Puzzles with numbers and things

[abextra]

I'll be off in the corner beating the living daylights out of myself for a while.

Take it easy and come out of the corner! Everyone here has made mistakes!

[elvaago]

Respotting the colours.

Right.

I'll be off in the corner beating the living daylights out of myself for a while.

[davis_greatest]

Congratulations, abextra!

Round 46

a) Every child recieves 6 sweets ( 6 balls into every pocket),

b) every child recieves 24 pennies (24 points in every pocket),

c) the maximum is on until the 15th red is potted, after that uncle Willie has to pot brown instead of black to make sure, that every child ends up with the same amount of money. The break will be 144, it's the biggest number under 147, which is divisible by 6 without a reminder.

d)I hope there are no typos and the table won't collapse

red into pocket 1
black ............. 3
red ............... 3
black ............. 5
red ............... 5
black ............. 2
red ............... 2
black ............. 4
red ............... 4
black ............. 6
red ............... 6
black ............. 1
red ............... 1
black ............. 3
red ............... 3
black ............. 5
red ............... 5
black ............. 2
red ............... 2
black ............. 4
red ............... 4
black ............. 6
red ............... 6
black ............. 1
red ............... 1
black ............. 3
red ............... 3
black ............. 5
red ............... 5
brown ............ 2
yellow ............ 4
green ............. 6
brown ............ 2
blue ............... 6
pink ............... 4
black ............. 1

Perfect answer! 1½ points. To be absolutely clear, the maximum is still on once the 15th red is potted - it is no longer on once Uncle Willie pots brown after the 15th red.

Good question, eh?


SO HERE IS THE SCOREBOARD AFTER ROUND 46 BUT BEFORE ROUND 43 (which snookersfun has almost but not quite solved):

snookersfun……………………….…..18½
abextra……………………………..…...10
Vidas……………………………………….8½
robert602…………………………………5
davis_greatest…………………..……5
elvaago...............................4
Semih_Sayginer.....................½

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

[davis_greatest]

Your balance is off after you pot 2 balls in pocket 4.
You have 1 in 1, 2 in 5, 2 in 4, 2 in 3 and 2 in 2, making 3 on the left, 6 on the right and one broken table.

(I've spent 3 hours on this problem already. I got as far as 6 sweets and 24 balls as well, but the rest is beyond me.)

abextra's balance is not off because, as she then points out, the colours will have been respotted. When the 2nd ball (a red) is potted into pocket 4, the black has already been removed from that pocket. So when that red is potted, there are 2 balls in the left-hand pockets and 3 in the right-hand pockets (and 2 reds in the baulk pockets and 2 in the top pockets) - so the table remains in balance!

Which leads to my next post...

[davis_greatest]

43: robo 7/11 chance???? My head is not quite here yet though....

and thanks Elvaago, I am moved

Welcome back snookersfun!

... I'm not quite sure what to make of this answer - the probability is correct but the player isn't!

[abextra]

Maybe you're right, elvaago, I'm not sure myself. Have you respotted the colours?

[elvaago]

Your balance is off after you pot 2 balls in pocket 4.
You have 1 in 1, 2 in 5, 2 in 4, 2 in 3 and 2 in 2, making 3 on the left, 6 on the right and one broken table.

(I've spent 3 hours on this problem already. I got as far as 6 sweets and 24 balls as well, but the rest is beyond me.)

[abextra]

Round 46

a) Every child recieves 6 sweets ( 6 balls into every pocket),

b) every child recieves 24 pennies (24 points in every pocket),

c) the maximum is on until the 15th red is potted, after that uncle Willie has to pot brown instead of black to make sure, that every child ends up with the same amount of money. The break will be 144, it's the biggest number under 147, which is divisible by 6 without a reminder.

d)I hope there are no typos and the table won't collapse

red into pocket 1
black ............. 3
red ............... 3
black ............. 5
red ............... 5
black ............. 2
red ............... 2
black ............. 4
red ............... 4
black ............. 6
red ............... 6
black ............. 1
red ............... 1
black ............. 3
red ............... 3
black ............. 5
red ............... 5
black ............. 2
red ............... 2
black ............. 4
red ............... 4
black ............. 6
red ............... 6
black ............. 1
red ............... 1
black ............. 3
red ............... 3
black ............. 5
red ............... 5
brown ............ 2
yellow ............ 4
green ............. 6
brown ............ 2
blue ............... 6
pink ............... 4
black ............. 1

[snookersfun]

Hey, thank you very much Abextra (no, haven't seen pictures, will do now) and April!

[April madness]

And belated happy birthday wishes, snookersfun.

[abextra]

Welcome back, snookersfun, I hope you enjoyed your vacation! Have you seen the pics of Charlie, Gordon and Oliver already?

[snookersfun]

43: robo 7/11 chance???? My head is not quite here yet though....

and thanks Elvaago, I am moved

[davis_greatest]

A nice little one for a Friday night…

ROUND 45 - Go Petal Go!

