Sports / Maths Puzzle(s)...

[Howard Somerset]

I can't see any way of getting the 9 round score to be anything less than an amazing 247. Bul I'll keep trying.

Nope

ok. I can now see a way. But not sure what the optimum score is. Back soon.

[Howard Somerset]

Dropped to 71 now. Might still be able to improve on that. Nice puzzle.

And now getting a bit more confident with 63

[Charlie Reams]

I trust that as you (sort of) have a maths degree, you're not counting 1 as prime.

[JimBentley]

I can make a 55, but I'm not sure that's optimal either.

[Howard Somerset]

I trust that as you (sort of) have a maths degree, you're not counting 1 as prime.

I'm quite sure Dinos doesn't regard 1 as prime. I believe we've discussed it from time to time.

I'm now lower than my previous attempt, but not quite as low as Jim.

[Hugh Binnie]

Ah, I understand now. >.< Can't beat Jim's score, but I'm working on it.

[Jimmy Gough]

Ah, I understand now. >.< Can't beat Jim's score, but I'm working on it.

Because you suck?

[Howard Somerset]

And now same as Jim.

[Gavin Chipper]

PUZZLE D1: GOLF

There's a 9-hole golf course near my house where I like to play, but my golf swing is very inconsistent. One day while totting up the scores after playing with my friend, he points out to me that my difference in scores on subsequent holes are all different prime numbers. What is the lowest possible 9-hole total score I could have achieved?

I've tried a few different things, and while I haven't come up with any sort of proof, I think it's 55. I used the lowest 8 prime numbers and tried to keep the shots on each hole down as much as I could. So 17 and 19 (the highest two) were either side of the highest number. Also 11 and 13 were either side of the second highest number.

1, 20, 3, 1, 14, 3, 8, 1, 4 (19, -17, -2, 13, -11, 5, -7, 3)

1, 20, 3, 14, 1, 8, 3, 1, 4 (19, -17, 11, -13, 7, -5, -2, 3)

[Dinos Sfyris]

1 is certainly not prime. End of.

Jim is the first person with the correct answer: 55

SOLUTION: The trick is to first list the lowest 8 prime numbers and fit them in as differences between scores while minimising deviation from the optimal score of 1. 19 is the highest prime difference that must be used so a score of 20 must be scored on one hole. It is then optimal to use the next largest prime (17) to bring this score down as low as possible. One such example of how you can achieve the optimal total score is:

Hole 1: 1
Hole 2: 20 (+19)
Hole 3: 3 (-17)
Hole 4: 1 (-2)
Hole 5: 14 (+13)
Hole 6: 3 (-11)
Hole 7: 8 (+5)
Hole 8: 1 (-7)
Hole 9: 4 (+3)

Howard has suggested I open the same puzzle up to a 18-hole scenario, which I'm more than happy to do.

[Derek Hazell]

Howard has suggested I open the same puzzle up to a 18-hole scenario, which I'm more than happy to do.

Shouldn't you be saving that sort of stamina for your last night of freedom?

[Matt Bayfield]

Well done to Dinos for adding a puzzle of his own (and well done to me because I matched Jim's correct answer). I'm especially pleased, as by the looks of Dinos's scorecard there, drinks are on him in the clubhouse for getting three holes-in-one!

Incidentally, anyone who has a puzzle which is similar in nature to those in this thread, then please post it here if you wish (although to avoid confusion, preferably let's only have one puzzle on the go at any one time, i.e. wait until the answer to the previous puzzle has been confirmed as correct).

ANSWER TO PUZZLE FIVE

The trick is to work out - not necessarily in this order - (a) which combinations of scores give a maximum of 11 shots at the full 10 pins, and then (b) to maximise the bonus points over the 10 frames. These are the key steps in achieving this:

(1) the first ball of frame 1, which is the only first ball counted singly in your total score (the first ball of all subsequent frames effectively counting "double", assuming the previous frame was closed) should always be as low a score as possible, i.e. 1.

It cannot be a zero as the next ball (i.e. frame 1 spare ball) would also be delivered at the full 10 pins, which would mean you end up with 12 balls at the full 10 pins rather than the maximum allowable 11 (unless you leave frame 10 open, which turns out to give a lower max score).

