Puzzles with numbers and things

[snookersfun]

Poor Vidas

Vidas, are you a boy? And is your name an Anagram of Davis?

[April madness]

Poor Vidas

[snookersfun]

I don't think it is easy to put up a total proof. I got so far that in order to transfer the 11 points from Oliver (-6,-6,+1), they will always end up in such a way, that players sitting in position 1,3,5 of him (the girls/Vidas) can end up with an uneven number of balls and only the 'boys' could have even number of balls. So, no chance for AG to end up with 0 balls.

[davis_greatest]

What I meant here was, if Oliver has 5 balls, he gives no balls away, so he has 5. If he has 7 or more balls, he gives 6 balls away, so he has a minimum of 1 ball. The only way he ends up with 0 balls is if he starts with 6 balls and the 3 people before him have 5 or less balls.

Unless I completely misread your puzzle, I think this is correct.

I think I see what you are saying. But Oliver can easily have 0 balls at some point. For example, snookersfun may start by passing balls anticlockwise (3 to robert602, 2 to April Madness, 1 to Oliver). Then Oliver will have 12 balls. He can then pass 6 clockwise (3 to April Madness, 2 to robert602, 1 to snookersfun), and then pass 6 balls again, and he'll have none left!

[davis_greatest]

Well, yes, it is!

[elvaago]

So it's utterly random? Oh my. In that case, I have utterly no clue!

[davis_greatest]

OK. After round 1 of swapping, when everyone (except Austrian_Girl) has given their 6 balls to the next 3 people, the result is like this.

Austrian Girl now has 6 balls, since the three people before her gave her 1, 2 and 3 balls. No one has less than 6 balls. So in each round after, everyone simply gives away 6 balls and then gets back 6 balls from the previous 3 people. After the first round, the number of balls with each person doesn't change.

I think you are assuming that everyone passes balls along, once in each "round". The game doesn't require this. Balls can be passed in either direction and there is nothing in the rules to prevent one player, for instance, obtaining 12 balls and then giving them all away in 2 goes while someone else has still not given away any balls.

In other words, there are no "rounds", and players can give balls away whenever and in whichever direction they like, so there is no way of knowing in advance how many balls any particular person will have at any particular time.

[elvaago]

Sorry, I don't understand this. He doesn't need to have 6 balls. Anyone can pass 6 balls along, provided he / she has AT LEAST 6 balls (otherwise he / she wouldn't have the 6 balls to pass!). And it doesn't matter how many balls the recipient has.

What I meant here was, if Oliver has 5 balls, he gives no balls away, so he has 5. If he has 7 or more balls, he gives 6 balls away, so he has a minimum of 1 ball. The only way he ends up with 0 balls is if he starts with 6 balls and the 3 people before him have 5 or less balls.

Unless I completely misread your puzzle, I think this is correct.

[davis_greatest]

I'm currently working out on paper (read: Excel) a real solution. I'll keep you posted.

OK - but I'd prefer an answer of no more than 3 or 4 simple sentences

[Edit - I posted this before reading your subsequent post, which I'm about to read now...]

[elvaago]

OK. After round 1 of swapping, when everyone (except Austrian_Girl) has given their 6 balls to the next 3 people, the result is like this.

Austrian Girl now has 6 balls, since the three people before her gave her 1, 2 and 3 balls. No one has less than 6 balls. So in each round after, everyone simply gives away 6 balls and then gets back 6 balls from the previous 3 people. After the first round, the number of balls with each person doesn't change.

[davis_greatest]

April madness is sitting directly opposite in a circle of 11 people. Because of that she's sitting an odd number of places away from Oliver. So he can never have just 6 balls with the 3 people in front of him having 5 or less.

Sorry, I don't understand this. He doesn't need to have 6 balls. Anyone can pass 6 balls along, provided he / she has AT LEAST 6 balls (otherwise he / she wouldn't have the 6 balls to pass!). And it doesn't matter how many balls the recipient has.

[elvaago]

I'm currently working out on paper (read: Excel) a real solution. I'll keep you posted.

[snookersfun]

Thought I tell you as I do that all the time....

[elvaago]

I got confused. I mean Austrial Girl, in that case. :-)

(I blame being hellishly nauseous in the morning for my lack of reading abilities!)

[snookersfun]

Austrian_Girl not April Elvaago and it is 10 people!