Puzzles with numbers and things

[Monique]

Hello lads and lasses...

I'm still alive. I got the messages from snookersdun and abextra on that addy I PMd you. But not Ben's . I have no idea why and I'm ... puzzled

[moglet]

R412, news flash!

My clever algorithm for testing if the triangle covers the blue spot works just fine, excel has an error in its final decimal place digit (15 places ), correct for the error and all that sweat looking for some other reason was wasted, well not quite, I learned quite a bit too. Definitely 1 in 4 for both two and three ball cases whatever the proportions of the table, even 10x1 or round.

[Monique]

Actually it's 1/4 for the three balls case for every shape that has a centre of symmetry
Shall I put my proof on the thread or do you guys want to think/discuss it further?

[moglet]

Hi, Moglet! Did you solve this question?

I found the answer - and I have to say it surprised me - but my reasoning was wrong, so I'm afraid it was just a fluke... probability is not for me.

abextra, while we wait for Monique's formal proof, why not post your "wrong reasoning", it can't be any worse than my original thinking that was so far off the mark you wouldn't believe.

[abextra]

abextra, while we wait for Monique's formal proof, why not post your "wrong reasoning", it can't be any worse than my original thinking that was so far off the mark you wouldn't believe.

Wanna have a good laugh?

Well, I had no problems with "two balls" case... but will leave it to someone else to answer and explain!

I drew some pictures for the "three balls" case, looking for such positions of the balls that the blue spot would be inside the triangle formed by these balls. I put the first two balls on their places (it doesn't matter where exactly they were) and drew a line from each ball through the blue spot. The third ball had to be on the other side of the blue spot in the triangle between these lines (yellow area in my pic).


I thought both the lines divide the table into halves, for both the lines there is 1/2 chance that the third ball finishes on the "right" side (areas with vertical lines and horizontal lines) and in total the chance that the third ball finishes in the "right" area (squared and yellow) is 1/2*1/2=1/4.

Well... this way was wrong... ... but the answer was correct, lol!
(Pime kana leidis tera, irw...)

[snookersfun]

...and the curious incident of the encompassing triangle continues blame excel!

Well, my attempt at explainig the probabilities was not much better. I tried to find possible areas for that third ball, much like abextra did, noticed that areas were between 0 and ½ and then stated boldly that all must even out to ¼ Of course Mon wanted more proof than that, but at that point moglet popped in with many other, over sudden quite sensible sounding and very intriguing, probabilities found by excel, saving me from looking further... and am afraid still don't have much more...

But here are my initial drawings incidentally for both cases.
quarter.bmp
a) for the two ball case it is easy to see that the 2nd ball can only be positioned in the quarter diagonally opposite the 1st ball, hence a probability of ¼.

b) the probability of the third ball depends on the angle formed between the first 2 balls and the blue ball spot and is between 0 and ½ of the table. The only thing I can add now is that while (for any given first ball) some angles might be favoured with bigger areas because of the form of the table, those same areas are obviously disfavoured for its mirror case or rather the 360 -angle.

waiting for Mon

oh, and great, after updating also this PC to {expletive} IE8, it won't show me my own bmps either help!!!
rolled back to IE7-much better now

[moglet]

This is the Excel column header for the .xls to calculate whether a random triangle would cover the blue spot:

[Monique]

Ok ... I can't draw pictures from here so bear with me and try to draw as I explain.
The first ball can be anywhere with the same probability. Let's put it somewhere and call it A and make it "fixed" there. The second ball again can be anywhere ... let's call it B. To make it easier for the drawing place them both in the "baulk" half of the table and rather close to each other. If O is the blue spot the lines AO and BO make a sort of "cone" and for the third ball C to enclose O in the triangle ABC, C must be in the "top" part of the cone as you all found out. Now the odds for that particular position of A and B is the ratio between the "half cone surface" and the full table surface and this varies from 1/2 to 0 depending on where B is relatively to A... as all of you found out aswell. Now to find out the odds for our question we have to make B "traveling" all over the table and calcultate the average of that ratio ...
As I didn't feel like calculating an " integrale de surface" (sorry don't know the english term and no dictionary here) I looked up for another way.
Consider B' the point symmetrical to B relatively to O. AO and B'O also form a cone. the angles AOB and AOB' sum up to 180 and the area covered by the two half cones is half of the table, so there average is 1/4. Therefore if we calculate the odds by taking the average of the particular odds for all positions of B we will find 1/4 because each point can be "grouped" with its symmetrical when covering the full table whem "summing" the odds.

Hope this makes sense ...

[abextra]

Round 414 - Guess the breaks.

"How did you play today?" asked Gwenny, when Charlie, Gordon and little Oliver came home from snookerclub.
"It was awful," said Oliver. "None of us made a maximum tonight!"
"Yeah, none of us even made a century!" added Charlie. "I'm glad I managed to pot the pink in the end of my highest break, otherwise it would have been as low as Oliver's highest break!"
"If I wouldn't miss the green in the end, my highest break would have been as high as Charlie's," said Gordon.
"So, what were your highest breaks then?" asked Gwenny.
"Too small to tell you," answered Oliver.
"OK, I can figure it out myself," was Gwenny sure of her math skills. "If you guys would add up your highest breaks and then add the number of coloured balls on the snooker table and then add the number of pockets on the snooker table and if you then would multiply the sum by 67 and tell me the last two digits of the number you got... then I would know!"
"The multiple ends with 73," said Charlie, Gordon and Oliver simultaneously after a second. And Gwenny told them their highest breaks.

What were the highest breaks and how did Gwenny know?

Answers by PM please.

My English is what it is... ... feel free to ask clarification if something seems to be too weird.

[abextra]

Round 414 solved perfectly by Monique, Snookersfun and Moglet. Well done everybody, congratulations and thank you!

The answer can come up on thread now.

[Monique]

Round 415 ... breaks

This is an open question as I don't have a definitive answer ... yet
So ... the apes have played some snooker and have made ALL possible breaks between them during their session. What is the minimum number of frames they have played?

[moglet]

R415

Umm, 148 to 155 need a free ball, it looks as if the option to use the free ball for lower breaks would reduce the number of frames needed, so is the freeball an option?

The definition of a break in the "rules" suggest 3 as the minimum break, otherwise 1 and 2 are possible?

Oh, and all possible break "values"? there are about 30 possible ways of scoring 9 points in a single visit........

[Monique]

We have a bid of 78 from Moglet and one of 77 from Abextra

This round can go directly on the thread. Let's open the discussion
Free balls allowed.

[abextra]

We have a bid of 78 from Moglet and one of 77 from Abextra

This round can go directly on the thread. Let's open the discussion
Free balls allowed.

Ohh... I see. My bid of 77 frames was without a free ball, breaks from 1 - 147.

EDIT: Can't think of a way to beat Moglet's 78.

R415
The definition of a break in the "rules" suggest 3 as the minimum break, otherwise 1 and 2 are possible?

What "rules"?

[moglet]

R415

I can see no advantage in using the free ball more than twice in any frame. I've assumed a minimum break of three as the official Rules of Snooker definition suggests in Section2,8 "...a number of pots in successive strokes..." However, there are two "spare" frames one of which could be used to score 1 off a freeball, unfortunately to score 2 we have to use at least 1 red, this would require a whole frame just to make the 2 pt score.

If my thinking is correct the table should be self explanatory: