple)
red, non-red (black, white, or purple), red
non-red (black, white, or purple), red, red

What is it that given one, you will have either two or more?

A couple

What seven letters did Old Mother Hubbard say when she opened her cupboard?

OICURMT

Zips: 2 Qors/hr
Kigs: 2.5 Qors/hr

What is it that given one, you will have either two or more?

Choice

Stop the clock! Mark wins the Master Wizard title.

What is it that given one, you will have either two or more?

"Choice"

Not content with that, he goes on with," Zips: 2 Qors/hr. Kigs: 2.5 Qors/hr "

How could he know that?
This problem can be solved algebraically: 10 (8k + 14l) = 510 and 12 (13k + 6l) = 492. You then isolate 1 variable in 1 equation, & substitute it in the other equation. Zips build 2 tors an hour, & Kigs build 2.5 Qors an hour.

Easy peasy - not!

However I am confident that no one can answer this;


If a juggler juggles 4 objects, how many total throws must he or she make before the objects are returned to their original positions (i.e. the original 2 objects in each hand)

The juggler starts out with 2 objects in each hand, & throws 1 object from 1 hand, then another object from the second hand, then the remaining object from the first hand, & so on.

Except for the first throw for each hand, there is a moment where the throwing hand no longer holds anything after each throw.


I am a 12 letter word.

A. My 2, 9, 10, 11, 12, 8 is a word that means the same as the word that is the name of a very well known search engine.

B. My 7, 5, 6, 4 is a famous philosopher.

C. My 10, 11, 3, 1, 8 is the snow leopard of Asia.

What am I?

8 throws

What am I?

A 12 letter word.

I can't find anything that fits with Uncia (snow leopard).

I've been going through the old puzzles in this topic. Here's an alternate answer for
"There are 7 1's 3 2's 2 3's 1 4's 1 5's 1 6's 2 7's 1 8's 1 9's and 1 0's in this sentence." (2/9/04)

There are 11 1's 2 2's 1 3's 1 4's 1 5's 1 6's 1 7's 1 8's 1 9's and 1 0's in this sentence.

Oh, had to translate uncia from Latin to English.

You want to be 20th in line.
Assuming 365 days, your probability of winning is about .03232.
Assuming 366 days (with 2/29 just as likely as any other), your probability of winning is about .03227.

If K is odd, then the population eventually becomes 1 (a killer) since killers are killed in pairs. Therefore, you are guaranteed to be killed.

If K is even, it's a matter of whether or not you get killed before the last two killers kill each other.

Worked it out for small numbers - noticed a pattern - didn't try to prove.

As before, if K is odd, probability of survival is 0.
If K is even, probability of survival is 1/(K+1).

Combining these gives:
P(survival) = (1-(K mod 2))/(K+1)

Interestingly, it doesn't matter how many pacifists there are.

I urge all visitors to read Mark's replies above - classic.

A town has two hospitals, one big and one small. Every day the big hospital delivers 1000 babies and the small hospital delivers 100 babies. There's a 50/50 chance of male or female on each birth.

Which hospital has a better chance of having the same number of boys as girls


Is a round-trip by airplane longer or shorter if there is wind blowing

Longer.
Let A=rate of Airplane
Let W=rate of Wind
Let D=one way Distance
Let T=total Time
After a bit of algebra you get:
T(no wind) = 2D/A
T(wind) = 2DA/(A^2 - W^2)
If W=0, these are equivalent.
Otherwise the latter is larger.

The smaller hospital will have the greater chance (by a factor of about 3).
Given N babies (N is even), the probability of having the same number is:

C(N,N/2)/(2^N)

where C(N,N/2) is the number of combinations of N things taken N/2 at a time [N*(N-1)/2].

Mark, what a pleasure it is to read your answers. I have now made it my life's ambition to find a topic that confounds you. I therefore, (to use a technical term)intend to crank it up a notch. :wink:



In a universe with the same physical laws, but which is mostly water with little bubbles in it, do the bubbles attract, repel, or what

Harry throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first.
If Harry now throws another dart at the board, aiming for the center, what is the probability that this third throw is also worse (i.e., farther from the center) than his first

Assume Harry's skilfulness is constant.

Confounding me won't be difficult!
BUBBLES
I don't have a clue (see, it wasn't difficult). I'm leaning toward repulsion (the bubbles, not me) since they are less dense than their surroundings. On the other hand, if the universe is finite, won't they all collect at the surface? Bubbles at the surface often seem to clump.

DARTS
Since I can't think of a different way to solve this, I'm going to take a simple approach.
Assuming equal likelihood of the rankings of the throws (closest, middle, farthest):
123, 132, 213, 231, 312, 321
we already know that 1 comes before 2.
That leaves
123, 132, 312
So the probability would be 2/3.
Of course, using this logic, it is equally likely that his last throw will be the closest, in the middle, or the farthest. This seems counterintuitive since the regions between the center and the first dart and the first dart and the second dart are finite, but the region beyond the second dart is infinite.
I'll stick with 2/3.