Sports / Maths Puzzle(s)...

[JimBentley]

Yep, 118 is as good as I can get, and I think Innis was first. Or maybe Bob, I'm unsure. (edit: OK, I'm more sober now and Innis had the first correct total, then Bob gave the first correct sequence - well done fellas!)

I did it with the following sequence (running total in brackets):

Red (1)
Black (8)
Red (9)
Pink (15)
Red (16)
Blue (21)
Red (22)
Blue (27)
Red (28)
Black (35)
Red (36)
Green (39)
Red (40)
Blue (45)
Red (46)
Blue (51)
Red (52)
Blue (57)
Red (58)
Black (65)
Red (66)
Green (69)
Red (70)
Black (77)
Red (78)
Pink (84)
Red (85)
Yellow (87)
Red (88)
Green (91)
Yellow (93)
Green (96)
Brown (100)
Blue (105)
Pink (111)
Black (118)

The biggest problem I found was fitting in the valuable 27 points for the final colours without falling into the 101-103-107-109 minefield. But this is just something I thought up while I was trying to go to sleep the other night, so I've no idea if that's the best solution. I was thinking of writing a brute force routine to spew out every possible sequence, which would be one way of proving it I suppose, but couldn't be arsed in the end (and I think Bob's justification is as good as any - it's definitely impossible to get to 106 or higher before the final six colours, so 118 probably is the best).

[Bob De Caux]

Seeing as we're now well onto the next puzzle, here are the answers to the poker questions:

i) The best I can find is for 25/44 cards to give you the win, which is 56.8%. One example is:

You: 9h, 10h
Opponent: 2c,2d

Board: 7h, 8h, 4s, 4c

You can now win with any 6 (4), 7 (3), 8 (3), 9 (3), 10 (3), J (4), or any other heart not mentioned apart from the 2h or 4h (5)
ii) There are at least two ways of making the probability of not losing 100%. My way was:

You: 5h, 6h
Opponent: 2d,7s

Board: 3h,4h,3c,4c

You win with any heart for a flush (9), any 2 or 7 for a straight (4), or any 5 or 6 for better two pairs (6) = 19/44 = 43%
You split the pot with any 3 or 4 for a full house on the board (4), or anything over 7 will have you both playing the board (21) = 25/44 = 57%

Mr Hulme also managed to get 100% with 2h 2s vs 2d 3d on a board of Ac Kc Ac Ks, but that only has a win percentage of 2.27% (the one remaining 2 in the pack), i.e. I like mine better!

Congrats to Matt M for part i) and Andrew H for part ii)

[Matt Morrison]

Thanks for that Bob. I did mean to ask for your other solution, so as to show I did still care despite moving on to the snooker!
Thanks for giving the solution then - it's genius, yet also so annoying that none of us could get near it, doesn't seem at all hard now that you showed us.
Great stuff.

[Matt Bayfield]

Thanks to Bob and Jim for their poker and snooker puzzles - exactly the sort of thing I enjoy solving (and it looks like others do, too). If anyone has any more, please share (though if you could wait until the next puzzle has been solved, that would be nice).

ANSWER TO PUZZLE SIX

Solution
========

The key here is to work out how many points each club must finish on. As the total number of letters in all 20 clubs' names is 231, then the total points scored by all clubs over the season must be a multiple of 231.

There are 380 games in a 20-club Premier League season (each club plays each of the other 19 clubs twice). As 3 points are awarded for a game which is won, and 2 (one to each club) for a draw, the maximum total number of points that can be scored in a season is 380 x 3 = 1140, which occurs when no games are drawn. The minimum number of 380 x 2 = 760 points occurs when all games are drawn.

The only multiple of 231 which falls between 760 is 1140, is 924, which is 4 x 231. Hence each club must finish on a points total equal to 4 times the number of letters in its name.

Furthermore, we can also work out how many draws there are in a season where 924 points are scored in total. As each drawn game reduces by 1 the number of points from the maximum of 1140, there must be 1140 - 924 = 216 drawn games in the season. Subtracting this from 380 tells us there are 164 won games.

Hence, if we add up the Win column in the league table, it will come to 164
If we add up the Loss column in the league table, it will also come to 164.
As a draw is recorded against both clubs, if we add up the Drawn column in the league table, it will come to 216 x 2 = 432.

To produce a possible league table, we can now divide the 432 drawn points amongst the 20 clubs. If we do this approximately evenly, then most clubs will have about 21 draws. Starting with 21 draws against each club's name, adjust the number of draws upwards so that the number of points total minus the number of points from draws is divisible by 3 for each club. Once this is done, you end up with 435 draws, 3 above the desired total, so you need to remove 3 draws from any club you choose. I have chosen to remove 3 draws from champions West Brom. Now calculate the number of wins (difference between points total and number of draws, then divided by 3) and number of losses (38 minus number of wins and draws) and you will end up with a table which might look like the one below:

Letters W D L Pts
WestBromwichAlbion 18 18 18 2 72
ManchesterUnited 16 14 22 2 64
TottenhamHotspur 16 14 22 2 64
BlackburnRovers 15 13 21 4 60
BoltonWanderers 15 13 21 4 60
NewcastleUnited 15 13 21 4 60
ManchesterCity 14 11 23 4 56
Middlesbrough 13 10 22 6 52
WestHamUnited 13 10 22 6 52
WiganAthletic 13 10 22 6 52
AstonVilla 10 6 22 10 40
Portsmouth 10 6 22 10 40
Sunderland 10 6 22 10 40
Liverpool 9 5 21 12 36
StokeCity 9 5 21 12 36
HullCity 8 3 23 12 32
Arsenal 7 2 22 14 28
Chelsea 7 2 22 14 28
Everton 7 2 22 14 28
Fulham 6 1 21 16 24


As for part (b) of the question, if 5 clubs manage to complete the season without drawing, then the 432 drawn points must be divided amongst the remaining 15 clubs. As 432/15 = 28.8, some clubs must draw at least 29 times. This is impossible since they only have 14 other clubs to draw against, both home and away, a total of 28 games.

