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bobscat Posted on 13/06/2008 15:39 | |

New fixture s fixed !!!
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Just watch when they come out , big four play each other on the same weekend near the end of the season , Another grand slam weekend or what ever .........all fixed for T.V ect | |

fatsuma Posted on 13/06/2008 15:44 | |

New fixture s fixed !!!
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Boro, Sunlan, Noocassle and presumably... Hull! | |

Homer_Southgate Posted on 13/06/2008 15:45 | |

New fixture s fixed !!!
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Where can i look at them? | |

Get_your_rat_out Posted on 13/06/2008 15:47 | |

New fixture s fixed !!!
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you must have a time machine they dont come out till next week | |

Boromart Posted on 13/06/2008 15:49 | |

New fixture s fixed !!!
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"big four play each other on the same weekend near the end of the season" -- easter weekend, I can guarentee it. The chances of the big 4 playing each other at the same time three years running 33.7 million to 1. I'll have a 10 on that! | |

atomicloonybin Posted on 13/06/2008 15:53 | |

New fixture s fixed !!!
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Of course it's fixed. If it was truly random, the odds say it should happen once every three seasons that the all the 'big four' play each other on a weekend. It happened twice last year, twice the year before. But why shouldn't the FA manipulate the fixtures? Each team has to play each other home and away anyway, so what difference does it make if Arsenal, Chelsea, Man U and Liverpool play each other one weekend? I once saw a documentary about the NFL. They had a few blokes armed with a bunch of discs each with a team on it putting them onto a chart, and all they did was manipulate the fixtures - 'that'll be a good game on week one' etc. Nothing random about it at all. As long as you're basically home one week, away the next, who cares? | |

mickbrown Posted on 13/06/2008 15:53 | |

New fixture s fixed !!!
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The Prem don't pretend that they are randomly drawn out of a hat do they? | |

the_throcking_man Posted on 13/06/2008 15:54 | |

New fixture s fixed !!!
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" It happened twice last year," That's because the return fixture is on the same date for every team | |

Tom_Fun Posted on 13/06/2008 16:00 | |

New fixture s fixed !!!
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By creating these 'grand slam weekends' the Premier League are advocating a 'big four' and promoting the idea that it's impossible for any other club to break in the top 4 (even though it's not). It stinks, no club should be valued more than others. | |

BenJammin Posted on 13/06/2008 16:01 | |

New fixture s fixed !!!
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Its not random, they don't claim it to be random so what is there to complain about? You play each team home and away over the season. | |

mickbrown Posted on 13/06/2008 16:02 | |

New fixture s fixed !!!
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It's bobbins, no doubt. The logical extension of this is Boro v Hull on Wednesday Bolton v Stoke on Thursday Dogpooh Midweek | |

Capybara Posted on 13/06/2008 16:05 | |

New fixture s fixed !!!
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alb is right that it should happen every three seasons or so, but if it does happen it will happen twice in the season, not once, because the way the fixtures are configured is that one set is played one weekend then the reverse of that is played on another. The fixtures are sort of random but with non-random elements built in - to avoid geographically close teams like Fulham and Chelsea being at home on the same day for instance - so it's easy enough to fix. But that probably means that some unpalatable options get spat out of the other end - look at our away fixtures at Portsmouth over the last few seasons, for instance. But that's not seen as mattering. | |

hewielewie Posted on 13/06/2008 16:09 | |

New fixture s fixed !!!
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It can't be random, you cannot have Man Utd and Man City playing at home on the same weekend because of policing issues, you mostly play someone from your area over Xmas, in the Championship Donny Rovers can't play at home when there is racing on etc etc. | |

Andy Posted on 13/06/2008 16:09 | |

New fixture s fixed !!!
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Boromart where did you find those odds? i wouldnt mind putting a quid on that! | |

Buddy2 Posted on 13/06/2008 16:13 | |

New fixture s fixed !!!
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Northampton can't play at home until October. Or has that changed since Allan Lamb retired? | |

Boromart Posted on 13/06/2008 16:23 | |

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Mick it used to be random, pre-sky...they just didn't inform any of us when it stopped being random. | |

Boromart Posted on 13/06/2008 16:32 | |

New fixture s fixed !!!
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it isn't a 1 in 3 chance of playing. There is a 1 in 19 chance of club a meeting club b. There is a 1 in 17 chance of club c meeting club d on the same weekend. 19 * 17 = 323 to 1. The following season would be an unconnected event, thus it's 323 (year 1) * 323 (323 year 2) = 1 in 104K For a third season running it is 104K * 323 = 33.7mill. It's something like that. It certainly isn't a 1 in 3 chance for year 1, and a 1 in 3 chance for year 2 giving a 1 in 9 chance of grand slam weekends on censecutive years. It does matter from a financial stand point. Certain weekends give a better financial return for the home team. Mainly holiday weekends. It is best to get a home fixture on boxing day for example because that is when the biggest crowds occur. But of course sky would like us all sat watching Manu/arse/chelsea/liverpoo live from OT/ES/SB/A on boxing day. It's more money for the big teams less for the rest AGAIN. | |

Boromart Posted on 13/06/2008 16:38 | |

New fixture s fixed !!!
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actually thinking about it the calculation will be a bit more complex as if team a play team b then when it comes to chosing team c and d many of there fixtures will have been picked, you only play everyone twice and that would affect the odds of the second selection. I guess it is closer to a 1 in 17 chance that they play each other at the same time. Still 17 * 17 = 1 in 289 chance of consecutive years or 84K for 3 years running. It's still daft odds. | |

Buddy2 Posted on 13/06/2008 16:42 | |

New fixture s fixed !!!
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There's a 3 in 19 chance of Club A meeting Club B, C or D, and a 1 in 17 chance of the other two meeting at the same time. So it's 323 to 3 or 107.66666 to 1. The other question is whether the following season is ANOTHER 107-1 and I don't think it is. That is ALWAYS the chance of four teams being "drawn" against each other on the same weekend, even if you don't have a summer break, so it should happen on average once every 107 fixture weekends or a bit less than three seasons. I think. | |

SmogOnTheTyne Posted on 13/06/2008 16:44 | |

New fixture s fixed !!!
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I think I heard, a few seasons ago, that the managers were allowed a say on a certain amount of games where they could say "change this fixture" "I dont want to play them on that day". So there is flexibility in the fixture list and it isnt set in stone. And Sky have money on the mind when it comes to grand slam weekends. Soon though they'll realise that those weekends are boring. The 2 that happened this season were total let downs. 4 boring games where there was too much too lose and were defended out. | |

Buddy2 Posted on 13/06/2008 16:44 | |

New fixture s fixed !!!
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Incidentally, I heard something on the radio last week which ALMOST convinced me that the people who say you should always switch your box are right. But not quite. | |

Capybara Posted on 13/06/2008 16:46 | |

New fixture s fixed !!!
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Firstly, fixtures have never been fully random. As long as I have watched the game teams have been able to specify teams who they didn't want to play at home on the same day as. And it used to be that you would always play a team close to you home and away over Christmas. And your statistics are seriously flawed there. You have taken the chances of it happening on any one specified weekend (the first weekend of the season, say), and not on one weekend (or two, actually) in a season of 38 weekends (or how ever many there are). And you have not taken account of the various possibilities for the other fixtures being used up as the season profresses. So, for example, if Liverpool played West Brom on the first day of the season, there would only be 18 options for teams they could play on the second Saturday of the season etc. All this brings the odds of it happening in any one season to about one in three when you do the maths properly, which I haven't. | |

Capybara Posted on 13/06/2008 16:48 | |

New fixture s fixed !!!
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Buddy's explanation is probably clearer. And better. | |