(2) Since your strike has the bonus of the next two balls effectively counting double (which will mean a score of 20 for that frame regardless of the number of pins knocked down with the first ball after your strike - we can assume the next frame is spared), the first ball of the frame after your strike should be as low as possible, i.e. 2, since you have already used 1 with your very first ball of frame one.

(3) To ensure you get only 11 shots at the full 10 pins, your "zero pins knocked down" ball must be the "extra" ball which you get the end of the 10th frame for closing the 10th frame.

Anyhow, if you apply all the above, you can have a scorecard like the one below. I have emboldened your eleven shots at all ten pins.

Frame One: 1 9 - score 20, running total 20
Two: 10 - score 20, running total 40
Three: 2 8 - score 13, running total 53
Four: 3 7 - score 14, running total 67
Five: 4 6 - score 15, running total 82
Six: 5 5 - score 16, running total 98
Seven: 6 4 - score 17, running total 115
Eight: 7 3 - score 18, running total 133
Nine: 8 2 - score 19, running total 152
Ten: 9 1 0 - score 10, final total 162

Answer: 162 (first correct answer: Hugh)


Now back to English Premier League football for the next puzzle. It's a two-parter.

PUZZLE SIX: FOOTBALL

Consider the 2008/9 Premier League (i.e. last season), composed of the 20 clubs listed below. Imagine that at the end of the season, by some mathematical fluke, every club finished on a points total which was directly proportional to the number of letters in the club’s name as spelled below (these spellings exclude spaces and additions such as “F.C.”).

(a) Complete a mathematically possible final league table for the 2008/9 season, including the following columns: club name, Wins, Draws, Losses, Points. There are multiple correct possibilities.

(b) What is the maximum number of clubs who could have finished the season without drawing any of their league matches?

Notes:

(i) For the purposes of this mathematical problem, you should ignore the real results from the 2008/9 season.

(ii) Assume that all fixtures were completed and no club had their points total adjusted (e.g. due to ineligible players, financial issues, etc.).

List of clubs (followed by number of letters in club's name):

WestBromwichAlbion 18
ManchesterUnited 16
TottenhamHotspur 16
BlackburnRovers 15
BoltonWanderers 15
NewcastleUnited 15
ManchesterCity 14
Middlesbrough 13
WestHamUnited 13
WiganAthletic 13
AstonVilla 10
Portsmouth 10
Sunderland 10
Liverpool 9
StokeCity 9
HullCity 8
Arsenal 7
Chelsea 7
Everton 7
Fulham 6

[Bob De Caux]

Not quite sure if I understand the question, but is this right for "a"?

P W D L Po
West Bromwich Albion 38 17 21 0 72
Manchester United 38 14 22 2 64
Tottenham Hotspur 38 14 22 2 64
Blackburn Rovers 38 13 21 4 60
Bolton Wanderers 38 13 21 4 60
Newcastle United 38 13 21 4 60
Manchester City 38 12 20 6 56
Middlesbrough 38 10 22 6 52
West Ham United 38 10 22 6 52
Wigan Athletic 38 10 22 6 52
Aston Villa 38 6 22 10 40
Portsmouth 38 6 22 10 40
Sunderland 38 6 22 10 40
Liverpool 38 5 21 12 36
Stoke City 38 5 21 12 36
Hull City 38 3 23 12 32
Arsenal 38 2 22 14 28
Chelsea 38 2 22 14 28
Everton 38 2 22 14 28
Fulham 38 1 21 16 24

Very pleased to see Spurs 3rd in the table, even if we are still behind Man U

[Gavin Chipper]

Howard has suggested I open the same puzzle up to a 18-hole scenario, which I'm more than happy to do.

I think the crowd have spoken with their silence. In case you didn't hear, they said "I'm not fucking doing that!"

[Howard Somerset]

I'll not attempt (a) because Bob seems to have that done.

My answer to (b) is 7 i.e. those with a multiple of three letters in their names.