Repeating this exercise for 4 clubs completing the season without drawing, we arrive at 432/16 = 27, which means that no club needs to draw more than 29 times (even bearing in mind that Fulham can draw no more than 24 times as they finish on only 24 points). As each club now has 15 other clubs it can draw against (home and away), I’m fairly sure that this is mathematically possible. A possible league table would be:

Letters W D L Pts
WestBromwichAlbion 18 24 0 14 72
ManchesterUnited 16 13 25 0 64
TottenhamHotspur 16 13 25 0 64
BlackburnRovers 15 20 0 18 60
BoltonWanderers 15 20 0 18 60
NewcastleUnited 15 20 0 18 60
ManchesterCity 14 10 26 2 56
Middlesbrough 13 8 28 2 52
WestHamUnited 13 8 28 2 52
WiganAthletic 13 8 28 2 52
AstonVilla 10 4 28 6 40
Portsmouth 10 4 28 6 40
Sunderland 10 4 28 6 40
Liverpool 9 3 27 8 36
StokeCity 9 3 27 8 36
HullCity 8 2 26 10 32
Arsenal 7 0 28 10 28
Chelsea 7 0 28 10 28
Everton 7 0 28 10 28
Fulham 6 0 24 14 24

So the answer is 4 clubs.

(Note: this is the only Premier League table in which West Brom will ever be at the top.)

Answer: (a) see league table above; (b) 4 clubs (first correct answer: Bob)


Now here's the most difficult puzzle I've got. I don't know the answer for certain, but I will give credit to anybody who can match, or beat, my solution.

PUZZLE SEVEN: SAILING (FINN CLASS)

Today’s teaser is on sailing and is based on the Finn class competition at the 2008 Olympic Games (although I’ve changed the number of sailors from 26 to 30, to make the maths easier). In case you’re interested, the Finn is a single-handed dinghy, and Ben Ainslie won gold for Team GB in this event in Beijing. However, if you’re not interested, don’t worry - you don’t need to know anything about sailing to solve this maths puzzle.


How points scoring works in the Finn class sailing

In the Finn class competition at the Olympics, there are 30 sailors. There are 11 races, comprising 10 qualifying races plus one “medal race”. For the 10 qualifying races, points are scored based on the finishing position in each race (1 for first place, 2 for second, down to 30 points for thirtieth place). At the end of the 10 races, each sailor drops his worst (HIGHEST) score, and his best 9 finishing positions are totalled to give his “qualifying score”. At this point, the top 10 sailors (i.e. the 10 sailors with the LOWEST qualifying scores) go forward to the “medal race”. The remaining sailors finish in 11th to 30th places.

For the 10 sailors competing in the medal race, points are awarded in the same way, but count double, so they earn 2 points for first place, 4 points for second, down to 20 points for tenth place. Points earned in the medal race are added to the points earned in the qualifying races, to give a final total.

Of the 10 sailors in the medal race, the sailor with the lowest final total score (i.e. qualifying score plus medal race points) wins the gold medal.

Puzzle

What is the maximum possible final score with which a sailor can win the gold medal?

Notes

(a) in the event of a tie on points (for 10th place) after the 10 qualifying races, the tie is broken by number of first places, number of second places, number of third places, and so on.

(b) in the event of a tie on points after the medal race, the winner is the sailor who finished in the best position in the medal race.

(c) You cannot drop your score in the medal race, even if it is a worse score than all of your qualifying races.

(d) For the purposes of the maths puzzle, it is assumed that no sailor starts on the wrong side of the start line, fails to finish, or is otherwise disqualified in any of the races. (In reality, a disqualification would lead to a score of 31 points for that race, i.e. one greater than the total number of sailors in the competition.)

(e) You cannot win the gold medal, or indeed finish higher than 11th, if you do not qualify for the medal race (regardless of your points total at the end of the qualifying races). The 10 sailors in the medal race take places 1 to 10.


Good luck!

[Gavin Chipper]

Just looked at this and got a possible initial answer of 470.

Edit - Ballsed it up. Was thinking 31 races.

Edit 2 - So that would make it 160

[Matt Bayfield]

Gavin: not correct, but a good attempt, which you can build on. Hint: The number you've given there (under your "Edit 2") would be the answer if all 10 of the scores from the 10 qualifying races counted towards a sailor's final score. However, if you read the problem again, this isn't the case. It looks like you've forgotten, or not noticed, that only 9 of a sailor's first 10 race scores count towards his final score, as he automatically drops the worst (i.e. HIGHEST) score from those 10 qualifying races.

Btw, I liked that your initial answer of 470 was, like Finn, the name of another sailing class raced at the Olympics. Although I suspect that may have been a coincidence...

[Gavin Chipper]

Thanks for pointing out my omission. My latest stab at the answer is 137.

[Gavin Chipper]

I will give my reasoning for my above answer.

We want the scores to be as high as possible so for people's worst scores that get dropped, we want them to be as good as possible. But obviously in every race someone has to be 30th, 29th etc. So the best thing would be to get as few people as possible to take each "hit". So we might have one person coming 30th in every race. However, that would ruin our other goal of trying to get everyone to have near average scores. So two people could have five 30ths and and finish at or near the top in the others to get the desired score. We would have pairs of people with all the positions from 16th-30th for their worst score.

If all scores counted, the average position would be 15.5, the average total score would be 155 and the total score of everyone would be 4650. But we're now subtracting from 4650 the integers from 16-30 twice, which hopefully gives us 3960 and an average of 132.

