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West Ham away first game


Yorkshire Saint

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There are 3 promoted teams, and 3 top 3 teams, the probability of one promoted club playing a top 3 club is (3/19*3/19*3/19) = 9/19, so only slightly below a 50/50 chance. So it's basically as likely to happen as not.

 

Unless my recall of probability is fcked of course.

 

Swansea got Citeh last season

Newcastle and WBA got Man U and chelsea

Birmingham got Man U

WBA got Arsenal

Birmingham got Chelsea

Sheff u got liverpool

 

thats 7 big teams in 6 seasons, more than double the statistical average.

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I remember a few years back when Liverpool fixtures came out a day early.... People even out bets on it and it was spot on

 

I suspect the fixtures have been done and just get reviewed today

 

 

No way, i thought someone made it all up at one minute to nine tomorrow morning

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I've heard we will be playing each premiership team home & away over the course of the season, put a tenner on that!

 

If I did that, it'd be just my luck somewhere like the Madejski stadium would be closed due to a health scare...

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http://www.bbc.co.uk/blogs/paulfletcher/2009/06/secrets_of_the_fixture_compute.html

 

It may be a few years old now, but above is a link to a BBC blog on how the fixtures are determined, and I presume hasn't changed much.

 

bloody hell

 

I wonder if we will be paired with reading

 

Saints are always paired with Pompey in the fixtures. It doesn't matter that they aren't in the same league as us, they want to make sure we don't both play at home on the same day.

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The fixture list isn't purely random, so you can't strictly look at it through probability.

 

I am aware of that, having posted the link to "how the fixtures are calculated" on here a number of times over the past 3 years.

 

It is however, a very decent estimate of how likely it is.

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Saints are always paired with Pompey in the fixtures. It doesn't matter that they aren't in the same league as us, they want to make sure we don't both play at home on the same day.

 

You're missing the point, which is that paired clubs who ARE rivals can't play each other on opening day, and that the teams who don't have local rivals in the same division still must be paired with someone to fill out the "derby week" fixtures. They don't go "right, Arsenal v Spurs, Liverpool v Everton, Manchester derby, we've got Reading, Saints, Swansea left over without derbies, oh, it doesn't matter who they play this week". They'd still pair us with someone.

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Swansea got Citeh last season

Newcastle and WBA got Man U and chelsea

Birmingham got Man U

WBA got Arsenal

Birmingham got Chelsea

Sheff u got liverpool

 

thats 7 big teams in 6 seasons, more than double the statistical average.

 

Were they all top 3 sides ?

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There are 3 promoted teams, and 3 top 3 teams, the probability of one promoted club playing a top 3 club is (3/19*3/19*3/19) = 9/19, so only slightly below a 50/50 chance. So it's basically as likely to happen as not.

 

Unless my recall of probability is fcked of course.

 

It isn't right at all, but provides an estimate that isn't too bad. The probabilities of these events occurring aren't independent (as you're assuming). Instead, we can quite easily enumerate the respective probabilities for all combinations of relevant outcomes as below (assuming equally likely trial outcomes, which is wrong, as we know certain clubs are paired, others would be controlled for police reasons etc., which in this case (if we are paired with Reading) increases the chance of a promoted side playing a top three side, as Reading and Saints can't play each other first day (and decreases the probability of Saints playing West Ham on the first day of the season) - which does deem that in the spoiler somewhat irrelevant, but if anyone cares, the maths is below (though probably wrong).

 

 

 

The probability of all three promoted clubs playing top three sides:

 

3/19 * 2/18 * 1/17 = 1/864

 

For any two promoted clubs to play top three clubs:

 

3/19 * 2/18 * 16/17 = 16/969

3/19 * 17/18 * 2/17 = 1/57

18/19 * 3/18 * 2/17 = 6/323

 

Summing these: the probability of any two promoted clubs playing clubs in the top three is 1/19.

 

Probability of only one promoted club playing a top three club:

 

3/19 * 16/18 * 15/17 = 40/323

16/19 * 3/18 * 15/17 = 40/323

16/19 * 15/18 * 3/17 = 40/323

 

Summing these: the probability of one promoted club playing a top three club is 120/323.

 

So, the total probability of at least one promoted club playing a top three club is the sum of these probabilities: 120/323 + 1/19 + 1/864 = 118961/279072 ~ 0.425

 

Because they're not independent events (that of a team playing a side in the top three), the probability of no promoted clubs playing top three clubs is ~ 0.575, which is quite a bit higher than your estimate.

 

I don't know why I've done this.

 

 

 

 

Actual probability of at least one promoted side playing a top three side is ~ 0.425.

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There are 3 promoted teams, and 3 top 3 teams, the probability of one promoted club playing a top 3 club is (3/19*3/19*3/19) = 9/19, so only slightly below a 50/50 chance. So it's basically as likely to happen as not.

 

Unless my recall of probability is fcked of course.

 

It is flawed, I'm afraid, and your recollection of more simple arithmetic is also a little off:

 

Probability

 

The equation you have quoted is for the 3 promoted teams to ALL be drawn against a top 3 team.