Very pleased to see Spurs 3rd in the table, even if we are still behind Man U

We're only listed behind MU alphabetically, which makes us level with them. Which reminds me of something I heard not long ago - the main motivation for the Arse changing their name from Woolwich Arsenal was not because they were leaving Woolwich, it was so they could appear first when the clubs were listed alphabetically.

[Howard Somerset]

I've now adjusted my answer to (b) to 5, because I don't think it can be done with 7.

And here is an alternative answer to (a) which is consistent with my answer for (b)

pl/w/d/l/ team name
17/21/0/72 West Brom
13/25/0/64 Man U
13/25/0/64 Tottenham
16/12/10/60 Blackburn
20/0/18/60 Bolton
20/0/18/60 Newcastle
9/29/0/56 Man C
7/31/0/52 Middlesbro
7/31/0/52 West Ham
7/31/0/52 Wigan
1/37/0/40 Villa
1/37/0/40 Pompey
1/37/0/40 Sunderland
12/0/26/36 Liverpool
12/0/26/36 Stoke
0/32/6/32 Hull
0/28/10/28 Arse
0/28/10/28 Chelsea
0/28/10/28 Everton
8/0/30/24 Fulham
Bolton, Newcastle, Liverpool, Stoke and Fulham are the five teams with no draws.


Later edit - if you want us to put in goals too, I'll happily arrange it so that Tottenham are ahead of ManU, and Arse are relegated.

[Bob De Caux]

I was going to put 7 when I submitted my first answer, but then realised it wasn't possible. I think I've managed 6 though, Howard.

P W D L F A Po GD
West Bromwich Albion 38 24 0 14 24 14 72 10
Tottenham Hotspur 38 13 25 0 31 0 64 31
Manchester United 38 13 25 0 13 0 64 13
Blackburn Rovers 38 20 0 18 22 18 60 4
Bolton Wanderers 38 20 0 18 21 18 60 3
Newcastle United 38 20 0 18 20 21 60 -1
Manchester City 38 9 29 0 9 0 56 9
Middlesbrough 38 7 31 0 9 0 52 9
West Ham United 38 7 31 0 8 0 52 8
Wigan Athletic 38 7 31 0 7 0 52 7
Aston Villa 38 1 37 0 2 0 40 2
Portsmouth 38 1 37 0 1 0 40 1
Sunderland 38 2 34 2 2 6 40 -4
Stoke City 38 0 36 2 0 2 36 -2
Liverpool 38 12 0 26 13 26 36 -13
Hull City 38 0 32 6 0 6 32 -6
Everton 38 0 28 10 0 10 28 -10
Chelsea 38 0 28 10 0 11 28 -11
Arsenal 38 0 28 10 0 28 28 -28
Fulham 38 8 0 30 8 30 24 -22

I've added goal difference for my own amusement. A pretty dull season was enlivened by Spurs thrashing Arsenal 5-0 home and away.

[Charlie Reams]

A pretty dull season was enlivened by Spurs thrashing Arsenal 5-0 home and away.

Also my prediction for the 2010/2011 season.

[Steve Durney]

A pretty dull season was enlivened by Spurs thrashing Arsenal 5-0 home and away.

Also my prediction for the 2010/2011 season.

The reverse would be a LOT more likely!

[Matt Bayfield]

For part (a), Bob is the first (and only) correct answer so far, with his post initially made at 7:35 pm (or possibly 6:35 pm, depending on how your clock is set) yesterday. Bob, the league table you gave in that post is valid.

As for part (b), all answers given so far are incorrect! The league tables which Howard and Bob have posted are both impossible, although I do appreciate Bob's addition of goal difference to ensure that the Arse go down! I won't yet say why Howard and Bob's tables are invalid, otherwise I'll pretty much be giving away the answer.

(Note: for those attempting part (b), to obtain full kudos, please either give a league table to justify your answer (as Bob/Howard have done), or state your logic.)

[Bob De Caux]

Doh! Can see why they are invalid. Serves me right for posting late at night. Will work on revised answer...

