Jump to content
BrainDen.com - Brain Teasers
  • 0


unreality
 Share

Question

Zarball Olympics

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. and figured out the probability of hats being returned to the proper heads in The Royal Zarball Spectators crisis.

You then arranged the Village Zarball Tournament to set up the brackets and byes to make the game as fair as possible in one task, and as skewed toward your buddy Perry as possible in another.

You then went on to master the betting process of The Zarball Champions...

and now...

in Realm year 1080...

the Olympics!

In the Olympics, Zarball matches are held with either red or blue balls- the color of the ball is random. When a game finishes, the Ball Lackey (BL - one of 8 Ball Lackeys) that oversaw that game runs the ball back to the equipment shed, where there are two bins: a bin for red balls and a bin for blue balls. There is also a green chute that incinerates the ball instantly. The red & blue bins are tall opaque cylinders. There is a notch every 2 ball spaces, so if a BL looks down into the bin, he can see whether there are an odd or even number of balls in the cylinder (if the number of balls approaches a certain limit, the cylinder incinerates the lowest 2 balls, keeping the even/odd count the same and not overfilling the bin) - but only the even/odd value, not how many balls there are exactly. Ie, the BL sees the top ball and sees whether there is a notch above it or if the notch is hidden, and determines if there is an even or odd number of balls in the bin. Balls cannot be removed once placed in a bin. The blue bin starts out with one blue ball and the red bin starts out with one red ball.

It's up to the BLs to end the Zarball event when all 8 of the BLs have overseen a game: one of the BLs must announce when all 8 have overseen at least one game. As many games of Zarball are played as needed until one BL announces correctly that all 8 have overseen - if a BL guesses incorrectly, all of the BLs are fired, so the guess must be 100% sure no matter what. Another thing: the games are random, the order is random, the ball color (red or blue) is random, and which BL oversees a game is all random. Also, the BLs cannot modify the equipment room in any other way cuz it may mess up other sports that use it- EXCEPT for the light switch in the equipment shed. It starts out at OFF and nobody except the BLs will use it for the duration, so it's fine to turn the switch ON or OFF as they please.

The BLs meet the night before to plan out their method to make the games go as swiftly as possible. The following day, they execute it perfectly... what was their strategy?

Link to comment
Share on other sites

12 answers to this question

Recommended Posts

  • 0

Okay, my fried brain (yum yum :P) doesn't like all the reading...

Well, actually, now there are a lot of questions I'd like to ask about this problem, but since you're probably asleep...I'll give possibilities

Question 1: Do they have to put the ball in the right bin? If not,

It's basically a binary code (with a little mixing of values).

Red bin, light, blue bin

The starting values (odd, light off, odd) are assigned the values (0,0,0), the values (even, light on, even) are assigned the values (1,1,1)

First BL puts ball in blue bin->(0,0,1), which is binary for the number 1, second BL (he knows he is the second because sees the binary for the number 1) and he puts the ball in the blue bin, and turns on the light -> (0,1,0), third BL puts in blue bin -> (0,1,1), fourth BL puts in red bin, and turns out light-> (1,0,1), fifth BL puts in blue bin -> (1,0,0) (so 4 and 5 are switched, but they know this ahead of time), sixth BL puts incinerates and turns on light->(1,1,0), seventh BL puts in blue bin->(1,1,1), eighth BL announces they're done. If a BL is called out more than one time, he just incinerates the ball and doesn't change the values.

If they do have to put them in the right bin, then,

Question: Do they know how many games have been played? I don't think it is possible if they don't, since

All the balls end up being the same color...the only thing they can change (or notice the change in) is the even/odd parity of one bin and the light switch, which makes it impossible to figure out how many have gone before them...

So assuming that they do know how many games have been played...

Actually, I don't know...been trying to work out the possibilites, haven't found one that works yet...need to sleep, will work on it later...

Link to comment
Share on other sites

  • 0

There's no reason that you can't put a ball into a different-colored bin, and if someone looks down, you can see the color of the ball. So it starts off at 1011 - or odd-red-odd-blue. Each person must do one status change to mark their passing and then just incinerate if it's their 2nd or 3rd etc visit, etc.

The number again: the digits are abcd. All are binary. A and C are the even/odd (0/1) statuses for the red and blue bins. B is the most recent color in the red bin (0=red, 1=blue) and D is the most recent color of the blue bin. So it starts out at 1011 = odd-red-odd-blue

If you come and see 1011 with the lightswitch off, you know you are the first to visit. Put your ball in the blue bin and leave the lightswitch off.

If you come in and see 1000 or 1001 with the switch off, you know you are the second BL. Put your ball in blue and turn the switch on.

If you come in and see 1010 or 1011 with the switch on, you know you are the third BL, and you must put your ball into the red bin and leave the switch on.

If you come in and see 0010, 0011, 0110 or 0111, and the switch on, you know you are the fourth, and you must incinerate your ball and then turn the switch off.

If you come in and see 0010, 0011, 0110 or 0111, with the switch off, you know you are the fifth, so you have to put your ball in the red bin and then leave the switch off.

If you come in and see 0000, 0100, 0001 or 0101, with the switch off, you know you are the sixth and you have to incinerate your ball and turn the switch on.

If you come in and see 0000, 0100, 0001 or 0101, with the switch on, you know you are the seventh, and you have to put your ball in red and leave the switch on.

If you enter and see 1000, 1001, 1100 or 1101 with the switch on, you know you are the eighth person. If you put your ball in blue and keep the switch on, the 9th person will, 3/4 of the time, be able to see 1010, 1110, or 1111, with the switch on, and will know that he/she is the 9th person. However if the 9th person sees 0100 or 0101 or 1011 (a 1/4 chance) with the switch on, the 9th person can't be sure if they are the 7th person or the 9th person, and it fails. Thus it only works with 9 people 3/4 of the time.

So if the 8th person has a blue ball (1/2) and sees 1000 or 1001 (1/2, thus the total chances of that are 1/4), they know that they can only send on 0100, 0101, or 1011, all of which have been used before with the switch both off and on. If the 8th person sees this, they can just call that all 8 have been there without fail. However, if the other 3/4 occurred, they must put their ball in blue and leave the switch on, and the 9th person will see this and call that all 9 have been there.

Link to comment
Share on other sites

  • 0

This seems to be too simple for our YS ... I second her "The Simple Solution" ... 2 3 is the right answer :P

Yeah...that's why I spent most of my time trying to figure out a way to do it with putting the ball in the right color bin...can't figure it out, though, I tried counting repeats, but I get stuck on the third step since I have to go back to a value I already used...most recently trying something with remainders...I don't really know if it's possible, might not be...XP

Link to comment
Share on other sites

  • 0

yeah it's probably not. You can change the even/odd status of one bin which you don't even get to choose (the ball color is random), and then there's the lightswitch, but even so it's probably impossible

I worded the riddle so as to allow putting a red ball in the blue bin and vice versa without actually stating that you could (it's just common sense ;D), that's sort of the 'catch'. Then I also made it clear that you can see the top ball in the bin very clearly, ie, to discern its color if you wanted

Which is why it may be possible to work with 9 Ball Lackeys, see my post 5 in this topic :P I'm curious if anyone else can do better than 3/4 to work with 9

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...