Boromart Posted on 13/06/2008 16:52 | |

New fixture s fixed !!!
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"There's a 3 in 19 chance of Club A meeting Club B, C or D, and a 1 in 17 chance of the other two meeting at the same time. So it's 323 to 3 or 107.66666 to 1." -- it is if your looking at week 1 fixtures ONLY....but when you come to week 2 then those odds reduce to 2 in 18 and 1 in 16.....because the same fixture cannot happen twice in a season. so week 3 then its 2 in 17 and 1 in 16.... week 4 then its 2 in 16 and 1 15 etc. Given that team a will meet team b at some point, then there is a 1 in 17 chance that team c and d will meet at the same time. "The other question is whether the following season is ANOTHER 107-1 and I don't think it is." -- it is an unrelated event with exactly the same odds. If I toss a coin it is 50-50 if it lands on heads. if I toss it again it is still 50-50 to be heads. However the odds of getting 2 consecutive heads is 1 in 4. Not 1 in 2. it could be H-H H-T T-H T-T 4 options and each with the same probability. You multiply the odds of seasons 1 by the odds of season 2. | |

goalscrounger Posted on 13/06/2008 16:59 | |

New fixture s fixed !!!
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You should always swap. | |

Boromart Posted on 13/06/2008 17:01 | |

New fixture s fixed !!!
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it might be clearer capy, but it's far more flawed than mine You beat me to it with the weekend 1 calculation. | |

Jonicama Posted on 13/06/2008 17:04 | |

New fixture s fixed !!!
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There is some seriously bad maths on this thread. First of all because of the pairing of games if it is going to happen twice a season or not at all. Then going with Buddys logic it can happen on a particualr weekend with a probability of around 1 in 100. There are 19 sets of two games. So actually down to about 1 in 5 for it to happen twice a season. So it should happen that we have two Grand Slam weekends in a season once every five years. If you want to takes the maths further then what is the chances it happens three seasons in a row - 1 in 125. Long long odds but not the millions to one some people are talking about. | |

borobadge Posted on 13/06/2008 17:06 | |

New fixtures fixed....!
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the fixture list is controlled. and works to a computer programme that sets in certain pre-requisites..... any fine tuning of the list can be adjusted once they have been viewd by amongst others, the t.v. companies, the police, the f.a., england managers, cup replays, the weather, bank holidays and fixture pile ups between teams in the same locality/city... there was a new one added last season, care to guess what it was ?..it helped us (the boro) | |

Boromart Posted on 13/06/2008 17:15 | |

New fixtures fixed....!
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Jonicama you seem to have joined the bad maths brigade, 1 in 5? Show me how you worked that one out . | |

AnitaWheater Posted on 13/06/2008 17:19 | |

New fixtures fixed....!
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Buddy's been listening to Melvyn! | |

Capybara Posted on 13/06/2008 17:28 | |

New fixtures fixed....!
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The Manchester City situation presumably, Mr badge? | |

borobadge Posted on 13/06/2008 17:41 | |

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exactly mr bara.. | |

Jonicama Posted on 13/06/2008 17:59 | |

New fixtures fixed....!
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Boromart Is it a whoosh for the 5 versus 5.6 - you need to keep it simple on this board | |

goalscrounger Posted on 13/06/2008 18:00 | |

New fixtures fixed....!
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Hmmm....a one in 5 chance that a particular two fixtures will be on at the same time?? Seems ridiculously small doesn't it? Well, lets look at this a different way and try and work out the probability of any two teams being drawn to play each other on a particular day. 20 teams: a, b, c, d, e, f, ........... Picking those 20 teams out for the first round of fixtures randomly will produce one of 20! (20 x 19 x 18 x 17 x .................. x 2 x 1) permutations. 20! = roughly 2,432,902,008,000,000,000 So, there are 2,432,902,008,000,000,000 unique different possibilities for the first set of fixtures. For teams a and b to be drawn together, we must imagine they are one "unit" called ab, so we now have 19 possible draws of ab, c, d, e, f, g, ........... This has 19! permutations which is 121,645,100,400,000,000 sets of fixtures However, it doesn't matter if we have ab or ba, so we have to multiply our answer by 2. This gives 243,290,200,800,000,000 possible random permutations of fixtures where teams a and b are drawn together. So, the probability of team a and team b meeting each other on any one given day is: (19! x 2) / 20! or 243,290,200,800,000,000 / 2,432,902,008,000,000,000 This is (believe it or not) a 1/10 chance. So the probability of team a and b being drawn together is 1/10 This means the probability of teams c and d being drawn together is ALSO 1/10 The probability of teams a & b AND c & d then being drawn together must be 1/100 (1/10 x 1/10) For ANY of the four teams to meet each other, then any permutation of a, b, c and d is acceptable. There are 24 (4!) ways in which the four teams could meet each other in a fixture list and each permutation has a probability of 1/100. Consequently, the probability of the big 4 being drawn together on any one occasion is a mind-blowingly large 24/100 (a 24% chance or a "one in 4.2" chance if you want to look at it a different way) Amazing?!! Not really. What are the chances that Boro v Villa and Everton v Man City are scheduled to be drawn on the same day? Exactly the same as the probability that Man Utd v Arsenal and Chelsea v Liverpool are. No one would think it "amazing" or "a fix" if the former set of fixtures comes out, hence there is nothing amazing about the latter. Right. That's that sorted. Time for a cold beer. | |

Capybara Posted on 13/06/2008 18:04 | |

New fixtures fixed....!
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Lovely. From a professional | |

goalscrounger Posted on 13/06/2008 18:06 | |

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...except I've just read that back and, logical though it sounds, it would imply that if you did a mock draw 100 times, 24 times you would get a, b, c and d drawn together and 76 times you wouldn't. That seems uncomfortably frequent for my "logical" mind to accept? So, what have I missed?? | |

Capybara Posted on 13/06/2008 18:19 | |

New fixtures fixed....!
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I was taking it for granted that the professional would get it right. Perhaps the calculation you have done is a different one ie the chances of it happening in drawing a single weekend of fixtures from 20 teams. What we are after is the chances of it happening in a season which means you have to discard certain possibilities as fixtures are used up. I was sure that when this happened last year some bright spark in one of the newspapers did the maths and it worked out that the chances of it happening in any one season was in the low single figures. | |

goalscrounger Posted on 13/06/2008 18:26 | |

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It might be right; it just seems uncomfortably large (matron). | |

bear66 Posted on 13/06/2008 19:05 | |

New fixtures fixed....!
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Can we play Man C four times next season? | |

Jonicama Posted on 13/06/2008 21:48 | |

New fixtures fixed....!
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Goalscrounger You're analysis is flawed. It starts to go wrong when you say the chances of two teams playing each other is one in ten. See my note which builds on Buddy's note. Should happen one every 5.6 seasons | |

Buddy2 Posted on 14/06/2008 09:15 | |

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Anita - I DID listen to that, but the person explaining it still didn't convince me. It was dealt with again on Material World the following week and the guy on the line from California, who apparently used to write Star Trek episodes, was the first to actually set my mind off down the "oh I see!" route, in that he explained that the act of the host opening a given box removes the randomness. I know scrounger has been telling me that for five years, and I still haven't fully got my head round it, but at least I'm wavering. | |

goalscrounger Posted on 14/06/2008 09:52 | |

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So what IS the probability that any two teams meet each other on any given day then? Nowhere in the above is an explicit method and answer provided. That was the bit of my calculations that I was convinced was the most rigid! | |

BenJammin Posted on 14/06/2008 10:00 | |

New fixtures fixed....!
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1 in 190 | |

Buddy2 Posted on 14/06/2008 10:11 | |

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I did say it was 1 in 19, which seems the most basic and obvious answer to me. Team A can play any one of 19 other teams. If you start to pin it down to "what are the chances of them playing each other on 'match weekend 6'", then presumably that's 1 in 361, as there are 19 available match weekends for it to happen. But that's not what we're interested in - it's really just the chance of it happening at the same time as a different specified fixture, bearing in mind that there are four possible participants. I do think Jonicama's logic might be a bit flawed as well, because I don't think you need any extra maths to get from once a season to twice a season - the one leads directly to the other without any other permutations, because fixture weekends are always reversed. | |

goalscrounger Posted on 14/06/2008 10:33 | |

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Benjammin, 1/190 is NOT the chances two team play each other (2/20 x 1/19). It's the odds those two teams are the first two drawn out of the hat as fixture number 1. They could be the last two teams left in the hat, the fourth fixture drawn etc. | |

goalscrounger Posted on 14/06/2008 10:41 | |

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Ok buddy, that makes sense, but does that assume team A is at home against team B only, or does that include team A being away to team B? If we assume that it covers both, then p(A vs B) = 1/19 and p(C v D) = 1/19 So p(A vs B & C vs D) = 1/19 x 1/19 which is 1/361 But, the above permutauion is not the only one that would satisfy all of the big 4 playing each other. There are 24 possible permutations, so we need to multiply the 1/361 by 24. This gives 24/361 which is roughly 1/15 (or just under 7%). This seems a much more "realistic" estimate than my first attempt (however, I am still not sure it is actually "correct" !!) | |

Buddy2 Posted on 14/06/2008 10:48 | |

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I hesitate to debate probability with a maths teacher tbh, as I never got my head round it properly, but: isn't p(A vs B & C vs D) 1/19 x 1/17, or 1 in 323? Because by the time you get to C vs D you've taken two teams out of the calculation. | |

Buddy2 Posted on 14/06/2008 10:55 | |

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And there's only three possible combinations of two fixtures which would create a "Grand Slam Sunday", ignoring home or away as they will be repeated: A v B C v D A v C B v D A v D B v C which gets us back to my 1/107.66666 calculation above. Boromart said you need to keep adding probabilities for the second weekend and the third weekend and so on, but this calculation only relates to the probability of it happening on ANY weekend, surely. The probability of weekend 1 is 1/29070, I think (190 x 153). | |

goalscrounger Posted on 14/06/2008 10:56 | |

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Yeah, you're probably right. But my gut feeling is that there are more nuances to this that we are including. As for my mathematical pedigree, I teach GCSE mathematics and no higher. The statistics we do barely scratches the surface, so there is no guarantee I am doing anything than talking B*******! I've always said I am a better teacher than I am a mathematician! My instincts still draw me back to my original permutations approach to finding the probability of two teams playing each other, although the fact it comes out as 1/10 is surprising but also comforting as there are ten fixtures and it is one of those ten. We need a statistician, not a mathematician! | |

bobscat Posted on 14/06/2008 11:00 | |

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What it all boils down to is , Sky pay the money so they call the shots , it fixed as simple as Just like the big four getting all the decisions in their favour to push them up the league , what benifit would sky get from having an couple of unfashionable sides contesting the title ..... certainly a loss of T.V Money from Asia etc.. | |

Buddy2 Posted on 14/06/2008 11:01 | |

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I understand where Jonicama is coming from now. We've got to 1/107.66666 (or 3/323), and at the point the fixture computer runs there are 19 possible dates for that to happen, which makes 57/323 or 1/5.6666666. So for two consecutive seasons it's 1/32.11111111, and 1/181.96 for three in a row. I am now off to Ladbrokes to ask for 182-1 against a Grand Slam weekend being in the fixtures when they come out next week. | |

goalscrounger Posted on 14/06/2008 11:03 | |

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But when you are dealing with permutations, you have to remember that they could come out in different orders, so A v B C v D is different to A v B D v C etc Also A v B C v D is different to C v D A v B In a more basic version, the probability that, when you toss a coin three times you get exactly one head isn't just HTT (1/8). It could be HTH or TTH so the correct answer is 3/8. On reflection, I think the best answer is this part from my original: "What are the chances that Boro v Villa and Everton v Man City are scheduled to be drawn on the same day? Exactly the same as the probability that Man Utd v Arsenal and Chelsea v Liverpool are. No one would think it "amazing" or "a fix" if the former set of fixtures comes out, hence there is nothing amazing about the latter." | |

Buddy2 Posted on 14/06/2008 11:09 | |

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Taking your point about the permutations, my calculation would come down to 12/323 or 1/26.92, which over the season would be 1/1.416666!!!! (That's an expression of surprise, not a factorial ) I'm not convinced you need to factor in the four variables to each fixture pairing though, because the earlier calculations allow for ANY combination of pairings - that's how we got to 3/323. | |

goalscrounger Posted on 14/06/2008 11:17 | |

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It depends which way you do it. You have, in effect, factored in the permutations by doing 1/19 then 1/17. I've tried, in my original approach, to work on one outcome only and then look at the permutations seperately. | |

goalscrounger Posted on 14/06/2008 12:01 | |

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Ok, thinking out loud. Let's try and crack the probability of two teams being drawn together rather than four to begin with: We have 20 teams. If I list them in a random order, there will be 20! rows to my list (i.e 2,432,902,008,176,640,000 possible combinations) If you want A and B drawing together, you must include them as a "unit" AB, which now gives you a list with only 19! rows (i.e 121,645,100,408,832,000 possible combinations) However, remember, this is only for AB, not for BA, so we should double the number of possibilities EXCEPT (and this is the "nuance" I missed earlier) this would could as a "success" any set of fixtures where A follows B or of the hat (or vice versa) so G v A B v M would be included! Obviously, we don't want to include combinations like this, so we must half the answer again. This leaves us with: 2,432,902,008,176,640,000 ways the fixtures could come out for one particular day. 121,645,100,408,832,000 ways that the fixtures could come out with two particular teams playing each other. This gives a probability of 1/20 for a set of fixtures to be generated to produce a given two teams to face each other (logical?!!) The probability of a different two teams playing each other is ALSO 1/20. So, p(A v B) = 1/20 and p(C v D) = 1/20 P(A v B and C v D) = 1/20 x 1/20 = 1/400 But, for any of the big 4 to face each other, it could be A v C and B v D etc. Last time I factored this into my calculations, I stated that there were 24 combinations of A,B,C and D playing each other (which there are), however, my original calculations have already included the different combinations of a particular fixture (i.e A v C and B v D as opposed to C v A and B v D etc), so I only need to factor in the number of different fixtures regarding teams, not the order they appear on the fixture list. As buddy stated earlier, there are only three of these: A v B, C v D A v C, B v D A v D, B v C so we need to multiply the 1/400 by 3 which gives 3/400. This is a "one in 133 and a third" chance (or 0.75%) which seems much more likely that any other. So, my conclusion is that the big four should be drawn to play each other roughly once every 133 fixtures on average (i.e once 3.5 seasons). I'm fairly happy that is now correct. I knew I had missed something but wasn't sure what until now. | |

goalscrounger Posted on 14/06/2008 12:03 | |

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Just read that back. What a way to spend a Saturday morning. What a nerd I am! Right, now to get back to calculating Pi to 6 billion places in hexadecimal. | |

Buddy2 Posted on 14/06/2008 12:52 | |

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Sorry to labour the point, but it seems to me that you're engaging in "showing the working" of basic logic, and then missing a bit out. Even if we go with your assertion that a particular fixture has a 1 in 20 chance (and I'm not sure I do, as I think that allows Team A to be drawn against themselves), you MUST have to take those two teams out of the calculation when you look for the chance of the second fixture, which would make it 1 in 18. To get to 1 in 400 you would have to allow A and B to be still in the hat when drawing out C and D. So if your starting point is 1 in 20, your 1 in 400 becomes 1 in 360, and the 3 possibilities bring it down to 1 in 120. Or, looked at another way, the only difference between your calculations and mine (to this point) is whether you start with 1 in 20 or 1 in 19. You then go on to make the same mistake I did about interpreting 1 in 133 (or 1 in 107 in my case) as representing "once every 3 and a bit years". The problem with that is that there are only 3 random events in that time, not 133, if you see what I mean. The computer only runs once a year, not after every round of fixtures. So your 1 in whatever it is is simply a chance for any given weekend in that particular season, which needs to be divided by 19 for the chance of it happening at ANY time in the season. As Boromart says, it has no bearing on the subsequent seasons, because you start again in the summer. | |

the_throcking_man Posted on 14/06/2008 12:57 | |

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So can someone list the amount times that the big 4 have played each other on the the same weekend since the premier league started? Bet it isn't many. you also have to take into account that spurs don't play at home the same weekend as Arsenal, same for manchester clubs, liverpool clubs,possibly villa and birmingham,sunderland and newcastle. | |

Robbo_89 Posted on 14/06/2008 13:08 | |

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It was a bit unfair that we had to travel to Birmingham and Portsmouth within 4 days between christmas and new year last year. I thought you were supposed to play someone closer over xmas. | |

goalscrounger Posted on 14/06/2008 13:45 | |

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Yeah, I've modelled it with 2, 4 and 6 teams and p(AvB) is always 1/(n-1) {where n = number of teams} So for 20 teams it's definitely 1/19 for AvB | |

Liamo Posted on 14/06/2008 13:57 | |

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Since it's already been established that the fixtures are not drawn at random, what is the point of trying to calculate what the odds would be if they were? Or has this one just taken on a life of its own? | |

Buddy2 Posted on 14/06/2008 14:01 | |

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You've never spoken to any mathematicians, have you Liam? Suppose it's just to prove that it's not millions to one even for it to happen three years running - even if it WERE random. | |

NeddySeagoon Posted on 14/06/2008 14:30 | |

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Odds of a playing b is 1/1 every season odds of c playing d on the same day = 1/17 Odds of a playing c 1/1 odds of b v d on the same day = 1/17 odds of a v d = 1/1 odds of b v c = 1/17 So odds of all on same day = 3/17 or once every 5.66 seasons | |

junkyard_angel Posted on 14/06/2008 14:48 | |

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Goalscrounger, is this part correct? "However, it doesn't matter if we have ab or ba, so we have to multiply our answer by 2. This gives 243,290,200,800,000,000 possible random permutations of fixtures where teams a and b are drawn together." If you can have a/b or b/a are the odds of this happening not shorter, so therefore should you not divide? | |

goalscrounger Posted on 14/06/2008 15:36 | |

New fixtures fixed....!
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I'm avoiding cutting the grass Liam! Right. Here's a whole new spin on it. I've modelled it with smaller numbers of teams and then found formulae that give us what we are after. The number of possible fixtures with n teams is n! So, if you listed all possibilities with only 4 teams, you'd get 24 possible combinations (4! = 4 x 3 x 2 x 1) This would include counting: AvB and CvD as different to CvD and AvB Also, it would count: AvB as different to BvA By modelling, I am certain that the number of times 2 teams are drawn against each other is n!/(n-1). {where n = number of teams] So, the probability of two teams being drawn together is: number of times the play each other / number of possible combinations = [n!/(n-1)] / n! = 1/(n-1) CONCLUSION 1 - This means buddy's claim that for 20 teams the probability of two being drawn together is 1/19 is spot on. The general formula is 1/(n-1) I've then gone on and looked at the number of combinations where AvB and CvD This was surprisingly basic. The number of combinations for AvB AND CvD is 2^(n/2) x (n/2)! Test it out. For 4 teams, there are 8 possible sets of fixtures with AvB and CvD 1) AvB, CvD 2) AvB, DvC 3) BvA, CvD 4) BvA, DvC 5) CvD, AvB 6) CvD, BvA 7) DvC, AvB 8) DvC, BvA The formula gives 2^(4/2) x (4/2)! = 2^2 x 2! = 4 x 2 = 8 It works for 6 teams too if you care to check?! CONCLUSION 2: So, the probability of A playing B and C playing D in any given set of fixtures with n teams is: [2^(n/2) x (n/2)!] / n! For a 20 team league, the probability of A playing B and C playing D is: [2^(20/2) x (20/2)! / 20! = [2^10 x 10!] / 20! = 0.00000000152734930856706 This is 1/654,729,075 This seems miniscule, but my formula DOES work (if you care to check by writing them all out?!!!!! :p) However, this still hasn't answered the actual question, which was the probability of the teams A,B,C and D playing each other in any order. Well, as established, the probability of a particular two teams meeting and the other two meeting at the same time (lets say A playing B and C playing D) is 1/654,729,075 All we need to do now is find out how many possible fixtures there are that satisfy A,B,C and D playing each other and we're there. buddy did this earlier. As well as 1) A playing B and C playing D we could have 2) A playing C and B playing D 3) A playing D and B playing C So CONCLUSION 3 The probability of any 4 teams meeting each other in a 20 team league is 3 times 1/654,729,075 which is 1/218,243,025 This seems MUCH more likely as an answer to me. The number of permutations is mind boggling with 20 teams. The only bit I'm NOT sure on is if it is 1/654,729,075 x 3 (as above) or 1/654,729,075 x 24 (which is all the permutations of A,B C and D meeting) This would bring the final probability down to roughly 1/27,280,378 So, it's either 1 in about 218 million or roughly 1 in 27 million. | |

goalscrounger Posted on 14/06/2008 15:39 | |

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junkyard, I was working out combinations, not probabilities at that stage. If you have ab, c, d as one combination then introduce ba, c, d as a different combination, you're doubling the number of possibilities. | |

goalscrounger Posted on 14/06/2008 15:40 | |

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I now know the meaning of "publish and be damned". Do your worst!! If fairly confident in conclusions 1 and 2 as I have hard evidence to test my theories, I'm just not sure on which approach to use for the final conclusion, although gut feeling is with buddy's "three possibles" approach. | |

NeddySeagoon Posted on 14/06/2008 15:48 | |

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GS, yer missed A v D and B v C there | |

Scrote Posted on 14/06/2008 15:53 | |

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two points 1. why not just email the FA? 2. | |

bobscat Posted on 14/06/2008 15:56 | |

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And the F.A SAY " Yes all the Fixtures are fixed coz we get shed loads of cash from Sky .............! " | |

goalscrounger Posted on 14/06/2008 16:00 | |

New fixtures fixed....!
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Anyone wanna come and cut my grass??! | |

icarus1965 Posted on 14/06/2008 16:44 | |

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top 4 sides tend to get easy games at the start and the end of each season... don,t tend to play like this man utd vs arsenal then liverpool vs man utd first two games of the new season one of the top 4 will play a promoted side or maybe two will | |

goalscrounger Posted on 14/06/2008 19:04 | |

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I'll be buggered if I let this fall off the front page!! | |

sixtyniner69 Posted on 14/06/2008 19:37 | |

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of cos it aint fixed each half of the season everyone plays everyone else once ( few exceptions ) then 2nd half of the season they do the same but reverse grounds. big events used to be local derbys ie boxing day ( was that fixed) now local derbys are mid season ( for us at least) would have liked away to chelski , manure both away to start with, cos we probably would lose these any way and the team can use them as their warm up ( cos we normally do not bother pre season). | |

Boromart Posted on 14/06/2008 20:04 | |

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goalscrounger good work. I'm quite pleased that my initial guestimate/rough working of "33.7 million to 1" in looking fairly near the mark . IO thin for a while I overcomplicated it and tied myself up in knots, but I'm happier with my initial rough estimate. | |

Buddy2 Posted on 15/06/2008 09:28 | |

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"CONCLUSION 1 - This means buddy's claim that for 20 teams the probability of two being drawn together is 1/19 is spot on. The general formula is 1/(n-1)" Which makes me repeat my earlier assertion that you are over-complicating common sense. No need for enormous factorials to work out that there are 19 possible opponents for 1 team in a 20 team group. I'm afraid you lose me in Conclusion 2. Because I don't understand factorials I don't know why you keep dividing by them. What I DO know is that before you get to Conclusion 2 you're missing half the possibilities, as Neddy says. We're not just interested in any combination of AvB and CvD, but also AvC and BvD, which makes 16 possibilities. I have tried, briefly, to rework your formula on that basis, but my instinct is that the phrase "/n!" will get in the way and I don't understand it well enough. For that reason I'm going to stay with my 1 in 5.66 per season. Boromart - if you look closely you will see that scrounger has not gone as far as 3 seasons, so his 27 million or whatever is your 323, I'm afraid. | |

Buddy2 Posted on 15/06/2008 09:34 | |

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To go back to your 16:32 post on Friday Boromart, the flaw is you're starting with 1/19 and then multiplying by 1/17, when you should be starting with 3/19 because A could play B, C or D to fulfil the requirement. That's why it's 1/107.6666 rather than 1/323. And you've still only done it for any weekend. Because the fixture computer only runs once, it has 19 possible dates to chuck out that 107.66666 chance, hence 1/5.6666666. | |

goalscrounger Posted on 15/06/2008 09:36 | |

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"What I DO know is that before you get to Conclusion 2 you're missing half the possibilities, as Neddy says. We're not just interested in any combination of AvB and CvD, but also AvC and BvD, which makes 16 possibilities" I've tackled it a different way buddy. I know you agree with me that p(AvB and CvD) is exactly the same as p(AvC and BvD) and also the same as p(AvD and BvC)? Consequently, if you know the probability of p(AvB and CvD) then, to find all three combos, you simply multiply your result by 3 (which is what I did in the last part of my epic). I derived my formulae from looking at the data and working backwards, not from some obscure theoretical viewpoint. Consequently, I wanted as small a data set as possible to work from. That's | |

Buddy2 Posted on 15/06/2008 09:37 | |

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scrounger? SCROUNGER?!?! Nurse, he's had a stroke.... | |

Buddy2 Posted on 15/06/2008 09:41 | |

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Assuming you ARE still breathing, I see your point, and my 16 in my last post should in fact be 24, because I didn't include AvD and BvC. Go on then, explain the division by 2 trillion for me? | |

goalscrounger Posted on 15/06/2008 09:47 | |

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Oooh, where did my last sentence go?? Yeah, 24 (3 sets of 8 combos) [2^(n/2) x (n/2)!] / n! The n! in the denominator is the number of possible fixtures for n teams. The formula is a calculating the probability of A playing B & C playing D, not just the number of times they could meet (the numerator gives that part). | |

Buddy2 Posted on 15/06/2008 10:10 | |

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Hmmm. Struggling to demonstrate it, but I instinctively don't like your formula, partly because I can't see that it restricts the number of teams involved. My brain is wanting at the very least a toasted muffin now though, I'll come back to it.... | |

goalscrounger Posted on 15/06/2008 10:27 | |

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Doesn't matter if I've only used one of the three permutations, so long as I multiply by three at the end. You've proved that yourself by stating that my 8 combos should be 24 (i.e 3 x 8). When I get my answer for my one set and multiply it by three, it covers all three sets, honest! Enjoy the muffin. | |

Buddy2 Posted on 15/06/2008 10:28 | |

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I've got the seed of an objection forming here, which I think is best demonstrated if you use a 5 team league. I think your formula will include ABCDE ABCED etc as valid combinations even though they're nonsense in footballing terms. That means that 20! is not a valid number in this calculation. Doesn't it? | |

NeddySeagoon Posted on 15/06/2008 10:47 | |

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that's what I love about math, it's such an exact science.... "So, it's either 1 in about 218 million or roughly 1 in 27 million." A margin of error somewhere around 180 million. Anyone who believes that the odds of picking a, b, c, d in any order out of a bag of twenty balls is 1:20m or 1:200m should be battered with an abacus!! ;o) The permutations are vast but in exactly 1/19 of those permutations a and b will be together. Of that nineteenth 1/17 will have c and d together. you're making it far too complicated. The odds of it happening on easter weekend are 1/5.666 X 1/38 = 1/212.8 | |

Buddy2 Posted on 15/06/2008 12:39 | |

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Wot he said ^^^^^^ but I'd still like to prove it even using scrounger's factorials | |

Muttley Posted on 15/06/2008 12:42 | |

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*jaw drops open* Err does this thread come with subtitles? Well done everyone, I think! | |

Big_Shot Posted on 15/06/2008 12:47 | |

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I don't mind the 'Grand Slam' weekends to be honest. Far better than some of the S***e fixtures they dished up in the Sunday 4pm slot in the 2nd half of last season. | |

Buddy2 Posted on 15/06/2008 13:01 | |

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OI SCROUNGER.... | |

Link: You done that grass yet? | |

BenJammin Posted on 15/06/2008 13:36 | |

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It seems fairly clear to me that NeddySeagoon is spot on. Man U will play Arsenal, Chelsea and Liverpool with absolute certainty in the first 19 games. It then becomeas a question of what are the odds that either:- a) Chelsea play Liverpool when Man U play Arsenal or b) Chelsea play Arsenal when Man U play Liverpool or c) Arsenal play Liverpool when Man U play Chelsea The odds of any of these happening are clearly 1 in 17 (without knowing what any of the other fixtures are) so the odds of any of the 3 happening are 3 in 17 or 1 in 5.6667. Obviously this is all assuming the 2nd 19 fixtures are mirror images of the first 19 and that the draw is randon which it clearly is not. Surely must be obvious to everyone that the odds of this happening is not one in hundreds of millions!!! | |

MKredleaderOne Posted on 15/06/2008 13:41 | |

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I agree it's shocking that the Premier League are going to make us play Arsenal Aston Villa Blackburn Bolton Chelsea Everton Fulham Hull Liverpool Man City Man Utd Newcastle Portsmouth Stoke Spurs Sunderland West Brom West Ham Wigan and home away at that. Disgusting. | |

Buddy2 Posted on 15/06/2008 13:42 | |

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That's four of us in a consenus now, around an answer we had at 5pm on Friday. I would still like somebody who can do it scrounger's way with an understanding of the factorials to get it down to the same figure though. | |

bobscat Posted on 15/06/2008 18:44 | |

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Come on lads only another 7 to go | |

goalscrounger Posted on 15/06/2008 23:09 | |

New fixture s fixed !!!
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"Anyone who believes that the odds of picking a, b, c, d in any order out of a bag of twenty balls is 1:20m or 1:200m should be battered with an abacus!! ;o)" I don't believe I (or anyone else) has said that??? "The permutations are vast but in exactly 1/19 of those permutations a and b will be together. Of that nineteenth 1/17 will have c and d together." Okay, that makes sense. If we check this with a 6 team model it works With 6 teams 1/5 should be a and b together. Of that fifth, 1/3 will also have c and d together. There are 720 permutations with 6 teams, 144 fixtures give a and b together (this is the predicted 1/5) and, of those 144, 48 of them also have c and d together (this, as predicted is a third of 144). However 48 out of 720 is a probability of 1/15. So, your calculation below, if altered for 6 teams, should also equal 1/15 for avb & cvd. If we multiply this by three it allows for the avc & bvd and also the avd & bvc combos, giving 1/5 chance of a,b,c and d meeting each other in a 6 team league. "The odds of it happening on easter weekend are 1/5.666 X 1/38 = 1/212.8" You don't mention explicitly if, by "it" you mean the grand slam or just avb & cvd? However, looking at your calculation: 1/5.66666 is 1/(your 17 teams shared by 3) 1/38 is your 1/ (your 19 teams multiplied by 2) Where you had 17 and 19 teams, the model has 3 and 5 teams. So, 1/(3 shared by 3) x 1/(5 multiplied by 2) should equal 1/15 (or 1/5 if you were talking of the full grand slam), but 1/1 x 1/10 = 1/10, so I'm afraid your calculation doesn't work. HOWEVER, you are right, it is getting very complicated and your 1/19 and 1/17 hypothesis holds water and is actually almost the answer. We could generalise this to: p(avb & cvd) = 1/(n-1) x 1/(n-3) {where n=number of teams in league} For a 4 team league, the probability (obviously) that a,b,c and d have a "grand slam" is 1 1/3 x 1/1 = 1/3 (using your hypothesis above) is the probability of avb & cvd. Multiply it by 3 to account for the other two possible head to head combos (i.e instead of avb & cvd it's avc & bvd or avd & bvc) and 1/3 x 3 = 1. IT WORKS FOR 4 TEAMS For 6 teams, I modelled that the probability of avb & cvd was 1/15 so the probability of the grand slam was 1/15 x 3 = 1/5 Using your hypothesis: 1/5 x 1/3 = 1/15 Multiply by 3 = 1/5 IT WORKS FOR SIX TEAMS! For an 8 team league, the probability of avb and cvd is 1/105, so the grand slam odds become 1/105 x 3 = 1/35 Again, use the hypothesis: 1/7 x 1/5 = 1/35 1/35 x 3 = 3/35 IT DOES NOT WORK FOR 8 TEAMS :-( For 10 teams, prob of avb and cvd is 1/945 (again from modelling) Using the hypothesis: 1/9 x 1/7 = 1/63 1/63 x 3 = 1/21 DOES NOT WORK However, I noticed something 4 teams (1/3 x 1/1) x 3 = 1 CORRECT 6 teams (1/5 x 1/3) x 3 = 1/5 CORRECT 8 teams (1/7 x 1/5) x 3 = 3/35 not 1/35 BUT (1/7 x 1/5 x 1/3) x 3 DOES = 1/35 CORRECT NOW! 10 teams (1/9 x 1/7) x 3 = 1/21 not 1/945 BUT (1/9 x 1/7 x 1/5 x 1/3) x 3 DOES = 1/945 CORRECT NOW! Therefore for 20 teams, it seems that: (1/19 x 1/17 x 1/15 x 1/13 x 1/11 x 1/9 x 1/7 x 1/5 x 1/3) x 3 will give the correct answer. This gives 1/218,243,025 which, I believe, was the answer I arrived at from a different viewpoint a few posts earlier. Again, people will now scream "oooh, far too small odds". To those people, I suggest you look at 8 or 10 teams and see what the odds of a grand slam are there. If, for ten teams the odds are 1/945, then odds of 1/212.8 or amazingly 1/5 are obviously far, far off the mark. Can any coders out there write a simulation and solve this once and for all??!! | |

goalscrounger Posted on 15/06/2008 23:25 | |

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Lol buddy. Did you post that "cry for help" on the stats site?!! Anyway, I'm certain my answer is correct now. | |

BenJammin Posted on 15/06/2008 23:25 | |

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The thing is, surely you can see that team a has to play team b and when that happens the odds of team c playing team d are 1 in 17. The odds therefore of a 'grand slam weekend' are lower than 1 in 17 (due to the other combinations, and are actually 3 in 17) and certainly lower than the 1 in 218 million you estimate. | |

buffaloboro Posted on 16/06/2008 02:54 | |

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why don't they make the Xmas games more user friendly ? ( travel times etc ) if this is the case ? | |

Buddy2 Posted on 16/06/2008 06:40 | |

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Not me guv, it was the first result for "fixture combination probability" on Google. I shall read your latest version later but I note with disappointment that it doesn't end up at 3 in 17 which is the correct answer | |

NeddySeagoon Posted on 16/06/2008 07:59 | |

New fixture s fixed !!!
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My humble apologies to anyone who’s bored with this – it’s doing my napper in and I have to have an answer! Anyone who believes that the odds of picking a, b, c, d in any order out of a bag of twenty balls is 1:20m or 1:200m should be battered with an abacus!! ;o)" I don't believe I (or anyone else) has said that??? YES, I BELIEVE YOU DID, AND YOU SAY IT AGAIN AT THE END OF THIS MESSAGE – DO YOU SERIOUSLY BELIEVE TAT THIS WILL ONLY HAPPEN ONCE EVERY HOWEVER MANY MILLION YEARS? "The permutations are vast but in exactly 1/19 of those permutations a and b will be together. Of that nineteenth 1/17 will have c and d together." Okay, that makes sense. If we check this with a 6 team model it works With 6 teams 1/5 should be a and b together. Of that fifth, 1/3 will also have c and d together. There are 720 permutations with 6 teams, 144 fixtures give a and b together (this is the predicted 1/5) and, of those 144, 48 of them also have c and d together (this, as predicted is a third of 144). However 48 out of 720 is a probability of 1/15. So, your calculation below, if altered for 6 teams, should also equal 1/15 for avb & cvd. If we multiply this by three it allows for the avc & bvd and also the avd & bvc combos, giving 1/5 chance of a,b,c and d meeting each other in a 6 team league. SORRY, MY ARGUMENT IS THAT AVB IS A 1/1 SHOT GUARANTEED EVERY YEAR AND THUS THE ODDS OF BVC ARE 1/3 SO THE ODDS OF THE FOUR TEAMS PLAYING EACH OTHER IN A SEASON ARE 3X 1/3 – A CERTAINTY "The odds of it happening on easter weekend are 1/5.666 X 1/38 = 1/212.8" You don't mention explicitly if, by "it" you mean the grand slam or just avb & cvd? However, looking at your calculation: YES I DID, 1/5.666 IS THE ODDS OF THE FOUR PLAYING THE SAME WEEKEND IN A SEASON IN ANY ORDER 1/5.66666 is 1/(your 17 teams shared by 3) 1/38 is your 1/ (your 19 teams multiplied by 2) –THERE ARE 38 GAMES SO THE ODDS OF ALL FOUR PLAYING AT EASTER ARE 1/5.66 X 1/38 Where you had 17 and 19 teams, the model has 3 and 5 teams. So, 1/(3 shared by 3) x 1/(5 multiplied by 2) should equal 1/15 (or 1/5 if you were talking of the full grand slam), but 1/1 x 1/10 = 1/10, so I'm afraid your calculation doesn't work. MY MODEL SAYS THAT THE ODDS OF THEM PLAYING IS ONE AND THE ODDS OF THAT HAPPENING ON ANY GIVEN WEEKEND IS 1/10 HOWEVER, you are right, it is getting very complicated and your 1/19 and 1/17 hypothesis holds water and is actually almost the answer. We could generalise this to: p(avb & cvd) = 1/(n-1) x 1/(n-3) {where n=number of teams in league} For a 4 team league, the probability (obviously) that a,b,c and d have a "grand slam" is 1 1/3 x 1/1 = 1/3 (using your hypothesis above) is the probability of avb & cvd. Multiply it by 3 to account for the other two possible head to head combos (i.e instead of avb & cvd it's avc & bvd or avd & bvc) and 1/3 x 3 = 1. IT WORKS FOR 4 TEAMS For 6 teams, I modelled that the probability of avb & cvd was 1/15 so the probability of the grand slam was 1/15 x 3 = 1/5 WRONG – THE ODDS OF AVB CVD IS 1/3 ETC ETC…..= 1/1 (Your model says it happens only once every five years – you must see that that is wrong?) Using your hypothesis: 1/5 x 1/3 = 1/15 Multiply by 3 = 1/5 IT WORKS FOR SIX TEAMS! For an 8 team league, the probability of avb and cvd is 1/105, so the grand slam odds become 1/105 x 3 = 1/35 Again, use the hypothesis: 1/7 x 1/5 = 1/35 1/35 x 3 = 3/35 IT DOES NOT WORK FOR 8 TEAMS :-( ODDS OF AVB CVD = 1/5 ODDS IN ANY ORDER = 3/5 (THAT IT’LL HAPPEN IN A SEASON) ODDS THAT IT’LL HAPPEN ON EASTER WEEKEND = 3/5 X 1/7 = 3/35 For 10 teams, prob of avb and cvd is 1/945 (again from modelling) Using the hypothesis: 1/9 x 1/7 = 1/63 1/63 x 3 = 1/21 DOES NOT WORK However, I noticed something 4 teams (1/3 x 1/1) x 3 = 1 CORRECT 6 teams (1/5 x 1/3) x 3 = 1/5 CORRECT 8 teams (1/7 x 1/5) x 3 = 3/35 not 1/35 BUT (1/7 x 1/5 x 1/3) x 3 DOES = 1/35 CORRECT NOW! 10 teams (1/9 x 1/7) x 3 = 1/21 not 1/945 BUT (1/9 x 1/7 x 1/5 x 1/3) x 3 DOES = 1/945 CORRECT NOW! Therefore for 20 teams, it seems that: (1/19 x 1/17 x 1/15 x 1/13 x 1/11 x 1/9 x 1/7 x 1/5 x 1/3) x 3 will give the correct answer. This gives 1/218,243,025 which, I believe, was the answer I arrived at from a different viewpoint a few posts earlier. (SEE, YOU DID SAY IT!!! ONCE EVERY 218 MILLION YEARS THEN EH?) Again, people will now scream "oooh, far too small odds". (THAT’D BE ME!)To those people, I suggest you look at 8 or 10 teams and see what the odds of a grand slam are there. If, for ten teams the odds are 1/945, then odds of 1/212.8 or amazingly 1/5 are obviously far, far off the mark. Can any coders out there write a simulation and solve this once and for all??!! ODDS FOR TEN TEAMS THAT CVD WILL OCCUR WITH AVB. = 1/7. ODDS THAT IT’LL HAPPEN IN ANY ORDER = 3/7, SO ROUGHLY EVERY OTHER YEAR, NOT ONCE EVERY THOUSAND YEARS! ODD FOR EASTER 3/7 X 1/18 = 3/126 OR 42:1 My appologies to all esp GS, I'm not shouting here, just using CAPS to differentiate my text from yours. Buddy, I wish to ffff I'd just googled it right at the start. I have bits of damaged brain dribbling out of my lug'oles | |

