Puzzles with numbers and things

[davis_greatest]

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.

Getting warm!

So, no chance for AG to end up with 0 balls.

She starts with 0 balls - so she can easily end with them, if no one ever passes her any!

[davis_greatest]

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

Yes, it is an anagram. If you put the D in 3rd place, the A in 4th, the V in first, the I in 2nd, and don't move the S, you get Vidas!

[snookersfun]

Getting warm!

good

She starts with 0 balls - so she can easily end with them, if no one ever passes her any!

That last part was of course one of my utter nonsenses as she is supposed to end up with 11 balls like all the others.

and about the Anagram part, did Vidas choose it because it is an Anagram, aaarrrrgh?

[snookersfun]

Quote:
Originally Posted by 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.

and it seems that there will be always a minimum of two boys with an even number of balls as well (except Oliver with the 0 balls eventually)

[davis_greatest]

...

and it seems that there will be always a minimum of two boys with an even number of balls as well (except Oliver with the 0 balls eventually)

How do you make that out? At the beginning, there are NO boys with an even number of balls (as everyone except poor austrian_girl has 11)!

Nevertheless, you are thinking about this in the right way...

[snookersfun]

How do you make that out? At the beginning, there are NO boys with an even number of balls (as everyone except poor austrian_girl has 11)!

Nevertheless, you are thinking about this in the right way...

ok, slowly again, in the beginning, only AG with 0 balls (even), everybody else 11 (uneven of course). After moving 11 balls (-6,-6,+1) from Oliver, he and at least another boy will have an even number of balls (or 5 even?) +AG still of course. Trying to rid now the other players of their excess balls always seems to leave the two extra evens (except the 0), and as they are in even and uneven positions in the ring, they will never 'combine' to two 11s.
and now, I'll give up anyway, as it is dinner time....

[The Statman]

For example, Oliver can pass 3 balls to April Madness, 2 to robert602 and 1 to snookersfun.

Or The Statman could pass 3 to austrian_girl, 2 to Semi_Sayginer, and 1 to snookersfun.

I don't mean to come across as an idiot (again), but it appears to me, from your example above, that the balls could be passed on in either direction – either Oliver or I could pass the balls on to end up with 1 to snookersfun.

But you do not say in what order the balls are passed. For example, once Oliver has finished sharing his balls out, does the next person in the circle do so? And turns are taken always in the same order going round the circle?

Assuming that, let's see.

Oliver has 11 balls (lucky chap). He passes his along to his left as described above. So he now has 5.

Then April hands out some of her 14 (now) balls; then Robert passes round some of his 16 (now) balls.

When it comes to snookersfu, I don't see why she cant decide to hand them out to her right, rather than her left. This would give Robert three, April two and Oliver 1.

If all the next people in the circle decide to follow snookersfun's lead by giving balls out to their right, then when it is next Oliver's turn to do the giving, he will have 6 balls and will end up with nothing.

Or am I missing something fundamental?

[davis_greatest]

I don't mean to come across as an idiot (again), but it appears to me, from your example above, that the balls could be passed on in either direction – either Oliver or I could pass the balls on to end up with 1 to snookersfun.

Yes, anyone can pass the balls in whichever direction he or she pleases.

But you do not say in what order the balls are passed. For example, once Oliver has finished sharing his balls out, does the next person in the circle do so? And turns are taken always in the same order going round the circle?

Anyone can pass the balls at any time, once someone else has finished. There is no specified order or taking turns.

....If all the next people in the circle decide to follow snookersfun's lead by giving balls out to their right, then when it is next Oliver's turn to do the giving, he will have 6 balls and will end up with nothing.

Or am I missing something fundamental?

Remember that we are not showing simply that Oliver cannot end up with no balls - we are showing that never can Oliver have no balls and everyone else have 11 each.

In your example, the others do not each have 11 balls.