It seems fairly logical to me that after the ten initial races we want all the top ten separated by two points each and then in the medal race they finish in reverse order to finish level on points. And to maximise their score we want to reduce the scores of the other 20, so 10-30 would ideally be level on points, with the last player qualifying for the medal race equal on points with everyone behind him.

117 + 119 + 121 + 123 + 125 + 127 + 129 + 131 + 133 + 135 * 21 gives a total of 3960. So this part of the maths works. Then Those in the medal race reverse the order from qualifying and all end up on 137.

I haven't worked out if these results actually are possible as yet. The totals add up, but it doesn't necessarily follow from that that you can come up with positions in the races to make this work. But it's a start.

Edit - I think it might not be possible to get the individuals to finish in the right places for this though.

[Matt Bayfield]

Gavin: very nice indeed. Your answer of 137 matches mine, and your logic is also the same as mine. And since you were wondering (in your "Edit" above), I can confirm that it is possible to allocate finishing positions to each sailor for each race, such that our answer is achieved. I'll post a possible combination of results when I write up the solution fully.

Just one further thing to consider though. If we tried to take the score one point higher to 138, that would require 30 fewer points to be dropped across the 30 sailors, i.e. on average, each sailor's worst score, would have to one position better (i.e. a smaller number). And although that looks impossible to achieve, I can't definitively prove it.

So that's what I mean about not being 100% certain about the correct answer for this one. But unless someone comes along with a list of results that beats yours (and mine), I think we can give you the credit for solving this puzzle.

Incidentally, did you find this one particularly challenging? When I tried it with my sample of quizzers last year (all degree-qualified in a science subject, though not mathematicians), I didn't get a single correct answer (although some people got close). So I assumed it was a fairly difficult puzzle for most people to get their heads around.

[Bob De Caux]

Here's a solution for 137. I used the same logic as you two guys by the looks of it. I may try to prove it's optimum if I get chance later, which is unlikely. Sorry about the large picture, I've no idea how to make it smaller!



Two notes. I added 1000 points to your total if you didn't qualify for the final. I also couldn't decide who should get tenth place as Sailors 1 and 30 are dead level! It's a very nice puzzle Matt. I didn't think proving a theoretical answer of 137 was too hard, but I had a lot more difficuly getting the boats in order to make it work!

[Gavin Chipper]

Just one further thing to consider though. If we tried to take the score one point higher to 138, that would require 30 fewer points to be dropped across the 30 sailors, i.e. on average, each sailor's worst score, would have to one position better (i.e. a smaller number). And although that looks impossible to achieve, I can't definitively prove it.

So that's what I mean about not being 100% certain about the correct answer for this one. But unless someone comes along with a list of results that beats yours (and mine), I think we can give you the credit for solving this puzzle.

I think it's probably impossible to increase the score of the winner. If you increase the average of the 30 in the 10 races to more than 132, I think you could only do that by having individual sailors with scores higher than 135 so it would still have the effect of dragging down the scores of the others. I might think about this though. Also the system of having the top 10 separated by 2 seems right to me and I don't think another system would be better.

Incidentally, did you find this one particularly challenging? When I tried it with my sample of quizzers last year (all degree-qualified in a science subject, though not mathematicians), I didn't get a single correct answer (although some people got close). So I assumed it was a fairly difficult puzzle for most people to get their heads around.

I quite enjoyed this puzzle. Because I didn't go as far as proving it but just went with what seemed right, I didn't find it that challenging to be honest. My answer was what I found quite intuitive and didn't take that long.

[Dinos Sfyris]

Here's another football puzzle, this time to do with half-time scores:

PUZZLE D2: Football
From the final score of a football match you can determine all the possible half-time scores ie If a game finishes 1-1 there are 4 possible half-time scores: 0-0, 1-0, 0-1 and 1-1.

I've just missed the latest Sheffield United vs Wednesday match. I'd placed a bet on what the full-time score would be so I asked my friend (who likes maths) what the score was.

He unhelpfully replies: If United had scored 3 more goals and Wednesday had scored just one more, there would be 3 times as many possible half time scores.

Identify all the possible full time scores.

[Bob De Caux]

Identify all the possible full time scores.

I think there are 3 scores that work.... they are (United first) 5-0, 2-1, 1-4

Reasoning:
For each score, the half time score possibilities (United first) are (U+1)*(W+1). Take the score 2-1 say. There are three half time options for U (0,1 or 2), two for W, so 3*2 half time possibilities.
To satisfy your friend's criteria mathematically: (U+4)(W+2) = 3*(U+1)(W+1) (3 times more possibilities if U is 3 greater and W is 1 greater)
Therefore:
(U+4).(W+1) = 3
(U+1) (W+2)

Which works for U=5,W=0 and U=2,W=1 and U=0,W=4

[Andrew Hulme]

This one is as much a sports question as a maths one:

What is the least amount of strokes a player can play and snooker in making a century break? (assuming no more than one ball is potted per shot)

[D Eadie]

26......?

[Ian Volante]

This one is as much a sports question as a maths one:

What is the least amount of strokes a player can play and snooker in making a century break? (assuming no more than one ball is potted per shot)

This one has to be 25.

12 reds followed by 12 blacks gives 96 points, although if with three of those shots, an extra red was potted, then the break would be 99, and the next pot, the 25th would be on the yellow to get to 101.

EDIT: I missed the latter stipulation, so the answer is therefore 27.

[Marc Meakin]

I think it is 25 strokes 10 reds 10 blacks yellow green brown blue pink= 100

[JimBentley]

I think it is 25 strokes 10 reds 10 blacks yellow green brown blue pink= 100

Haha, bastard, I was about to post that. Pretty sure it's correct.