[Bob De Caux]

It can't be done with 5, so here is my answer for 4

P W D L Po
West Bromwich Albion 38 24 0 14 72
Tottenham Hotspur 38 13 25 0 64
Manchester United 38 13 25 0 64
Blackburn Rovers 38 20 0 18 60
Bolton Wanderers 38 20 0 18 60
Newcastle United 38 11 27 0 60
Manchester City 38 10 26 2 56
Middlesbrough 38 8 28 2 52
West Ham United 38 8 28 2 52
Wigan Athletic 38 8 28 2 52
Aston Villa 38 4 28 6 40
Portsmouth 38 4 28 6 40
Sunderland 38 4 28 6 40
Stoke City 38 3 27 8 36
Liverpool 38 3 27 8 36
Hull City 38 2 26 10 32
Everton 38 1 25 12 28
Chelsea 38 0 28 10 28
Arsenal 38 0 28 10 28
Fulham 38 8 0 30 24

Brief explanation:
There are a maximum of 1140 points available (if every game is won/lost). For all points tallies to be proportional to letters, you need a constant to multiple them by: 5 is too big (not enough points available), 3 is too small (would need too many draws) so the constant is 4.

That means you need 924 points in total, implying 216 draws (432 in the table as each draw will shown twice). For 5 teams to have no draws, then the max draws available would be if every other team drew = 15*28 = 420. However you need 432, so 5 teams on 0 draws is impossible.

[Howard Somerset]

I'm not at home at the moment, so don't have access to the speadsheet which I used to complie my earlier table. I'll have another look later when I get home - not sure just when that'll be.

It's a great problem.

[Matt Bayfield]

Bob: perfect answer and explanation - well done.

[Bob De Caux]

Cheers Matt, great question. Here's a little puzzle I concocted which hopefully fits in here (although it's not easy to explain to non poker players). I will retract if deemed unsuitable to the thread!

For this puzzle you need to know the rules of Texas Hold 'Em Poker (explained briefly below for those who don't). However the puzzle is about card probabilities rather than any tactics, betting, etc. If it's not clear or you need further info, let me know!

PUZZLE B1: POKER (Texas Hold 'Em)

Imagine you are playing against 1 other player at Texas Hold 'Em poker.
In a hand against them you both find yourselves all in after the turn card.
You both reveal your hole cards and see that as things stand, you are losing.
Given the above and that there is one card to come, what cards would give the highest probability:
i) of you winning the whole pot (several solutions)
ii) of you not losing the whole pot (1 solution I think!)
You need to give your hole cards, your opponents hole cards and the four cards down on the table

Example
You have Kh,Jh. Your opponent has 4c,7c
On the table are 2d,3c,7d,10d
Your opponent currently leads with a pair of 7s
You can win with a J (for pair of Js) or a K (for a pair of Ks)
There are 3 Jacks and 3 Kings left, and 44 available cards
So winning prob = 6/44 = 13.64%
You can do a lot better!

Rules (Texas Hold 'Em)
Each player gets dealt two cards each face down (the hole cards)
Three cards are then dealt face up on the table (the flop)
A further card is dealt face up (the turn card)
One final card is dealt face up (the river)
Using any five of the two cards in their hand and the five on the table (so any combination of 5 from 7), the player must make the highest possible poker hand in the rankings, which are as follows from best to worst:
Straight Flush, Four of a Kind, Full House, Flush, Straight, Three of a Kind, Two Pairs, Pair, High Card
All 5 cards count, so for example if you have just a pair of 6s, your hand would be the 6s and the next three highest cards in order. Higher is always better (Ace can be high or low)
To see who has the better hand, you work through all 5 cards in order until one hand is better
e.g 6,6,K,3,2 is better than 6,6,J,10,9, as the third best card here becomes the tiebreaker (King is higher than Jack)
The best hand wins the pot. If you both have the same hand, you split the pot 50:50

Notes
"All in" means that you both have all your money in the middle, at which point you both reveal your cards - any previous betting or how you got to be all in are irrelevant
Assume a standard 52 card deck which was shuffled before the hand
The "pot" is the total money available to win on the hand

[Matt Bayfield]

Bob - looks like a great puzzle. That'll keep me (and no doubt, others round here) busy for a while!