Petal "the rocket" Ebdon is practising potting balls, in front of her close friend and mentor Rowena O'Sullivan. Petal starts thinking about her first pot at 6pm on a Friday night, and is playing rather slowly. However, spurred on by Rowena, Petal plays more and more quickly. In fact, she takes 10% less time on each shot than on her previous one. (So, if one shot takes 10 seconds, the next one will take 9 seconds.) She finishes at 9 o'clock on Saturday morning, having played through the night and having potted an infinite number of balls!

When does Petal complete her first pot?

elvaago answered this, but I never really posted a full explanation, so here it is.

Suppose that Petal takes X hours for her first shot.

Her 2nd shot takes 0.9X hours, the 3rd shot takes 0.9^2 X hours, the 4th takes 0.9^3 X hours, etc.

We know that Petal started at 6pm on Friday and finished at 9a.m. on Saturday, so the total time for all the shots was 15 hours.

Therefore
(1 + 0.9 + 0.9^2 + 0.9^3 + ....)X = 15 ..................... [Equation 1]

Multiply Equation 1 by 0.9 and you get

(0.9 + 0.9^2 + 0.9^3 + ....)X = 15 x 0.9 ..................... [Equation 2]

Subtract Equation 2 from Equation 1 and get

X = 15 x 0.1 = 1.5
(all the terms in the brackets cancel except for the first "1", so you are left with just 1 in the brackets)

So Petal took 1½ hours for the first shot, completing it at 7.30pm.

[davis_greatest]

Round 46 update

I have decided to make this a little easier, and I will award 1 point for anyone whose answer meets all the conditions except the condition that Uncle Willie stays on for a maximum for as long as possible.

Your answer must still involve Uncle Willie giving each child the same number of sweets and same amount of money, and must still involve him giving away as much money as possible, and must still show that the table will not collapse. Just, you will now get the point even if your solution does not have Uncle Willie staying on for a maximum for as long as possible.

If your answer the original question fully, including having Uncle Willie staying on for a maximum for as long as possible (and explain that this MUST be the longest possible), then you will get 1½ points.

[davis_greatest]

Round 46 - Kind Uncle Wille (and reminder of Round 43 "Sweet 16")

Here’s a variation on a problem I once posed on the BBC site a while back. No cheating!...


ROUND 46 - Kind Uncle Willie

One Christmas, Uncle Willie is showing his 6 young nephews and nieces how to play snooker, on his table at home. He has made many maximum breaks on that table, helped in part by the pockets being the size of large buckets – so large, in fact, that whenever he pots a ball into the pocket, he never moves it out to another pocket (he only removes the ball if it is a colour that needs respotting).

Unfortunately, having been played on so often and having suffered so many maximum breaks, the table has become rather rickety and unstable.

In fact, if ever the difference between
(i) the total number of balls lying in the top pockets
and
(ii) the total number of balls lying in the bottom pockets
is more than one ball, the table will collapse and Christmas will be ruined.

The table will also collapse if ever the difference between
(i) the total number of balls lying in the pockets on the left hand side of the table
and
(ii) the total number of balls lying in the right-hand pockets
is more than one ball. It is rather a rickety table!


While teaching the children about snooker, Uncle Willie also wants to teach them about mental arithmetic, so he tells each of the 6 children to stand by a pocket and count the points of the balls potted in that pocket. Once he has finished, he will give each child a penny for every point scored in that child’s pocket. He will also give each child one sweet for every ball that enters that child’s pocket.

Uncle Willie wants to be fair and makes sure that each child will end up with the same number of sweets and the same amount of money at the end. Being very generous, he also wants to give the children as much money as possible, and he knocks in a total clearance! No surprise there.

Because he needs to give each child the same amount of money, on this rare occasion he does not make his break a 147 maximum. However, wishing to maintain his reputation, he makes sure that the maximum looks on for as long as possible.

For the point, you must answer all four parts:

(a) how many sweets does each child receive?
(b) how much money does each child receive?
(c) at what point is the maximum no longer on?

and, the more difficult bit, to prove you have solved it -

(d) give an example of how the balls could be potted, i.e. explaining into which pocket each ball could be pocketed. For instance, let the “yellow” pocket be Pocket 1, the “green” pocket be Pocket 2, and continue clockwise around the table (so the top pockets are pockets 4 and 5). Then, you might say:

Red into Pocket 1
Black (or whatever) into Pocket 1 (or whatever – in fact, the table would collapse if you started this way)
…
finishing...
Pink into Pocket 3 (or whatever)
Black into Pocket 6 (or whatever)


==================================================

And a reminder that Round 43 is still live:

ROUND 43 - Sweet 16 (reminder of the question)

I was watching the Grand Prix final on Astronomicus, a best-of-17-frames match. Robotsdaughter had taken a 4-1 lead against Cantcope when I had to stop watching. I knew at that point that Cantcope would have a 52% chance of winning each future frame (and Robotsdaughter a 48% chance), and that each frame would be independent of all the others.

The following day, my friend asked me "Did you see the final last night? It was a thrilling deciding frame. 9-8, on the black!"

Given that information, who was more likely to have won the penultimate (16th) frame and what is the chance that that player did so?