[Marc Meakin]

I knew it straight away but was having trouble concealing the answer

[Ian Volante]

I think it is 25 strokes 10 reds 10 blacks yellow green brown blue pink= 100

Haha, bastard, I was about to post that. Pretty sure it's correct.

There are 15 reds to get rid of before you can start on the colours!

[JimBentley]

I think it is 25 strokes 10 reds 10 blacks yellow green brown blue pink= 100

Haha, bastard, I was about to post that. Pretty sure it's correct.

There are 15 reds to get rid of before you can start on the colours!

Not if your opponent's already potted five of them.

[Ian Volante]

Haha, bastard, I was about to post that. Pretty sure it's correct.

There are 15 reds to get rid of before you can start on the colours!

Not if your opponent's already potted five of them.

Damn you lateral thinkers!

[Marc Meakin]

So to clear up any loose ends player A breaks off and flukes a red he then goes on to pot a colour then a further 4 reds and colours before missing his next red.
Player B comes to the table and clears up to the pink to score 100 this takes 25 strokes.

[Andrew Hulme]

Basically im gonna level with you. That was far too easy for you chaps mainly because i messed up the question. Slightly trickier, same question, but this time you can pot two reds with ONE of the shots.

[Ian Volante]

Basically im gonna level with you. That was far too easy for you chaps mainly because i messed up the question. Slightly trickier, same question, but this time you can pot two reds with ONE of the shots.

Same answer I reckon, although now an extra way of doing it in 26!

[Marc Meakin]

Basically im gonna level with you. That was far too easy for you chaps mainly because i messed up the question. Slightly trickier, same question, but this time you can pot two reds with ONE of the shots.

The answer is 24 moves : 10 reds (9 strokes as the the first 2 reds were potted at the first shot)9 blacks(9 stokes) and all the colours (6 strokes)

[Matt Bayfield]

As an aside, you can make a break of 99 in just 22 shots, using this logic:

Player A fouls off the break, leaves a free ball.
Player B pots his nominated extra red, plus one of the 15 reds, in the same shot. Follows with black. 2 strokes, 9 pts.
For his next 14 shots, he pots 2 reds in the same stroke, followed by a black. 14 strokes, 63 further points.
He then takes the 6 colours in sequence.
Total points scored: 99

[Martin Smith]

He'd run out of reds after 8 shots playing that way.

[Kevin Davis]

Only allowed to pot two reds in a shot once? (Answer concealed in colour #E4F4FE below; same colour as background.)
24. 1+7+2+7+1+7+1+7+1+7+1+7+1+7+1+7+1+7+2+3+4+5+6+7. At no point does it say the player pots the FIRST red; he instead manages a century clearance. It's assumed that five reds had been previously potted. The player then pots the remaining ten reds in nine shots, with nine accompanying blacks for a break of 73, before downing the colours in sequence (yellow for 75, green - 78, brown - 82, blue - 87, pink - 93, black - 100.

[Bob De Caux]

He'd run out of reds after 8 shots playing that way.

He'd run out of reds after 16 shots Martin - 8 double reds (including the free ball) and 8 blacks = 72. Plus the 6 colours is 99 in 22 shots.

Basically im gonna level with you. That was far too easy for you chaps mainly because i messed up the question. Slightly trickier, same question, but this time you can pot two reds with ONE of the shots.

I can do this in 23 with a loose interpretation of the sentence above

Opponent makes a break of 6 reds and 6 blacks (48), then leaves you snookered. You continue to foul until your opponent is 93 ahead. He/she then fouls to leave you with a free ball. You pot your free ball and another red with the same shot (this is the slightly dubious bit - I'm saying this is not two reds in one shot as one of them is a colour!), then the black. You then pot your two reds in one shots and the black, before 6 more red/blacks. This puts you on 66 after 16 shots. You sink the 6 colours to tie the frame at 93-93, then polish off the respotted black with your 23rd shot to complete a break of 100.

[Marc Meakin]

FWIW a respotted black does not add to a persons break.
Not least because you have to toss for who goes first.

[Marc Meakin]

Only allowed to pot two reds in a shot once? (Answer concealed in colour #E4F4FE below; same colour as background.)
24. 1+7+2+7+1+7+1+7+1+7+1+7+1+7+1+7+1+7+2+3+4+5+6+7. At no point does it say the player pots the FIRST red; he instead manages a century clearance. It's assumed that five reds had been previously potted. The player then pots the remaining ten reds in nine shots, with nine accompanying blacks for a break of 73, before downing the colours in sequence (yellow for 75, green - 78, brown - 82, blue - 87, pink - 93, black - 100.

Nice solution(same as mine see above)

[Matt Bayfield]

ANSWER TO PUZZLE SEVEN

It is easiest to consider this problem in two stages. First - part (a) - let's ignore the requirement for each sailor's worst score to be dropped, and then - part (b) - we'll make the problem more complex by including the dropped scores...
Part (a) - no dropping of scores

To maximise the winning sailor's score, it makes sense that after the Medal race, the total points available are distributed as evenly as possible between all sailors. So for 30 sailors and 10 preliminary races, there are 10 * (30 + 29 + 28 + ... + 3 + 2 + 1) points available, i.e. 4650 points. There are then a further 2 * (10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 110 points that can be gained in the Medal Race, giving a total of 4760 points.
Now if you divide the total of 4760 points, by the 30 sailors, you get 158.67, so this would suggest the winning score would be somewhere in this region. There are then two criteria we have to satisfy when dividing these points: (1) all scores must be integers, and (2) the top ten sailors after the Preliminary races (i.e. those with lowest scores) must take the 110 points from the Medal Race, which means that one of those 10 sailors has a 20 added to their score, another has an 18, and so on.
It hopefully doesn't take too much of a leap of faith from there, to realise that for total scores after the Medal race to be as "flat" as possible across all 30 sailors, then before the Medal Race, the 30 sailors should be on the following number of points:

• Sailors in 10th through 30th places after Preliminary Races: n points (the 10th sailor being decided out of these 21 sailors by the "most wins", "most second places", etc, tie-break criterion).
  Sailor in 9th place: (n - 2) points
  Sailor in 8th place: (n - 4) points
  And you get the idea, down to Sailor in 1st place: (n - 18) points

You can hopefully see where we're going here... the top 10 sailors will go on to finish the Medal Race in the reverse positions to their positions after the Preliminary Races, thus meaning they all end level on (n + 2) points, and the winner is the sailor in 10th place after the preliminary races, as he won the Medal Race. Okay?
Anyhow, a little bit of algebra (knowing that 20 sailors finish on n points, and the top finish on (n + 2) points, and that total points are 4760), allows you to calculate n and distribute the total of 4760 points. And in this case, n = 158, so the winning sailor finishes on 160 points. I've included an example table of finishing positions below. You'll notice that the winning sailor is Sailor #10.

Code: Select all

    Sailor number ----->   #1  #2  #3  #4  #5  #6  #7  #8  #9 #10 #11 #12 #13 #14 #15 #16 #17 #18 #19 #20 #21 #22 #23 #24 #25 #26 #27 #28 #29 #30  Total
                  Race 1    1   3   2   5   4   6   7   8   9  30  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  10    465
                  Race 2   30  29  28  27  26  25  24  23  22   1  20  19  18  17  16  15  14  13  12  11  10   9   8   7   6   5   4   3   2  21    465
                  Race 3    1   3   6   2   5   4   7   8   9  30  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  10    465
                  Race 4   30  29  28  27  26  25  24  23  22   1  20  19  18  17  16  15  14  13  12  11  10   9   8   7   6   5   4   3   2  21    465
                  Race 5    5   4   9   2   1   7   6   8   3  30  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  10    465
                  Race 6   30  29  28  27  26  25  24  23  22   1  20  19  18  17  16  15  14  13  12  11  10   9   8   7   6   5   4   3   2  21    465
                  Race 7    6   5   4   1   3   2   7   8   9  30  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  10    465
                  Race 8   30  29  28  27  26  25  24  22  23   1  20  19  18  17  16  15  14  13  12  11  10   9   8   7   6   5   4   3   2  21    465
                  Race 9    6   9   8   3   5   4   1   2   7  30  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  10    465
                 Race 10    1   2   3  25  26  27  28  29  30   4  23  22  21  20  19  18  17  16  15  14  13  12  11  10   9   8   7   6   5  24    465

    Total after 10 races  140 142 144 146 148 150 152 154 156 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158 158   4650

              Medal Race   20  18  16  14  12  10   8   6   4   2                                                                                    110

    Total after 11 races  160 160 160 160 160 160 160 160 160 160                                                                                   4760

[/size]


Part (b) - no dropping of scores

This is slightly trickier. The same principles apply as for part (a) regarding sharing the points (i.e. 21 sailors on p points after the preliminary races, and the other 9 on (p - 2), (p - 4), ... , (p - 18) points. The complication introduced by dropping each sailor's worst score is that we no longer have the full 4760 points to spread around.
To maximise the winning sailor's score, we want there to be as many points as possible to be spread around, so we want to drop the lowest scores we can. Now, we can do this by allocating certain sailors to have lots of high scoring races, e.g. if two sailors each finish 30th in five of the races, then there will only be two scores of 30 dropped, rather than the 10 scores of 30 if the preliminary race positions were spread more evenly. The trick is to work out how to balance this with the requirement that all sailors spread points as evenly as possible. And the best way that I was able to work it out, is to start by assuming that Sailor 1 is a bit bipolar, and he finishes 1st in 5 preliminary races, and 30th in 5 preliminary races. Then Sailor 2 finishes 2nd in 5 races, and 29th in 5 races. Sailor 3 finishes 3rd in 5 races, then 28th in 5 races. And so on, down to Sailor 30, who finishes 1st in the 5 races that Sailor 1 finishes 30th in, and 30th in the 5 races that Sailor 1 finishes 1st in. This means that the scores dropped are 2 x 30, 2 x 29, 2 x 28, ..., 2 x 16, 2 x 15, i.e. a total of 690 points are dropped.

Now, I concede it might be possible to do better than 690 by a few points... but you would need to find a score-dropping pattern which does better by 30 points (i.e. drops 720 points), for it to change the overall answer to this puzzle. And I'm reasonably confident - albeit not 100% sure - that it can't be done. Unless someone proves otherwise.
So, if we drop 690 points from our original total of 4760 points, and solve for p in the same way we solved for n in part (a) (except this time there are 4760 - 690 = 3960 points to share), we get p = 135. Unfortunately, the score-dropping pattern above has all sailors on different scores after their worst scores are dropped, and not in the pattern we need (21 sailors on 135 points, 1 sailor on 133, 1 sailor on 131, 1 sailor on 129, ... , 1 sailor on 117). So you have to do a little bit of manual jigging about with the finsihing positions per race... but if you play around enough, eventually you can prove there's a possible combination of finishing positions. Just like this one:

Code: Select all

    Sailor number ----->   #1  #2  #3  #4  #5  #6  #7  #8  #9 #10 #11 #12 #13 #14 #15 #16 #17 #18 #19 #20 #21 #22 #23 #24 #25 #26 #27 #28 #29 #30  Total
                  Race 1    1   2   3   4   5  11   7   8   6  10   9  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30    465
                  Race 2   30  29  28  27  26  25  24  23  22  21  20  19  18  17  16  15  14  13   4   1  12   9   8   7   6   5  11   3   2  10    465
                  Race 3    1   2   3   4  11   6   7   8   9  10  13  14   5  12  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30    465
                  Race 4   30  29  28  27  26  25  24  23  22  21  20  19  18  17  16  15  14   2  10  13   9  11   8   7   6   5   4   3  12   1    465
                  Race 5    1   2   3  11   5   6   7   9   8  10  12   4  14  16  15  13  17  18  19  20  21  22  23  24  25  26  27  28  29  30    465
                  Race 6   30  29  28  27  26  25  24  23  22  21  20  19  18  17  16  15   8  10  14  13   3   7  12   9   6   5   4  11   2   1    465
                  Race 7    1  10  12   4   5   6  11   9   8   7   2   3  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30    465
                  Race 8   30  29  28  27  26  25  24  23  22  21  20  19  18   9  16  14  15  17  12   7  13  11   5   8   6  10   4   3   1   2    465
                  Race 9   11   3   2   4   5   6   7   9  16  14   1  12   8  13  10  15  17  18  19  20  21  22  23  24  25  26  27  28  29  30    465
                 Race 10   30  29  28  27  26  25  24  23  22  21  20  19  18  17  15  16  12  13   7   5  14   9  10   8  11   6   4   3   2   1    465

    Total after 10 races  165 164 163 162 161 160 159 158 157 156 137 140 143 146 149 151 148 145 142 139 156 157 158 159 160 161 162 163 164 165   4650
             Worst score   30  29  28  27  26  25  24  23  22  21  20  19  18  17  16  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30    690
Total after 9 best races  135 135 135 135 135 135 135 135 135 135 117 121 125 129 133 135 131 127 123 119 135 135 135 135 135 135 135 135 135 135   3960
              Medal Race    2                                      20  16  12   8   4       6  10  14  18                                            110

    Total after 11 races  137                                     137 137 137 137 137     137 137 137 137

[/size]

So there you have it. Sailor #1 wins, on 137 points.[/color]

Answer: 137 points (first correct answer: Gavin)


And I'm afraid that's it for the puzzles I can think of for now. If I dream up any new ones that aren't trivially easy, I'll post them here. Meanwhile, anyone who wants to share a puzzle of their own, go ahead...

[Gavin Chipper]

Matt, I was thinking about your sailing puzzle and it might be possible to increase the average score by the one needed, but I think it would have the effect of putting some of the sailors' scores out of the range needed.

Instead of simply having two sailors with five scores of 30, two with five of 29 etc., I think you can increase some of the "bad" scores for some of the sailors. You would still have two sailors with five each for positions 27 to 30 but after that, you can start giving six and later more. Someone could have six scores of 26, for example, and still get away with it as long as they do well enough on the others. So, if each number or letter represents a sailor and their worst scores we could have for example:

26. 111111/2222
25. 22/333333/44
24. 4444/555555/
23. 666666/7777
22. 77/888888/99
21. 99999/00000
20. 00/AAAAAAA/B
19. BBBBBB/CCCC
18. CCC/DDDDDDD/
17. EEEEEEEE/FF
16. FFFFFF/GGGG
15. GGGGG/HHHHH
14. HHHH/IIIIII
13. IIII/JJJJJJ
12. JJJJ/KKKKKK
11. KKKK/LLLLLL
10. LLLL

So in terms of what we are taking out of the final score, based on number of sailors with each score as their worst we have 4650 - 2*30 - 2*29 - 2*28 - 2*27 - 2*26 - 2*25 - 1*24 - 2*23 - 2*22 - 1*21 - 2*20 - 1*19 - 1*18 - 2*17 - 1*16 - 1*15 - 1*14 - 1*13 - 1*12 - 1*11 = 3993. This gives an average of 133.1 which is enough with 0.1 to spare (or 3 points of the total)! However, if we are using the same system as before, the minimum score for a sailor would have to be 118. But sailor L ends up on 95! OK, there is some room for manoeuvre as we clawed back three more points than necessary but an average score of 13 for a sailor would put them on 117 and since it needs to be 118 at the very least, every sailor would need to have a worst score of 14th or worse. I can't see that there would be a way of pushing the scores back while keeping an average of 133.

[Matt Bayfield]

Gavin: yup, you've explained that better than I could have done. I does seem unlikely that you could rearrange race positions suitably within your proposed better score-dropping pattern, but I imagine it would take a computer, operated by someone seriously mathematical, to prove it either way.

[David Williams]

What is the lowest possible total points scored in a completed frame of snooker? All balls potted, including the black.

[Clive Brooker]

What is the lowest possible total points scored in a completed frame of snooker? All balls potted, including the black.

I think it's 31.

[Marc Meakin]

It has to be42, 15 reds plus all the colours

[Marc Meakin]

What is the lowest possible total points scored in a completed frame of snooker? All balls potted, including the black.

I think it's 31.

This answer could be correct only in this scenario Player breaks off miraculously potting all 15 reds and the white giving 4 away then the remaining colours are potted (a further 27 points)

[Matt Morrison]

What is the lowest possible total points scored in a completed frame of snooker? All balls potted, including the black.

I think it's 31.

This is the answer I'd go for, even though, if you're thinking about it the way I'm thinking about it, its statistical possibility belies its complete and utter physical unlikelihood.
For this to happen, all 15 reds would have to be potted at the same time as the white (or indeed a low value colour) gets potted - so that the 15 reds stay down, and only 4 points (for the foul) have been awarded, to which you can add the 27 points from the colours. Same as what you were thinking Clive?

[Clive Brooker]

Yes, that was my method. Clearly fantasy of course.

This reminds me of a somewhat less fanciful scenario which I'll post in the questions you've always wanted answered thread.

[David Williams]

Yes, that was my method. Clearly fantasy of course.

But correct.

[David Roe]

If we really want to nitpick, the question didn't say the final black just had to be potted.
Just that all balls had to be potted, including the black.