 

Also as Ludwig points out you have also ignored the fact that knowing the result of the first draw affects the result of the second draw (since there are now only 18 teams left to select from, and only 2 left to be paired with), so what you have described is "All three promoted teams get drawn against sides from the top three where the top side selected each time is replaced back into the pot so that a top side can be selected more than once". To be fair it's probably a reasonable simplification to make and won't make a huge difference, so I won't mark you down for that though you have missed out on bonus points.

 

Arithmetic

 

(3/19*3/19*3/19) = 9/19

 

Very wrong I'm afraid.

 

If this were true and I tossed a fair coin 3 times, the probability of it giving heads all 3 times would be 1/2 * 1/2 * 1/2 = 3/2 or 150% and all mathematics would instantly break down!

 

The result you would be looking for (bear in mind that the question this actually answering is not the one you intended) is 27/6859 or 0.39% (ie 0.0039).

 

As an aside, the answer you got to satisfies the equation (3/19 + 3/19 + 3/19) = 9/19. This would answer the question "What is the probability of Southampton (only) being drawn against a team in the top 3 or a team in the next 3 (ie 4 to 6) or a team in the next 3 (ie 7 to 9)?" which simplifies to "What is the probability of Southampton (only) being drawn against a team in the top 9?" - hence the 9 out of 19 equation.

 

 

It isn't right at all, but provides an estimate that isn't too bad. The probabilities of these events occurring aren't independent (as you're assuming). Instead, we can quite easily enumerate the respective probabilities for all combinations of relevant outcomes as below (assuming equally likely trial outcomes, which is wrong, as we know certain clubs are paired, others would be controlled for police reasons etc., which in this case (if we are paired with Reading) increases the chance of a promoted side playing a top three side, as Reading and Saints can't play each other first day (and decreases the probability of Saints playing West Ham on the first day of the season) - which does deem that in the spoiler somewhat irrelevant, but if anyone cares, the maths is below (though probably wrong).

 

 

 

The probability of all three promoted clubs playing top three sides:

 

3/19 * 2/18 * 1/17 = 1/864

 

For any two promoted clubs to play top three clubs:

 

3/19 * 2/18 * 16/17 = 16/969

3/19 * 17/18 * 2/17 = 1/57

18/19 * 3/18 * 2/17 = 6/323

 

Summing these: the probability of any two promoted clubs playing clubs in the top three is 1/19.

 

Probability of only one promoted club playing a top three club:

 

3/19 * 16/18 * 15/17 = 40/323

16/19 * 3/18 * 15/17 = 40/323

16/19 * 15/18 * 3/17 = 40/323

 

Summing these: the probability of one promoted club playing a top three club is 120/323.

 

So, the total probability of at least one promoted club playing a top three club is the sum of these probabilities: 120/323 + 1/19 + 1/864 = 118961/279072 ~ 0.425

 

Because they're not independent events (that of a team playing a side in the top three), the probability of no promoted clubs playing top three clubs is ~ 0.575, which is quite a bit higher than your estimate.

 

I don't know why I've done this.

 

 

 

 

Actual probability of at least one promoted side playing a top three side is ~ 0.425.

 

Wrong also, young whippersnapper, back to Kumbakonam for you!

 

Your flaw is here:

 

For any two promoted clubs to play top three clubs:

 

3/19 * 2/18 * 16/17 = 16/969

3/19 * 17/18 * 2/17 = 1/57

18/19 * 3/18 * 2/17 = 6/323

 

It should of course be

 

3/19 * 2/18 * 16/17 = 16/969

3/19 * 16/18 * 2/17 = 16/969

16/19 * 3/18 * 2/17 = 16/969

 

Or you can more simply note that there are 3 ways to get 2 from 3 (since the one that isn't successful can come first, second or third only), and so multiply any one of them by 3 (you should note from the numerators and denominators that they are entirely commutative).

 

 

 

Further, you have made a rather inexcusable arithmetic error in

 

3/19 * 2/18 * 1/17 = 1/864

 

Of course it should equal 1/969.

 

 

 

 

So correcting your formula the probability of at least one promoted team playing a top 3 team is:

 

1/969 + 3 * (16/969) + 3 * (40/323) = 409/969

 

Which equals 42.2% (or 0.422).

 

 

Now I know that this is virtually the same number you quoted (and not even too far away from The9's answer, however arrived at), but I'm sure that one of the unwritten rules of the internet is that if you correct someone you have to be 100% correct yourself (the reason I will get very nervous once I submit this post ;))!

 

 

 

 

...of course what you should all have done is simply find out the probability of none of the promoted sides plays a top 3 side (16/19 * 15/18 * 14/17 = 560/969) and then subtracted this from 1 to get the probability of this NOT happening. So (1 - 560/969) = 409/969 as per the result above.

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(though probably wrong)

 

;)

 

Wrong also, young whippersnapper, back to Kumbakonam for you!

 

doesn't fit in this margin but i can show: zorn's lemma undecidable, well-ordering principle true, axiom of choice false. ;)

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