[unreality]

You played Three Games of Zarball and had to win two in a row... you made the right choices (but unfortunately lost).

Your second chance for freedom came with the Five Games of Zarball, in which you had to win at least one of three combinations. Again, you lost in a mistake up against the King.

However you won your freedom, and deserved it, in the intense Royal Zarball Tournament, where you made your way to victory.

Now you have taken the King's offer as the Supreme Dignifiably Appointed Royal Zarball Trainer of Excellence.

First, it would help if I explained what zarball really is:

As you know, it's a 1-on-1 game of agility, strength, speed, accuracy and skill. It's played on the Zarball Arena on the palace roof, and the Arena is a circle of soft and spongy grass, well-tended by the Royal Gardeners, and high up on the palace roof where it gets a lot of sun, but is cut often by the Royal Gardeners to keep it perfect, just as the King likes it. The circular Zarball Arena is a perfect circle, with a radius of 20 feet. Around it's circumference is a chest-high woven fence where the front-row spectators, coaches, referees, ball lackeys and score-keepers stand. There are a ring of seats and bleachers surrounding the Arena after that.

At the center of the Arena is a hole with a diameter of 1 foot 2 inches, thus a radius of 7 inches, and since the center of the hole corresponds with the center of the circle, takes up the first 7 inches of the radius of the Arena.

Why is the hole 1 foot 2 inches in diameter? Because the zarballs are 1 foot in diameter, and they go in the hole. Back to that in a sec.

There are two players in zarball. The Red Player and the Blue Player. There are three balls: red, blue and green. If a red ball is scored, the Red Player gets 1 point. If a blue ball is scored, the Blue Player gets 1 point. And if a green ball is scored, 1 point is given to the player with the least amount of points- and if it is a tie, no points are given.

So how do you score a ball? By throwing it in the central hole of course ;D One of the big reasons that the Zarball Arena is on the palace roof is that there are ball lackeys in the Catch Room, as it's called, directly underneath the Zarball Arena, who take the balls that are scored (they fall through the hole from the Arena above to a basket in the Catch Room placed underneath the hole). The ball lackeys then slot the ball back in a device which spins it around the circle and pops it out randomly from underneath the fence at the edge of the circle, where the ball quickly rolls back into play. The three balls are initially placed at a random angle and a random radius from 2 feet to 18 feet from the center of the Arena.

There are other rules as to how you can take the ball and score it in the hole, but those aren't important to this riddle. What is important is how you win: (of course that's important) When you get 10 points. Yep, easy as that.

That should be important in the future, but right now, focus your attention on the five spectators standing at the fence. The King, Queen, Prince, Duke and Earl. They're the first here to see the match between you, Supreme Dignifiably Appointed Royal Zarball Trainer of Excellence, and your prime pupil.

Unfortunately there's a gust of wind on the palace roof, and the five spectators, all wearing identical hats rooting for your pupil to win (gee, thanks ), and the hats are blown off of the palace roof and scrambled randomly, by the wind of course, by the time they hit the ground in the palace gardens.

No matter, though, the ball lackeys quickly run down, grab the hats, and get back up before there's any Royal Fuss. Not sure whose hat is whose, they hand them out to the King, Queen, Prince, Duke and Earl randomly.

a) What are the chances that at least one royal spectator got their hat back?

b) What are the chances that EXACTLY ONE person gets their hat back?

c) What about exactly two people?

d) And finally, what are the chances that MORE THAN TWO of the spectators will get their hats back?

[Guest]

No matter, though, the ball lackeys quickly run down, grab the hats, and get back up before there's any Royal Fuss. Not sure whose hat is whose, they hand them out to the King, Queen, Prince, Duke and Earl randomly.

a) What are the chances that at least one royal spectator got their hat back?

b) What are the chances that EXACTLY ONE person gets their hat back?

c) What about exactly two people?

d) And finally, what are the chances that MORE THAN TWO of the spectators will get their hats back?