So, first shot, pot all 15 reds and the black, 7 points away.
Then, opponent pots 5 colours to the pink and misses the black. The game's over.

Total score 7 + 20 = 27.

[Gavin Chipper]

If we really want to nitpick, the question didn't say the final black just had to be potted.
Just that all balls had to be potted, including the black.

So, first shot, pot all 15 reds and the black, 7 points away.
Then, opponent pots 5 colours to the pink and misses the black. The game's over.

Total score 7 + 20 = 27.

Presumably a frame can be conceded at any point. Player 1 pots every ball off the break, finds himself 7-0 down and concedes.

[David Roe]

[post deleted, I misread your post above]

[Gavin Chipper]

If we really want to nitpick, the question didn't say the final black just had to be potted.
Just that all balls had to be potted, including the black.

So, first shot, pot all 15 reds and the black, 7 points away.
Then, opponent pots 5 colours to the pink and misses the black. The game's over.

Total score 7 + 20 = 27.

Presumably a frame can be conceded at any point. Player 1 pots every ball off the break, finds himself 7-0 down and concedes.

Also, in a free ball situation, we could probably both knock one point off in each of our scenarios. (6-0 down after potting all the balls since the black can be a red).

[Howard Somerset]

Here's a puzzle which might fit this thread. Not mine, I admit; it's the Teaser in today's Sunday Times. Quite a neat one, and I think I've got the solution.

PM, or answer in white, if you wish.

Our football league has six teams, A, B, C, D, E and F, who play each other once, earning 3 points for a win, 1 for a draw. Of last season's 20 goals, the best were in Primadona's hat-trick in C's derby against D. At one stage, when C and D had played all their games, but A and B each still had two in hand, the league order was A, B, C, D, E, F (goal differences separate teams tying on points). But A's morale was shattered because, in the end, C won the league (the number of goals scored also being needed for the final decision.).
What were the scores in C's matches? (C v A, C v B, C v D, C v E, C v F)?

Although the Sunday Times only asks for C's five games, I'll extend it to asking for the full set of scores.

[Bob De Caux]

Here are the first 11 games and corresponding table:
0 A v B 0
3 C v D 3
0 E v F 0
1 A v C 0
1 D v F 0
2 A v D 0
1 C v E 0
1 B v D 0
2 C v F 0
1 B v C 0
1 D v E 0

P W D L F A Po GD
A 3 2 1 0 3 0 7 3
B 3 2 1 0 2 0 7 2
C 5 2 1 2 6 5 7 1
D 5 2 1 2 5 6 7 -1
E 3 0 1 2 0 2 1 -2
F 3 0 1 2 0 3 1 -3


and here are the last four games and final table
0 A v E 1
0 B v F 1
0 A v F 1
0 B v E 1

P W D L F A Po GD
C 5 2 1 2 6 5 7 1
A 5 2 1 2 3 2 7 1
E 5 2 1 2 2 2 7 0
B 5 2 1 2 2 2 7 0
D 5 2 1 2 5 6 7 -1
F 5 2 1 2 2 3 7 -1

E finish above B because of the result between the two

[Howard Somerset]

Your analysis agrees entirely with mine, Bob.

(with the sole exception of the last line, as we were not given the fourth criterion for separating teams)

[Bob De Caux]

Your analysis agrees entirely with mine, Bob.

(with the sole exception of the last line, as we were not given the fourth criterion for separating teams)

You're right Howard - just wanted to use something other than alphabetical order!

[Dinos Sfyris]

Bump! I've got an extension for my original golf puzzle:

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?

Extension: Now try the same puzzle where the difference in scores on subsequent holes are all different positive square numbers and see what's the best 9-hole score you can get.

NB zero does not count as it is not positive.

[Marc Meakin]

Bump! I've got an extension for my original golf puzzle:

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?

Extension: Now try the same puzzle where the difference in scores on subsequent holes are all different positive square numbers and see what's the best 9-hole score you can get.

NB zero does not count as it is not positive.

What about fractions?

[Dinos Sfyris]

Fractions? I don't follow.

[Matt Bayfield]

As a first guess:

Scores on holes 1 through 9 are: 5 1 2 66 17 1 37 12 3, total for the round = 144.

It may be possible to do better.

[Hugh Binnie]

What about fractions?

Square numbers are squares of integers.

[Dinos Sfyris]

As a first guess:

Scores on holes 1 through 9 are: 5 1 2 66 17 1 37 12 3, total for the round = 144.

It may be possible to do better.

I'm pretty sure that's the best possible solution. Nice one, Matt. It's a little bit counter-intuitive because to minimise the total score it seems obvious to go: 1, 65, 16... which always yields a higher total score.

[Matt Bayfield]

Thank you to Dinos for his latest golf puzzle - finally I got one right!

And for those still hungry for hot sports/math action, here's a little teaser which I dreamed up over the weekend. I was trying to come up with a more general solution, for more rounds in the knockout competition, and more teams which could play derbies... but when trying to solve this, I found the maths so difficult, I had to simplify the puzzle to the version I’ve set below. You’ll see what I mean...

PUZZLE EIGHT: FOOTBALL (although the maths could be for almost any sport, quite frankly)

Consider the (English) FA Cup for the season 2012/3. Owing to some strange squad rotation policies by Premier League clubs, the following eight teams have progressed to the quarter-finals:

• Arsenal
  Blackburn Rovers
  Chelsea
  Doncaster Rovers
  Everton
  Fulham
  Grimsby Town
  Hull City

Excitingly, for the first time in Cup history, these last eight teams are of exactly equal standard, and thus in any possible match-up, either team is equally as likely to win the tie. The draw for the quarter-finals has not yet been made.

What is the probability of a London derby (i.e. match between two London teams) taking place in the remainder of the 2012/3 FA Cup? Express your answer as a fraction.