[davis_greatest]

ok, slowly again, in the beginning, only AG with 0 balls (even), everybody else 11 (uneven of course). After moving 11 balls (-6,-6,+1) from Oliver, he and at least another boy will have an even number of balls (or 5 even?) +AG still of course. Trying to rid now the other players of their excess balls always seems to leave the two extra evens (except the 0), and as they are in even and uneven positions in the ring, they will never 'combine' to two 11s.
and now, I'll give up anyway, as it is dinner time....

um...

[trying to understand this and determine whether it is brilliant or a load of balls ]

[snookersfun]

um...

[trying to understand this and determine whether it is brilliant or a load of balls ]

I think in a second I shall delete all my posts concerning this round...

[Vidas]

and about the Anagram part, did Vidas choose it because it is an Anagram, aaarrrrgh?

yeah, because of that too ...my own name is different,
But as April pointed out, Vidas is real Lithuanian name, for example former head of LBF was Vidas Končius (of course you never heard of him, but anyway)

..Going clockwise from Oliver, we have April Madness, robert602, snookersfun, Semih_Sayginer, austrian_girl, The Statman, Vidas, elvaago, and abextra (and then we arrive back at Oliver).
...
Explain why, no matter how long they play, it is impossible for there ever to be a time when Oliver has no balls and everyone else each has 11 balls.

About the problem itself - I think I got a solution, it is similar to snookersfun's, but slightly different. (it's about odd and even number of balls as well):
-Oliver, robert602, Semih_Sayginer, The Statman, elvaago have in total 55 balls (odd number)
-April Madness, snookersfun, austrian_girl, Vidas, abextra have in total 44 balls (even number)
If Oliver has 0, the rest of his group each has 11 - it's 44 in total.
Each time someone passes the balls, opposite group gets 4 balls (+3, +1).
So the first group always have odd number of balls, (44 is impossible).

[davis_greatest]

Congratulations, Vidas

... and Charlie sends his apologies for making you sit in a girl's position (there weren't enough boys to go round). Expressing your answer in a very slightly different way (with Vidas playing a girl):

At the start, 4 of the 5 girls have 11 balls each, so the girls have 44 balls in total. Each time a boy passes balls, he passes 3+1 = 4 balls to girls; and when a girl passes balls, she passes 4 to boys. So the girls always have an even number of balls in total and can never finish with 55 balls, which they would need for all 5 girls to have 11 each.

[davis_greatest]

Updated scoreboard

1 point to Vidas, and ½ to snookersfun, for her latest PM (which I couldn't really understand, but at least it mentioned odd and even numbers )


SO HERE IS THE SCOREBOARD AFTER ROUND 52

snookersfun……………………….…..21½
abextra……………………………..…...11
Vidas……………………………………….10½
davis_greatest…………………..……6
robert602…………………………………6
elvaago...............................4
The Statman……………………..……2
Semih_Sayginer.....................1½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[davis_greatest]

Round 53 - Ape Neighbours

Recall that in round 52, Oliver, my pet orang utan was sitting in a circle during his birthday party with 9 of his friends from this forum. For the next game, Charlie the chimpanzee and Gordon the gorilla decide to join in, so they each squeeze in somewhere into the circle, their positions chosen at random. (So there are now 12 in the circle.)

What is the chance that at least two apes are sitting next to each other? (not counting forum members as apes)

[elvaago]

It's early morning. Time for elvaago to make an idiot out of himself again with his feeble math skills!

There's already one ape sitting in the circle. There are 10 people in the circle, so there's 10 available spots for the second ape to sit. Sitting on the left or on the right of Oliver means there's two apes in a row, so the chance of the first ape joining sitting next to Oliver is 2 in 10 or 1 in 5 or 0.2.

If the first ape that joins isn't sitting next to Oliver, then there's two apes sitting in the circle. This makes the number of available spots for two apes next to eachother four, with eleven seating spots available. That's a chance of 4 out of 11 or 0.363636.

So I guess the chance is 1 out of 5 * 4 out of 11 = 0.2 * 0.363636 = 0.0727272?