NeddySeagoon Posted on 16/06/2008 08:01 | |

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I think i've just about earned the right to be post # 100 | |

littlejimmy Posted on 16/06/2008 08:18 | |

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This is much worse than any train-spotting thread. Does it really matter? | |

Buddy Posted on 16/06/2008 08:27 | |

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It's maths lj, of course it doesn't bloody matter. Cheaper than Dr Kawasaki's Learn To Count or whatever it's called though. | |

Capybara Posted on 16/06/2008 09:22 | |

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Bah! I just read through all that lot and was about to post that Buddy will be saying there's a use for calculus next, and he's gone and trumped me on the last post by saying none of it matters ..... | |

Buddy2 Posted on 16/06/2008 09:23 | |

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Nothing like a good "Bah!" to start the week. | |

Jonicama Posted on 16/06/2008 09:58 | |

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Back on the Board after the weekend and pleased to see this is running and running. To add fuel to the fire I'm still with the 1 in 5.66 (see my post of 17.59 on Friday) for two Grand Slam weekends in a season. Ps - I did do a Maths degree 20 years ago but I've forgotton most of it now. | |

Capybara Posted on 16/06/2008 10:19 | |

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So, now we know, has the 3/5/7/17/33.7m-1 shot come off again? | |

Capybara Posted on 16/06/2008 10:27 | |

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Unravelling this lot, apparently not: "United face top four sides Liverpool and Chelsea away on successive weekends, 13th September and 20th September, before also visiting Arsenal on 8th November. The Red Devils are at home to Chelsea on 10th January 10, play host to Liverpool on 14th March and Arsenal visit Old Trafford on 16th May, on the penultimate weekend of the season. Chelsea then face Liverpool at home on 25th October and Arsenal at home on 29th November. The Blues visit Liverpool on 31st January, Arsenal on 9th May and are away to Sunderland on the final day of the season. Arsenal's fixtures against fellow top four side Liverpool are at home on 20th December and away on 18th April." Guess they must have fixed them to make it look as if they are not being fixed ..... | |

goalscrounger Posted on 16/06/2008 11:39 | |

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I am at work, so haven't had time to digest what you have written Neddy (I barely get time to digest my lunch in this place!) but what I notice straight away is that you are talking about the probability of the grand slam happening in A YEAR, where as I am calculating the probability of it happening in A FIXTURE. We are answering a slightly different question to each other here. The only definitive answer we will get is if some coder on here can set a little app running to simulate this event and record the results. Any offers? | |

Buddy2 Posted on 16/06/2008 11:40 | |

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I reckon it's your modelling that's out scrounger. Not sure why yet though. Writing out the formula p(avb & cvd) = 1/(n-1) x 1/(n-3) for a progressively hugher number of teams gives the following: 4 teams: 1/3 5 teams: 1/8 6 teams: 1/15 (you were OK to this point) 7 teams: 1/24 8 teams: 1/35 9 teams: 1/48 10 teams: 1/63 I know mathematicians love patterns, and the bottom part of the fraction (denominator? can't remember...) is increasing by an odd-numbered pattern. ie 3 + 5 =8 8 + 7 =15 15 + 9 =24 24 + 11 =35 35 + 13 =48 48 + 15 =63 If you follow that to 20 teams it ends up at 323, which is what we've been saying all along. Then you multiply by 3 to get the 1/107.6666 and divide by 19 possible weekends to get 1/5.66666. Still! | |

Capybara Posted on 16/06/2008 11:50 | |

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Anyroad, it hasn't happened this time. So it must be more likely in 2009-10, right? | |

the_throcking_man Posted on 16/06/2008 11:51 | |

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put it in the conspiracy bag along with the hot and cold balls in the fa cup draw. | |

Buddy2 Posted on 16/06/2008 11:51 | |

New fixtures fixed....!
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Neddy - couple of basic figures errors in your last paragraph but the bit that really concerns me is the division by 19 to get to a specific weekend. I think that (in the case of 20 teams) 3/323 IS the probability of a specific weekend, and that's why it's 57/323 for "at any time during the season". | |

Buddy2 Posted on 16/06/2008 11:52 | |

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Capy - stop it | |

speckyget Posted on 16/06/2008 11:53 | |

New fixtures fixed....!
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No Grand Slam Sunday(s)???? Soccer is dead. | |

NeddySeagoon Posted on 16/06/2008 12:08 | |

New fixtures fixed....!
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Sorry fellers, the odds of it happening in a fixture is exactly what I've said. That is, the odds of it happening in a season divided by the number of fixtures 1/5.666 x 1/38 = 1/215 You're both starting from the wrong place. For the six team model, you both say that the odds of ab is 1/5, the odds of bc is 1/3 therefore the odds of this happening in a season is 1/15, then you multiply this by three to cover the other permutations to arrive at 1/5 as the odds for any combination happening in a season. OK, I'll have ten quid on that. I say it's a stone cold cert every season Here's your proof. In your six team league, the fixtures are out and a plays b the first week, a plays c the second week and a plays d the third week. Write em down and fill in the other fixtures below them. In week three it's impossible to avoid having b and c in one of the other fixtures. I'm in the middle of the Rhub Al Khali with eff all better to do. What's your excuse? [;o)] | |

Buddy2 Posted on 16/06/2008 12:20 | |

New fixtures fixed....!
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Agree totally with you then, but that's not what your last para of your last post said (albeit adjusted to 10 teams). With one exception - you only need to multiply your 1/5.66666 by 1/19, not 1/38, because the fixtures are always reversed. | |

goalscrounger Posted on 16/06/2008 12:22 | |

New fixtures fixed....!
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Neddy, my figures aren't per SEASON, they're per ROUND (i.e with a random set of fixtures there is a 1/5 chance of a grand slam on any given day). With 6 teams there are 5 fixtures and 5 x 1/5 = 1, so we agree that it's a cert over the season. Oh, and you can get passed week 3 without the slam if you're careful: 1) ab ce df 2) ac bf de 3) ad be cf | |

NeddySeagoon Posted on 16/06/2008 12:56 | |

New fixtures fixed....!
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This is hard work! In a six team league The odds of a playing b (not the grandslam) completely at random are 1/5, however a will play b (odds 1/1) so the odds of cd on that day are 1/3...... Ah Ferk it....I give up. You're right, once every two hundred and eighteen million games. And I owe you a tenner. I'm going to game number 139,645,818 I'll givt it yer on the concourse over a beer. | |

