Puzzles with numbers and things

[davis_greatest]

They're all tricks, semih. The trick to solving them is seeing which trick is being pulled each time.

[abextra]

I think, Cantcope is more likely to win the 16th frame and the chance he did so is much more than 52%, maybe somewhere around 70% or 80%, but this is just a guess, caused by the score 1-4 after 5th frame. I'll give up...

[davis_greatest]

Interesting that no one has snapped this point up yet. I'll give you a clue... you don't need a computer, nor a calculator, nor even paper!

[elvaago]

I'm going to go with 52% for Canteloupe. I mean Cantcope.

[davis_greatest]

I'm going to go with 52% for Canteloupe. I mean Cantcope.

Good to see someone else having a go but.... it's not 52%!

[davis_greatest]

Rounds 43 and 44 - "Sweet 16" and "Baskets of balls"

Here is a reminder of round 43, which someone I am sure will be able to solve in a few seconds for an easy point if looking at it in the right way, and a new question for round 44.

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?

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

ROUND 44 - Baskets of balls

My 3 pet apes, Gordon, Oliver and Charlie, whom you should all know and who have now posted their photos in the Members' Gallery, take an ordinary set of snooker balls (without the cue ball).

Each of my apes has a basket, and they share the balls among the 3 baskets so that each ape has the same number of balls in his basket and so that the total value of the balls (red = 1, yellow = 2 etc) is the same in each basket.

They also do this in such a way as to give Gordon, being the smallest, the least number of red balls possible.

How many red balls in Gordon's basket (and explain why)?

[elvaago]

There are 15+2+3+4+5+6+7=42 points total. So each ape has 14 points.
There are 21 balls in total. So each ape has 7 balls.

Take the black ball, 7. You still need 6 balls for a total of 14 points. They can't all be reds, because then you only have 13 points. So one of them has to be yellow. So one ape has black + yellow + 5 reds = 14
Next take pink. You need 6 more balls for a total of 14. They can't all be reds because then you only have 12 points. So let's say one of them is green. So the other ape has pink + green + 5 reds = 14 points.
That leaves five reds for the youngest ape. He has brown + blue + 5 reds for a total of 14 points.

My method includes figuring out the largest amount of red balls possible for the two oldest apes, leaving the least number of possible reds for the smallest ape.

[davis_greatest]

Congratulations, elvaago!

That's good enough for me. With 15 reds among 3 apes, if one ape had fewer than 5 reds, that would mean that another ape would have to have more than 5 reds. But if one ape had 6 reds, he would need to get 8 points from the seventh ball (to total 14 points) which is impossible; and if he had 7 reds, he would only have 7 points (rather than 14). So, no ape can have more than 5 reds, and hence no ape can have fewer than 5 either.

SO HERE IS THE SCOREBOARD AFTER ROUND 44 BUT BEFORE ROUND 43:

snookersfun……………………….…..18½
Vidas……………………………………….8½
abextra……………………………..…...8½
robert602…………………………………5
davis_greatest…………………..……5
elvaago...............................3
Semih_Sayginer.....................½

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

[elvaago]

I would like to thank my grade 11 and 12 math teacher, snookersfun for providing inspiration and of course davis_greatest for giving us these wonderful puzzles to break our brains over. :-)

[davis_greatest]

Rounds 43 and 45 - "Sweet 16" and "Go Petal Go!"

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?


(And don't forget round 43 - message 561 above. Crying out to award a point!)

[elvaago]

5400 seconds after 6 PM. ;-)

[davis_greatest]

5400 seconds after 6 PM. ;-)

Yes! I actually meant "at what time" (I know that "when" was a bit ambiguous). Tell us what time it was, elvaago, and I'll award you the point!

[elvaago]

7:30 PM Friday evening.

[davis_greatest]

elvaago moving up the scoreboard

7:30 PM Friday evening.

Indeed! Petal takes 1½ hours (90 minutes) for the first shot, 81 minutes for the second.... getting progressively faster until she has played an infinite number of shots and finishes 15 hours after starting.

SO HERE IS THE SCOREBOARD AFTER ROUND 45 BUT BEFORE ROUND 43:

snookersfun……………………….…..18½
Vidas……………………………………….8½
abextra……………………………..…...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]

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?