Notes
1) For the severely geographically-challenged among you, the London teams in the above list are Arsenal, Chelsea and Fulham.
2) The pairings for each round of the FA Cup are drawn randomly, usually by two ex-footballers pulling balls out of a covered bowl in one of the suites at Wembley Stadium.

[Bob De Caux]

What is the probability of a London derby (i.e. match between two London teams) taking place in the remainder of the 2012/3 FA Cup? Express your answer as a fraction.

I make it 17/28. Working available on request!

[Howard Somerset]

Agree with Bob.

[Matt Bayfield]

And I agree with Bob and Howard. Well done, chaps.

ANSWER TO PUZZLE EIGHT

I solved this puzzle by considering a knockout-style draw sheet from the quarter-finals onwards. In the quarter-finals there are 8 spaces, which I numbered 1 through 8, such that the team in space 1 plays the team in space 2, the team in space 3 plays the team in space 4, the team in space 5 plays the team in space 6, and the team in space 7 plays the team in space 8. In the semi-finals, the winner of the teams in spaces 1 & 2 plays the winner of the teams in spaces 3 & 4, and the winner of the teams in spaces 5 & 6 plays the winner of the teams in spaces 7 & 8.

Now, there are 8! (this notation means “8 factorial”, i.e. 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) ways to allocate the 8 teams into the 8 spaces. However, noting that any drawsheet can be re-drawn in an equivalent form such that any team of your choice appears in space 1, you can simplify matters by fixing one of the London teams in space 1. I have chosen to fix Arsenal in space 1.

With Arsenal fixed in space 1, there are 7! permutations of (ways of allocating) the other 7 teams to the other 7 spaces.

In 6! of these permutations, Chelsea are in space 2 and play Arsenal in the quarter-finals.
In a further 6! of these permutations, Fulham are in space 2 and play Arsenal in the quarter-finals.
In a further 6 x 5! = 6! of these permutations, neither Chelsea nor Fulham is in space 2, but Chelsea and Fulham play each other in the quarter-finals (i.e. Chelsea space 3, Fulham space 4, there are 5! permutations of the other 5 teams; Chelsea space 4, Fulham space 3, there are 5! permutations of the other 5 teams; Chelsea space 5, Fulham space 6, there are 5! permutations of the other 5 teams; Chelsea space 6, Fulham space 5, there are 5! permutations of the other 5 teams; Chelsea space 7, Fulham space 8, there are 5! permutations of the other 5 teams; Chelsea space 8, Fulham space 7, there are 5! permutations of the other 5 teams).

Thus, there are 3 x 6! permutations where there is a London derby in the quarter-finals. Or putting it another way, the probability of a quarter-final London derby is 3 x 6! / 7! = 3/7.


Now, if there is NOT a London derby in the first round, then the three London teams must be in different quarter-finals. The probability of this is 1 – 3/7 = 4/7. What is perhaps not immediately obvious is that in all these permutations, the RELATIVE positions of the London teams are EXACTLY EQUIVALENT, i.e. two teams will be in one half of the draw and could meet each other in a semi-final, and one team will be in the other half and can’t meet the other two London teams until the final.

If you can’t see this straight away, then consider Arsenal fixed in space 1 again. These are the possible permutations:

Chelsea in space 3 or 4, Fulham in space 5, 6, 7 or 8: total permutations = 2 x 4 x 5!
Fulham in space 3 or 4, Chelsea in space 5, 6, 7 or 8: total permutations = 2 x 4 x 5!
Chelsea in space 5 or 6, Fulham in space 7 or 8: total permutations = 2 x 2 x 5!
Fulham in space 5 or 6, Chelsea in space 7 or 8: total permutations = 2 x 2 x 5!
Total permutations = 24 x 5!
Thus probability of all 3 London teams being in different quarter-finals = 24 x 5! / 7! = 4/7.

Right. Now for some easy maths. When two London teams are drawn together in the quarter-finals (3/7 of the time), the probability of at least one London derby in the remainder of the competition is clearly 1.

When two London teams are not drawn together in the quarter-finals (4/7 of the time), the probability of a London derby is calculated as follows:

Assume that Arsenal are in the top half of the draw, and Chelsea and Fulham are in the bottom half. Remember, all other permutations are equivalent so the probability of a derby will be the same.)

p(Chelsea play Fulham) = p(Chelsea win qtr-final) x p(Fulham win qtr-final) = 1/2 x 1/2 = 1/4.
p(Chelsea play Arsenal, but do not play Fulham) = p(Chelsea get to final) x p(Arsenal get to final) x p(Fulham lose in qtr-final) = (1/2 x 1/2) x (1/2 x 1/2) x 1/2 = 1/32.
p(Fulham play Arsenal, but do not play Chelsea) = p(Fulham get to final) x p(Arsenal get to final) x p(Chelsea lose in qtr-final) = (1/2 x 1/2) x (1/2 x 1/2) x 1/2 = 1/32.
Adding these three terms, p(London derby) = 1/4 + 1/32 + 1/32 = 5/16.


And we’re nearly there. The overall probability of a London derby must then be p(London derby given that London teams are paired in qtr-finals) + p(London derby given that no London teams are paired in qtr-finals). And this is:

(3/7 x 1) + (4/7 x 5/16) = 17/28.

From A-Level Maths, I vaguely remember there are fancy symbols in probability with bars in, or something like that, to denote conditional probability, which I think is what we’ve calculated here. So I’m sure someone could have written all the above in about a tenth of the characters. And if anyone fancies trying to work out a more general solution, for the last 2-to-the-power-m teams in a draw, and n London teams, etc, then I’d love to hear, but since you’re clearly better at maths than I am, I won’t be checking your work!


Answer: 17/28 (first correct answer: Bob)