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R. 208 update:
Monique solved this one first as well
Congratulations!!!
Monique solved this one first as well
Congratulations!!!
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Round 209 - Clever Charlie ... nonplussed Barry
This will probably be trivial for some experts ... I hope it will be fun for the others!
Yesterday evening, Gordon was in top form and in a mood for intensive practice. For the whole evening he played frame after frame, total clearance after total clearance! In fact he was in a pattern of three: first he played reds and colours in ascending order - so red yellow, red green, red brown ... red black, red yellow ... until the final colours - then he played them in descending order - red black, red pink, ... red yellow, red black ... and so on - then a maximum total clearance. He did that until the club closed! By that time Oliver was fast asleep, Charlie was gone to the banana bar and Barry had lost the count!
Charlie was supposed to keep the score. Of course he got bored! So he made himself a tool to follow Gordon's progress easily. He took a ribbon of paper: 1 cm wide and 54 cm long. He glued the extremities so he got a sort of ring. He then started colouring squares - 1 cm x 1 cm - on the ribbon according to Gordon's progress, each square being immediately adjacent to the previous one. Much to Barry's astonishment he managed to record the three clearances of the pattern, without tearing the ribbon apart or turning the ring inside out! And when he reached the final black of the third clearance, he coloured the last square neatly adjacent to the first one! So he was ready fo follow Gordon's progress in the next round ...
What's the trick? Please explain this to poor Barry ...
Well, when Gordon finally got tired and Charlie was at the banana bar, a fascinated Barry decided to study this wonderfull ring more closely! So he took a pair of cissors and cut the ring along the middle line, in the length. So he thoutgh he would leave a ring to Charlie - half a cm wide but never mind - and have one for himself. Not so! Poor Barry was very frustrated! What happened?
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That was REALLY too easy for the experts! Congratulations to abextra!
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REALLY too easy! Well done abextra, D_G and Snookersfun!
Next answers on the thread please ...Comment
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shall move on, while we wait for someone to post answer to R209, with
Round 210 - The Entertainer
Dave Stevis likes to keep the crowds entertained. He knows that close finishes are good for viewers, good for sponsorship, and good for the nerves. So he often likes to let his opponents win twice as many frames as he, before he turns on the style to close out the match.
He is confident that he can do this, because he knows that he has a 95% chance of winning any given frame if he tries. (Luck can play a part and there is always a 5% chance that a chimpanzee may distract him and cause him to lose.)
For example, when playing Rollie O'Sonneyman in Wembley Zoo one year, he let Rollie go 8-4 ahead in a best-of-19 match. When playing Fergal "A Good Heart These Days is Hard to Find" O'Sharkey today, he let O'Sharkey go 4-2 ahead in a best-of-9 match.
Naturally, Dave came back to win both matches, but which was he more likely to win, and what were his chances of avoiding catastrophic chimpanzee distractions and winning each match?
Answers on the thread please..."If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.Comment
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I think he has 95,561945781% chances to win Rollie and 85,7375 to win Fergal.
This is a bit counterintuitive but results from the fact that each frame with Fergal is deciding while with Rollie he has a one frame cushion...Comment
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Congratulations, Monique! Exactly right!Originally Posted by Monique
And if it weren't for the chimpanzees, the chances would have been 100%.
Still, he won both matches anyway, of course.
"If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.Comment
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Against Fergal; its 0.95^3. But wouldn't you have to take into account the 2 frames to get to 4-2?
As for the match against Rollie should this answer be less than 0.95? He either wins 6-0 (0.95^6) or 6-1 (0.05 x 0.95^6)?Comment
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Regarding Fergal, he MUST win the three next frames (this is independent of the past - the situation would be the same if he was, say, 8-6 down in a best of 17) so yes 0.95^3Originally Posted by dantuck_7
Regarding Rollie the situation is a bit more complex ...
when 8-4 down he must win 6 frames of possibly 7, so:
EITHER he looses the next frame (0.05 probability) and he MUST subsequently win the six next frames OR he wins the next frame - is then 8-5 down - and has to win 5 frames of possibly 6 ...
Hence the formula
A2*(A1*A1*A1*A1*A1*A1) + A1*(A2*(A1*A1*A1*A1*A1) + A1*(A2*(A1*A1*A1*A1) + A1*(A2*(A1*A1*A1) + A1*(A2*(A1*A1) + A1*(A2*A1 +A1))))) where A1=0.95 (probability of win) and A2=0.05 (probability of loss)
Hopes this makes sense!
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Yes, chance of beating Fergal is 95% ^ 3 as above = 85.7375%
Chance of beating Rollie is
(a) Probability Dave Stevis wins 10-8 (i.e. Stevis winning 6 frames in a row)
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(b) Probability Dave Stevis wins 10-9 (i.e. Rollie wins any one of the next 6 frames and Stevis wins 6 other frames)
For (a), the chance is 95% ^ 6
For (b), there are 6 possible frames that Rollie could win (because if he didn't win any of the next 6, from being 8-4 up, he would lose 8-10 as (a) above) - so the chance is
6 x 5% x 95% ^ 6.
Thus (a) + (b) = (1+ 6x5%) x 95% ^ 6 = 1.3 x 95% ^ 6 = about 95.56% ... (interesting, slightly more than his chance of winning each frame individually)."If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.Comment
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Round 211 - Slappy birthday
You are playing an unusual variation of snooker on your friend Damon Grott's birthday. You have many balls of each colour (yellow, green, brown, blue, pink, black) and you have to put 15 colours (instead of reds) in the triangle.
The first row, of course, has 1 ball; the second row 2 balls;... the fifth row 5 balls.
The sum of points in each row of the triangle must be one point more or less than the sum in the adjacent row. For example, if one row adds up to 10, the next row must add to 9 or 11.
For every pair of touching balls in the triangle that are of different colours, you may give Grott one nice hard slap.
Find whatever arrangement of balls you want to give Grott as many slaps as you can.
Any answers before 21:00 UK time tonight, by Private Message please. Then, after 21:00, please post direct on this thread. (You can show yellow as "2", green as "3" etc...)"If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.Comment
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Any chance of extending this? I've only just seen it.....how about until 9pm tomorrow?Comment
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OK first bid ...
00007
000.26
000342
00.2233
0023222
for 24 slaps ...
for a start
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Congratulations, Monique. And congratulations to abextra, who also sent these answers, both also of 24 slaps.
...........7.............................7
........3.. 5..........................2..6
.......2..4..3......................4..3..2
.....3..3..2..2...................3..2..2..3
...2..2..2..3..2...............2..2..3..2..2"If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.Comment
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