I am by no means a probability expert, nor do I even like probability, but here goes nothing...

a) At least 1, or 1 - P(Nobody getting theirs):

1 - 1024/3125 = 2101/3125 = .67232

b) Exactly 1:

5 (256/3125) = 256/625 = .4096

c) Exactly 2:

2 (64/625) = 128/625 = .2048

d) At least 3:

10(16/3125) + 5(4/3125) + 1/3125 = 181/3125 = .05792

[Guest]

a) 4/5 (96/120)

b) 3/8 (45/120)

c) 1/6 (20/120)

d) 11/120

Edited February 21, 2008 by losenotloose

[bonanova]

Can't believe you'd post a shaggy dog puzzle ...

120 ways to give back the hats.

p[0] = 53/120 = .44166...

p[1] = 42/120 = .350 <-------------p[exactly one] (b)

p[2] = 17/120 = .1416... <----- p[exactly 2] [c]

p[3] = 7/120 = .0583...

p[4] = 0 of course

p[5] = 1/120 = .0083...

[a] - p[at least one] = [42+17+7+1 = 67]/120 = .5583...

[d] - p[more than 2] = [7+1 = 8]/120 = .066...

[unreality]

I guess nobody discovered the catch then ... I could have given this riddle out with 1000 spectators, and would you all have been able to figure out the solutions using the same methods?

hang with me here, I am leading somewhere:

5! is 120, so there are 120 possible combinations of the hats and the five spectators

of those, exactly 44 result in no hat being given back to its previous owner

yes, I could make a list of the entire sample space... but if there were 10 spectators? Would I want to list 3,628,800 (10!) different possibilities? no way

Just do 120/e (e as in the transcendental exponential identity, 2.718281828459045...), which gives you 44.14553294..., of which the integer part is 44. That's how many possible permutations there are where no hat is given back to its previous owner

(120-44)/120 = 76/120 = .633333333 chance that at least one person gets their hat back

(n!-integer(n!/e))/n!

that was [a]

- [d] was intended to trick everyone's minds, lol, I dont even have answers for those, though I could *probably* figure em out if necessary lol

still think it's a shaggy dog puzzle, bonanova? ;D

[Guest]

I've always been terrible at these combinatorial counting puzzles, but I'm not sure where I went wrong (as usual). Could either of you guys explain your method a little more, or point me at the name of the technique you use.

I basically did the same thing bonanova did (lay out the number of ways to get exactly x hats on the right heads). I came up with different numbers for 1-3 though.

p[5]=1

p[4]=0

p[3]=10

p[2]=20

p[1]=45

p[0]=44

Edited February 21, 2008 by losenotloose

[Guest]

I already see I made a typo.

I wrote 96 instead of 76.

a) 19/30 (76/120)

b) 3/8 (45/120)

c) 1/6 (20/120)

d) 11/120

[unreality]

76/120 is correct for [a]

[bonanova]

a: 0.6333

b: 0.3750

c: 0.1667

d: 0.0917

120 combinations: 5 x 4 x 3 x 2 x 1.

Let R[p] = number of correct assignments for p out of 5 hats

Let W[q] = number of ways q hats can all be wrong

Let A[n] = R[n] x W[5-n] = total number of ways exactly n hats can be correct

Let p[n] = A[n]/120 = probability exactly n hats are correct.

R[2] = 10: 1 2 x x x, 1 x 3 x x, 1 x x 4 x, 1 x x x 5, x 2 3 x x, x 2 x 4 x, x 2 x x 5, x x 3 4 x, x x 3 x 4, x x x 4 5.
R[3] = 10: 1 2 3 x x, 1 2 x 4 x, 1 2 x x 5, 1 x 3 4 x, 1 x 3 x 5, 1 x x 4 5, x 2 3 4 x, x 2 3 x 5, x 2 x 4 5, x x 3 4 5.
R[4] = 0
R[5] = 1: 1 2 3 4 5.

W[1] = 0
W[2] = 1: 2 1
W[3] = 2: 3 1 2, 2 3 1
W[4] = 9: 2 1 4 3, 2 3 4 1, 2 4 1 3, 3 1 4 2, 3 4 1 2, 3 4 2 1, 4 1 2 3, 4 3 1 2, 4 3 2 1

R[1] =  5: 1 x x x x, x 2 x x x, x x 3 x x, x x x 4 x, x x x x 5.

N[0] = 44 *see below -- p[0] = 44/120 = .3667

N[1] = 5x9 = 45 -- p[1] = 45/120 = .3750 <-- (b)

N[2] = 10x2 = 20 -- p[2] = 20/120 = .1667 <-- [c]

N[3] = 10x1 = 10 -- p[3] = 10/120 = .0833

N[4] = 0x0 = 0 -- p[4] = 0/120 = .0000

N[5] = 1x1 = 1 -- p[5] = 1/120 = .0083

*N[0] = 120 - [45+20+10+0+1 = 76] = 44

N[1] + N[2] + N[3] + N[4] + N[5] = 76 -- p[at least 1] = 76/120 = .6333 <-- (a)

N[3] + N[4] + N[5] = 11 -- p[more than 2] = 11/120 = .0917 <-- (d)

[unreality]

Yep you got em now, bonanova! ;D so nobody got the trick with e?

[bonanova]

Yep you got em now, bonanova! ;D so nobody got the trick with e?

Yeah, explain that for us...

[unreality]

5! = 1 * 2 * 3 * 4 * 5 = 120

120/e = 44 (there's an infinite string of decimals following it but that doesn't matter)

44/120 are the number of combinations where no hat is given back to its previous owner.

it works with other numbers, such as 10! or 17!, etc

just lop off the decimal part and keep only the integer

1/e = 0.something

so there are 0 combinations to give back the hats without giving one to its previous owner if there is only 1 person, obviously

2/e = .7something, which would be 0... even though with 2 hats there would be 1 combination where neither would get his hat back. So it's not perfect, though it's a pretty good ratio between e and what we're looking for

though since we wanted AT LEAST ONE then we had to subtract that from 1

(n!-integer(n!/e))/n!

for the 5 people:

(120-44)/120 = 76/120 = .6333333 chance that at least one person gets their hat back

[unreality]

So yeah. I hope that was clear

[bonanova]

Yeah, I guess. Wondering why, tho.

[unreality]

Same ;D

looking at it:

integer(n!/e)/n! = permutations where no hat is given back to its previous owner. Doesn't work with 1 or 2 hat-owners, works with 3 or more though.

consider it NOT using the integer function (which is there, to simplify it cuz of probability)

that would make:

(n!/e)/n!

which is:

n!/(e*n!)

which is:

1/e

however it's not 1/e every single time with 5 hats or with 10 hats, it varies slightly. Because of the taking-the-integer-only part.

So anyway, consider it in pure probability's sake, 1/e chance that nobody gets their hat back.

After some searching the internet, I found a section of an article on wikipedia that explains the property. Check this out:

Bernoulli trials

(read the section on Bernoulli trials)

The integer-taking part approximates e's limit to the limit with the slots-machine example

[bonanova]

Thanks. I kept reading to this.

[unreality]

Yeah. The hat check problem. I didnt know it was on wikipedia, I saw it elsewhere. It's a good'un of course e is cool and all, especially with e^(theta•i) = cos(theta) + i•sin(theta) equalling the special case e^(pi*i) = -1

even that's not as cool as the golden ratio and its much neglected other root of the quadratic equation that forms it (x^2-x-1=0), which is its negative inverse and also 1-phi, and shares its amazing properties, most of which I figured out on my own, others I had to look up. But the golden ratio is amazing... you think a few cool properties are amazing- but then you soon realize they are just specific cases of its much bigger and more awesome properties, mostly connected to the Fibonacci numbers... yeah lol I love the golden ratio. And the Fibonacci numbers of course, which go hand in hand with phi

*exhales long breath* hehe i'm a bit overenthusiasitic about that i have such nerdy pasttimes that nobody knows about lol

but yeah.