Puzzles with numbers and things

[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)

[elvaago]

7:30 PM Friday evening.

[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]

5400 seconds after 6 PM. ;-)

[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]

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]

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]

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]

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)?

[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%!

[elvaago]

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

[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!

[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]

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

[Semih_Sayginer]

yeah, so it wasnt another trick

just kick me next time. i dont mind.

so, we actually have to work something out? ill give that a miss then.