Puzzles with numbers and things

[davis_greatest]

The six apes have spent the last day or two with a stopwatch actually testing the times to first reach one particular pocket after the instant Oliver struck the ball, of course they are very dedicated and did not mind doing it many, many, times to see how long it took to reach each particular pocket for the first time, on average.

This is what they found, yellow and pink each took 33 seconds, green and blue 24 seconds, brown 27 seconds and Oliver got the ball back after 30 seconds.

Congratulations, then, to moglet, snookersfun and Monique who have all found the answer - albeit through different means - to round 391! Indeed, the answers are (a) 33 (b) 24 (c) 27 (d) 24 (e) 33 and (f) 30 seconds.

Labelling the pockets as described in post 3431 and following the approach I outlined gives the equations:

A = 5 + ⅓A + ⅓B + ⅓C
B = 5 + ⅓A + ⅓B
C = 5 + ⅔A

which we rearrange as

2A = 15+B+C
2B = 15+A
3C = 15+2A

which we can solve (3 equations in 3 unknowns) as:

A=33, B=24, C=27

Using an identical method we can find the average time between Oliver's shots. When he plays a shot, it takes 5 seconds and there is a ⅔ chance that it goes to a B pocket (green or blue) and a ⅓ chance that it goes to the C pocket (brown, opposite).

So the average time between Oliver's shots is 5 + ⅔B+ ⅓C = 5+16+9 = 30 seconds.

They didn't seem in the least surprised that sometimes, but not very often, it could take 40 or more shots to get to a destination pocket, which ever one it might have been!

Yes, we can calculate how likely it is to require 40 shots or more, if we so wish. It will take 40 shots or more from Oliver striking the ball to reach:
(a) the yellow pocket, with probability 0.0710% (same for pink)
(b) the green pocket, with probability 0.0479% (same for blue)
(c) the brown pocket, with probability 0.0572%
(d) Oliver again, with probability 0.0617%

[moglet]

R391

Round 391 update - only correct answer so far, via PM, is from snookersfun (maybe with a teeny bit of help ) ... a reminder that answers can go on the thread. A hint - apart from part (f), the answers are not multiples of 5 seconds.

The six apes have spent the last day or two with a stopwatch actually testing the times to first reach one particular pocket after the instant Oliver struck the ball, of course they are very dedicated and did not mind doing it many, many, times to see how long it took to reach each particular pocket for the first time, on average.

This is what they found, yellow and pink each took 33 seconds, green and blue 24 seconds, brown 27 seconds and Oliver got the ball back after 30 seconds.

They didn't seem in the least surprised that sometimes, but not very often, it could take 40 or more shots to get to a destination pocket, which ever one it might have been!

[Monique]

R392

Solved by Abextra as well now! Congratulations!

Still open and more imaginative setups for Barry's are welcome.

[davis_greatest]

R. 389: Mon has the greens right as well now (actually since yesterday, sorry for being so late), hurray!

and with that congratulations to d_g, moglet and Monique on solving this round. We still need one solver at least to put his/her diagram/solution up

This was my original diagram - click it to enlarge. Just pick any pocket and mark it DESTINATION - then for each node moving farther out, add the numbers of the closer nodes to DESTINATION that can be reached from that node. Essentially in the manner that you described, except that only one DESTINATION pocket is needed as this gives the answer at once (using symmetry) for the numbers of paths that reach any pocket from any other (i.e. 5, 41 or 121 if travelling 1, 2 or 3 pockets in either direction).

[snookersfun]

R. 389 and 393 updates

R. 389: Mon has the greens right as well now (actually since yesterday, sorry for being so late), hurray!

and with that congratulations to d_g, moglet and Monique on solving this round. We still need one solver at least to put his/her diagram/solution up


R. 393: d_g and Mon solved this one easily. Will give a bit more time, as e.g. I know moglet is stuck in the fascinating problem of random walks or rather shots

[Jacko]

...and here it is! lol

Actually, I'm struggling to work out the answer myself! If you can be dyslexic with numbers then I have it!

[abextra]

I vaguely recall that PMs are limited untill you post a minimal number of contributions. Maybe I'm wrong but ... maybe a mod could confirm?

Oh, I think it's true - 10 posts needed, I guess, just one more, Jacko!

Welcome on the thread and nice homepage!

[Monique]

I vaguely recall that PMs are limited untill you post a minimal number of contributions. Maybe I'm wrong but ... maybe a mod could confirm?

BTW ... I'm not sulking your question Jacko, it's nice to have new people on the thread, but I know this puzzle, so it would be unfair to participate.

Welcome!

[Jacko]

I have looked in there but cant see anything about PM's ?????

[abextra]

How do I activate my PM's? Cant find the section to do this.

My Panel - Edit Options - Enable Private Messaging - Save Changes. :snooker:

[Jacko]

How do I activate my PM's? Cant find the section to do this.

[moglet]

Hello Jacko, welcome to the thread

Your PM is disabled, how should we answer your puzzle?

[Jacko]

Here's a puzzle for you all....

3 snooker players go into a hotel.
The receptionist said the room was £30, so each man paid £10 and went to the room.
A while later, the receptionist realised the room was only £25,
so she sent the porter to the 3 guys' room with £5.
On the way, the porter couldn't figure out how to split £5 evenly between 3 men, so he gave each man £1 and kept the other £2 for himself.
This meant that the 3 men each paid £9 for the room, which is a total of £27.
Add to that the £2 the porter kept and the total is £29.
Where is the other pound?

[Monique]

R392 update

Snookersfun and D_G both solved this one ... Well done!
It stays open and ... it's easy!!!

[snookersfun]

update on R. 389:

Mon already yesterday solved the yellow and the brown paths. Well done! Slight problems with the green still

With that we should probably close this round and solutions for green and brown can go up on the thread now.

and here another short one (we might have done something similar, but none the less...)

R.393: separating

Charlie, Oliver and Gordon by now had many practise hours on their hexagonal table. They come to the conclusion that the final colour clearance is too easy as it is, with all the colours one after each other around the table; the colour spots actually lying on a smaller inner hexagon (points A,C,E,G,I,K on moglet's graphic from my previous round). Charlie's plan is to separate at least pink and black by not having them on neighbouring spots, and see if that will show a difference in play.

So, how many different ways are there to do that? Out of how many different ways to arrange all colours on those spots without those constrictions?