And a quick attempt to give a first answer to part (ii):

My hole cards: 2C 2D
Opponent's hole cards: 2H 3H
Community cards: AC AD AH AS
Probability of not losing hand: 43/44, as all of the remaining 44 unseen cards except 2S will mean both players play the board and split the pot.

[Matt Morrison]

For the first part (most winning river cards), something like:

Me: 9h 10h
Him: 2s Ad
Board to turn: Jh Qh 4s 5s

This gives any of 9 hearts to win via flush, any of the 3 nines or 3 tens to win via pair, also 3 other eights and 3 other kings (not four of each as we've counted the 8h and Kh already) to win via straight.
... a total of 21 winning river cards... 21/44 = 47.7%

That's a start anyway.

[Matt Bayfield]

And my attempt at part (i):

My logic: suppose my opponent is winning with something like a small pair, against my high card. I can then win if I hit a flush, straight, or higher pair on the river. I can max my number of outs for a flush at (9 outs) by having my hole cards suited, and exactly 2 of the community cards of the same suit, with my opponent having no cards of that suit. I maximise my number of additional outs via a straight if, between my hole cards and the community cards, there are 2 face values which give me a straight - assume all of the necessary cards are not in my opponent's hand (8 cards, but only 6 extra outs which will not also give a flush). I can then have up to 6 further outs if I form highest pair, again assuming my opponent does not have any of these cards. Thus total outs = 9 + 6 + 6 = 21, giving probability 21/42 of winning the pot.

One example of this:

My hole cards: 8C 7C
Opponent's hole cards: 2H 2D
Community cards: 10C 9C 4S 3S
My 21 outs to win the pot: AC 2C 3C 4C 5C 6C JC QC KC 6H 6S 6D JH JS JD 8H 8S 8D 7H 7S 7D

[Edit: Matt M has beaten me by 1 minute to post what is basically the same solution. I shouldn't have typed so much in my post. ]

[Matt Morrison]

Nice one Matt, note you said /42 rather than /44, but yeah, the same thing.
The 3rd and 4th cards I used on my example board can be almost anything at all, so long as they don't produce a flush or a straight for us too early, or give him his high-card pair.

[Matt Morrison]

[Edit: Matt M has beaten me by 1 minute to post what is basically the same solution. I shouldn't have typed so much in my post. ]

Yeah, that'll teach you to be so damn helpful and informative! If it makes you feel any better, I went and made a cup of coffee between writing it and posting it as my gran wanted a quick chat, just glad I got back to the PC quick enough for it not to be the other way round, me damning you for the same timing issues!

[Matt Bayfield]

Oops... typo. Yup, obviously I meant to put /44 rather than /42.

Although I suppose it could be /42 if the dealer accidentally flashed two cards whilst burning, or something. Somehow I think we're supposed to ignore that possibility, for the purposes of this puzzle.

[Bob De Caux]

Great, quick work guys. Luckily, although they're good answers, it's possible to do better on both i) and ii) so you haven't ruined my fun just yet! (I did come up with this puzzle myself so it's possible I'm wrong, but I've checked it and feel pretty comfortable).

For i) you can get over 50% and
for ii) you can get 100%

[Matt Morrison]

Ok, good news, get to have another look!

For (i), how about...

Me: 9h Th
Him: 2s 2d
Board: 7h 8h As Ad

This time we've got the 9 hearts for a flush again, 3 nines and 3 tens for winning pair, 3 sixes and 3 jacks for a straight. But in addition we now have 3 sevens and 3 eights that will cause two pair to appear on the board, thus destroying his low pair.

So 9 + 6 + 6 + 6 = 27/44 = 61.36%

Correct?

[Bob De Caux]

Correct?

Erm, sort of! Your cards give you the correct answer, but your calculation is wrong! (you don't have as many outs as that)

[Bob De Caux]

Just to clarify then:

- Matt M has the right answer (in terms of the cards) for part i, but the wrong probability (I'm reluctant to say you've got it right until you get the correct number of outs!)
- Matt B has the best answer so far for part ii (97.7%), but it can be beaten

[Matt Morrison]

Me: 9h Th
Him: 2s 2d
Board: 7h 8h As Ad

This time we've got the 9 hearts for a flush again, 3 nines and 3 tens for winning pair, 3 sixes and 3 jacks for a straight. But in addition we now have 3 sevens and 3 eights that will cause two pair to appear on the board, thus destroying his low pair.