Rochey Posted on 16/06/2008 13:12 | |

New fixtures fixed....!
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How come there is no new year fixtures this season? | |

goalscrounger Posted on 16/06/2008 14:44 | |

New fixtures fixed....!
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Strange that isn't it Rochey. I wonder what the probability of that would be? :p Neddy, your problem isn't necessarily that you're right or wrong, it's just that you haven't come up with anything that has got everyone else convinced. Neither have I. You're convinced you're right, I'm convinced I am but neither of us can prove it via the rather limited means of a messageboard that doesn't even allow bold type as an emphasis, never mind any more elaborate methods of explanation. The "this is hard work" attitude that suggests that you are some sort of acolyte dealing with simpletons doesn't help either. Perhaps you are not explaining it well? Perhaps you are wrong? Indeed, you stated above that you wouldn't be able to get the the third fixture without a grand slam and I proved you could, no one is infallible. TBH, I'm kind of at saturation point on this and am getting involved when I should be using my brain for work-related thinking at present. Still, it's been bloody good fun and a very interesting journey so far. So, toys back in the pram please and keep going until you either convince someone else on here that you're right or until you are convinced someone else on here is right. | |

bear66 Posted on 16/06/2008 15:10 | |

New fixtures fixed....!
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Just watch when they come out , big four play each other on the same weekend near the end of the season . . . well they haven't fixed them very well then . . . . Is the Man U vs Chelsea CL final in June next year? | |

Buddy Posted on 16/06/2008 15:20 | |

New fixtures fixed....!
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*raises hand* I'm sort of convinced he's right, as are a few others, with the exception that he went down the wrong path with a ten-team league at 8am today. You're right about the "this is hard work" though. Anyway, can you paste up your modelling iterations with 8 teams? I want to see where you're going wrong | |

skiprat Posted on 16/06/2008 15:47 | |

New fixtures fixed....!
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There's no New Years fixtures because it falls on a Thursday and the following Saturday will be FA Cup 3rd round. So glad I skipped to the bottom of this thread to see if anything interesting was occuring because it's certainly not happening throughout. GEEKS! | |

Buddy Posted on 16/06/2008 15:58 | |

New fixtures fixed....!
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There was an interview on the radio a couple of months back with a writer from The Simpsons who was saying that they tried to get as much mathematical stuff into the show as possible because they were fed up with the prevailing national view that an ability to multiply two numbers together or write down a fraction was tantamount to witchcraft. | |

Scrote Posted on 16/06/2008 16:06 | |

New fixtures fixed....!
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the actual published fixtures just prove that the whole league is fixed may 16th will be grand slam sunday and the top four at the time will be man U, arsenal, chelsea and blackburn you heard it here first... | |

the_throcking_man Posted on 16/06/2008 16:08 | |

New fixtures fixed....!
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goalscrounger just wasted 10 years of his life researching this | |

Scrote Posted on 16/06/2008 16:21 | |

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you mean that ten years ago he worked out that the probability of this thread appearing was 1 and set about formulating his reply? fook me hes a clever bastad you got the lottery numbers for wednesday gs? | |

boro74 Posted on 16/06/2008 17:30 | |

New fixtures fixed....!
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All I ever use maths for is to work out how many doubles you get from a certain number of race horses. For example, four horses gives six doubles, five horses gives ten doubles and so on. You work it out 4*3/2 =6, 5*4/2=10 and so on. Applying this principle, you have 6 different possibilities of the big 4 playing each other. If you take one of these fixtures,e.g. Manu -Arsenal it leaves you with 18 other teams that can play each other. That's 18*17/2=153. So there are 153 possible fixtures, only one of them, Chelsea- Liverpool, gives us the grand slam. So when Manu play Arsenal the chances that Chelsea play Liverpool on the same day are 153 to 1. With 6 possibilties of the big 4 playing each other this makes the odds of any grand slam combination 153 to 6. Or 1 in 25.5. I thank you. Now somebody tell me where I've gone wrong... | |

goalscrounger Posted on 16/06/2008 20:25 | |

New fixtures fixed....!
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Buddy2 Posted on 16/06/2008 20:28 | |

New fixtures fixed....!
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The difference with your horses is that C and D might not happen, whereas it will always happen in a league season, which is why Neddy starts his calculations with 1/1. | |

Jonicama Posted on 16/06/2008 22:56 | |

New fixtures fixed....!
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Boro74 You're talking rubbish. If Man U play Arsenal then Chelsea must play someone else. They can play one of the other 17 teams. The chance, once it is fixed that Man U are playing Arsenal, that Chelsea play Liverpool is 1 in 17. Not 1 in 153. I'll have one final go at setting this out. Credit to Neddy as I use some of his explanation as it makes it a bit clearer. It seems to be agreed that we have 19 "pairs" of matches. So we're only looking at the chance of a grand slam weekend in the first half of the season (ie the first 19 games). In the first 19 games Man u will play Arsenal with a probability of 1 (we're not bothered about venue). When that happens then Chelsea will play some other team. There is a 1 in 17 chance it will be Liverpool. So the chance of a "Man U v Arsenal & Chelsea v Liverpool" grand slam weekend (ignoring venues) in the first half of the season is 1 in 17. It follows that there is also a 1 in 17 chance of the other two combinations (1) "Man U v Chelsea & Arsenal v Liverpool" and (2) "Man U v Liverpool and Arseanl v Chelsea". It follows that there is a 3 in 17 chance of any of the 3 possible grand slam weekends in the first half of the season. This leads to the 1 in 5.66 chance. QED I can't make it any clearer. | |

zaphod Posted on 17/06/2008 04:35 | |

New fixtures fixed....!
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Am I right in thinking that the fixture list is drawn up with 38 rounds of fixtures independently of dates and then a separate decision is made regarding the dates each round is played? That would mean that if a Grand Slam occurred in any one round (effectively 2 rounds), then the weekend it happened could be decided upon. Under this scenario the probabilities are just reduced to the occurrence of a Grand Slam at any time & in any round and then the date could be fixed later. | |

NeddySeagoon Posted on 17/06/2008 15:01 | |

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quod erat demonstrandum indeed, nicely put Jon. Appologies again to GS, I was just trying to get out of the thread in a light hearted way - if it sounded like I was throwing the toys out or being patronising, it was unintentional. AND the effin thing was still bugging me so try this 20 fixtures - abcd with anything else = !17 permutations abcd in any order with anything else = 3 X !17 perms Take out the three fixtures we've just used there must be 17 remaining possibilities, so the permutation for any one of them must also be !17. That's 17 X !17 so the odds of any big four weekend become; 3 X !17 divided by 17 X !17 Cancel the !17's = 3/17 = 1/5.666 N Seagoon Emeritus Proffessor of sums Planet Zog | |

boro74 Posted on 17/06/2008 17:48 | |

New fixtures fixed....!
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"Boro74 You're talking rubbish." Steady on Jon boy. That was my attempt to bring a bit of light-heartedness to this thread. Goalscrounger laughed! Great thread, by the way. | |

bobscat Posted on 17/06/2008 19:24 | |

New fixtures fixed....!
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Just logged on today & carnt believe that this has so many posts ....... Never thought I'd start a tread that would get us in to the 100 club.........Cheers lads.............[(:.)] | |

Buddy Posted on 02/12/2008 12:12 | |

New fixtures fixed....!
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Did we all agree on this in the end then? | |

Capybara Posted on 02/12/2008 12:20 | |

New fixtures fixed....!
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Well, the fact that it hasn't happened this season means it is almost bound to happen next. Or have I already done that one? | |

Buddy Posted on 02/12/2008 12:26 | |

New fixtures fixed....!
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You most certainly have. I think I've said this before as well, but I'll condense it into a slogan: "Catflap Can Seriously Damage Your Employment". | |