So 9 + 6 + 6 + 6 = 27/44 = 61.36%

Matt M has the right answer (in terms of the cards) for part i, but the wrong probability (I'm reluctant to say you've got it right until you get the correct number of outs!)

Righto then Bob. Sorry it took a while for me to come back to this.

Yeah just realised the 2 of hearts flush is no good, as he hits his full house. So just need to remove that out, making 26/44 = 59.09%.

We good now?

[Bob De Caux]

We good now?

Very close but...you still have one out too many (similar to the one you just mentioned)

[Matt Morrison]

We good now?

Very close but...you still have one out too many (similar to the one you just mentioned)

Ah course. The Ah, so 25/44 = 56.8%! Got there finally.

[Bob De Caux]

Nice one Matt - correct answer! Just part ii to go...

[David O'Donnell]

Nice one Matt - correct answer! Just part ii to go...

I got the same percentage as the guys for part 1 but assume that the situation you're taking about has to be a one-outer or roughly 1/44.

I knew I had a one-outer from play where a guy completely sucked out on me so here it is:

me: Jh8d
Villain: 2h2s

Board and Turn: JdTd9dAh

Villain could only hit the 2 of clubs with a chance of 2.27% and no chance of a tie

Edit: Just realised that I have the same percentage as Matt - can there be less than one out?

Edit 2: You crafty bastard, just re-read the question, a locked hand gives you the highest percentage.

[Bob De Caux]

Hey DOD

Can't be a lock hand, as one of the conditions is that you are behind after the turn! You're ahead in your one-outer as well (although that it a sick, sick beat which would have me putting my fist through the screen), so that's not what I'm looking for. For part ii you're not trying to guarantee that you win the hand, just that you don't lose the hand. It's not about Villain trying to draw out on you as he's already ahead.

[David O'Donnell]

Hey DOD

Can't be a lock hand, as one of the conditions is that you are behind after the turn! You're ahead in your one-outer as well (although that it a sick, sick beat which would have me putting my fist through the screen), so that's not what I'm looking for. For part ii you're not trying to guarantee that you win the hand, just that you don't lose the hand. It's not about Villain trying to draw out on you as he's already ahead.

LOL. That beat was so brutal that I had to laugh.

I was only trying to answer the second part there, for the first part I had the hero having some manner of open ended flush draw - was something like 56.81%.

I have obviously misread part two though as I missed that you were still losing after the turn. Will have another look - cheers for posting these Bob!

[Andrew Hulme]

Guess I'm supposed to be the resident expert on poker but I have to confess I'm racking my brains and I've no idea about part 2. Will keep trying

[Andrew Hulme]

Actually...


We have: 2d2s
Opponent Has: 2c3c
Board Reads: AdAcKsKd

[Bob De Caux]

Nice one Andrew, guess there must be (at least) two solutions as that's not the one I had! Would be interesting to see if anyone can get my original one as it has similar characteristics (a double paired board) but a much higher win percentage (43.18% to your 2.27%). Anyway, well done!

[Matt Bayfield]

Not that it matters now, but I eventually got the same answer as Andrew for part (ii)...

Nice puzzle!

[JimBentley]

Hope you don't mind if I post a quick snooker one, Matt - I think it's quite easy, so I don't think it'll go unsolved for long. I should add that I've no idea how to prove that the answer I have is optimal, but I'm fairly sure it is.

Ronnie O’Sullivan is playing A. N. Other in the first round of the World Championships. Mr. Other isn’t a very good player, so Ronnie finds himself 9 frames to 0 up and needs just one more to progress to the next round. Bored by the lack of competition, Ronnie’s already won some of the frames left-handed, one-handed and blindfolded, so for the final frame he decides to restrict himself even further: in making a break, at no point is the break allowed to add up to a prime number.

For instance, after his first red he can’t take a yellow, as 1 + 2 = 3, which is prime. He could take a green, but then wouldn’t be able to go for another red as 1 + 3 + 1 = 5, which is prime. And so on…

Assuming Ronnie breaks off in the final frame and flukes a red (to remove any complications that a free ball might bring), what is the highest break he can make?

[Matt Morrison]

Hope you don't mind if I post a quick snooker one, Matt - I think it's quite easy, so I don't think it'll go unsolved for long. I should add that I've no idea how to prove that the answer I have is optimal, but I'm fairly sure it is.

On first attempt I've got a break of 95. Obviously, not being clever like you folk, I'm only looking at it from an interest in snooker and not maths; not only do I have no idea how you would prove it's optimal (like Jim), I doubt it even is optimal (unlike Jim). But it's a start. I'll have another look once Jim tells me it's a rubbish attempt.

[JimBentley]

Hope you don't mind if I post a quick snooker one, Matt - I think it's quite easy, so I don't think it'll go unsolved for long. I should add that I've no idea how to prove that the answer I have is optimal, but I'm fairly sure it is.

On first attempt I've got a break of 95. Obviously, not being clever like you folk, I'm only looking at it from an interest in snooker and not maths; not only do I have no idea how you would prove it's optimal (like Jim), I doubt it even is optimal (unlike Jim). But it's a start. I'll have another look once Jim tells me it's a rubbish attempt.

Heh, it's not rubbish at all, but it is some way short of what I've got. I'm off out for the night now and expect a solution by the time I return (although I will be inebriated so I might not recognise it as such).

[Innis Carson]

I ended up with 105 after a few stabs in the dark, and can't seem to beat it. Don't have any proper justification for it though, could still easily not be optimal.

[Matt Morrison]

Pfft, no idea. Yet another puzzle I'll just have to sit back and wait for the answer to! It's a hard life hanging around with your clever fucks sometimes

[Andrew Hulme]

I can get 117 Jim. Any good?

[Innis Carson]

Seeing Andrew's answer gave me an idea and I think I've got this time: 118. The highest (I think) that you can get with just the 15 reds+colours is 100, but that gets you stuck on 105. Working down from there, 91 was the highest score you could go on to clear the table with, giving you 118. There definitely is way of getting to 91 in the first place but I forgot it. That's the best unless I've mucked it up somewhere along the line.

[Bob De Caux]

You can get to 91 and then pot the colours to give 118. 1,8,9,14,15,21,22,27,28,35,36,39,40,45,46,51,52,57,58,63,64,69,70,77,78,81,82,84,85,91,

Would be surprised if this is not the best as the next number after 91 that allows you to pot Yellow-Black safely is 106, which is a long way off...

[Matt Morrison]

I'm such an idiot. Totally forgot you could pot the colours after you pot 15 reds and colours! So yeah, I've had 118 all along too. Just rather than potting a black to get 95 before clearing up the colours, I now pot a lowly green to move from 88 to 91, and the colours take it up to 118. Told you I wasn't clever like what you are.

[Hugh Binnie]

Bloody hate twin primes now.

[Gavin Chipper]

I've got 124 which seems to be different from the rest of you but I'll check.

Edit: Actually there might be a prime in there.

Edit: so I'm now in the 118 camp.

[Ian Volante]

I got 124 on my first work through, but I suspect with a bit of fiddling down below, there could be a better solution.

And I've checked my solution, and I can't find any primes in the path.

Second edit: Scrub that, I found a prime, and now with a clean solution I'm at 121.

[Gavin Chipper]

I got 124 on my first work through, but I suspect with a bit of fiddling down below, there could be a better solution.

And I've checked my solution, and I can't find any primes in the path.

I got the same as you but found that 97 was prime. Not that you've made the same mistake as me.

[Ian Volante]

I got 124 on my first work through, but I suspect with a bit of fiddling down below, there could be a better solution.

And I've checked my solution, and I can't find any primes in the path.

I got the same as you but found that 97 was prime. Not that you've made the same mistake as me.

Of course not, however see above for my improvement!

Shiiiit, actually I'm at 118 too