Any mathematician's in the house?

[de-fence]

What are the odds of team A having got drawn against team B in the FA Cup 3rd round where there are 64 teams in the hat? For example what were the statistical odds of Man City getting drawn against Man Utd home or away at any point during the draw?Was having a discussion with a mate who studies Maths and I thought it would be simply 1/63 but he was offering up this theory that you had to multiply that by two because either team could be drawn first. Got a feeling he may be right but is there anyone who can offer a comprehensive answer? Cheers.

Edited 8 December, 2011 by de-fence

[Verbal]

Mathematicians no, but grammarians - plenty.

[aintforever]

if its home or away then 1/63 has to be right as Man City had 63 possible opponents.

[Whitey Grandad]

I make it 1 in 63. All outcomes are equally probable. As far as Man City (for example) are concerned there are 63 other teams they could meet, all equally likely.

 

 

 

I'm still thinking about it.

[Badger]

Just a thought, but the odds of team A being drawn are 1/64.

 

The odds of team B being drawn are 1/63.

 

So is the probability of A v B , is 1/ 64 X 63 ( that is 1 in 4032* ).

 

Probably not that simple

 

* ammended for error - misread calculator - FAIL

Edited 7 December, 2011 by Badger

[Badger]

if its home or away then 1/63 has to be right as Man City had 63 possible opponents.

 

but before a ball was drawn Man City had a 1 in 64 chance of being drawn.

[Whitey Grandad]

Just a thought, but the odds of team A being drawn are 1/64.

 

The odds of team B being drawn are 1/63.

 

So is the probability of A v B , is 1/ 64 X 63 ( that is 1 in 1032 ).

 

Probably not that simple

There are 32 games and team A could be any one of those. The team A could be either Man C or Man U so that doubles the chances so divide your figure by 64

[aintforever]

Just a thought, but the odds of team A being drawn are 1/64.

 

The odds of team B being drawn are 1/63.

 

So is the probability of A v B , is 1/ 64 X 63 ( that is 1 in 1032 ).

 

Probably not that simple

 

That's wrong, the odds in A being drawn is 1 as all balls are drawn out the bag.

 

The odds in A being the first out of the bag is 1/64 but there are 64 different times it can be drawn out the bag and still be paired with ball B so it cancels out.

Edited 7 December, 2011 by aintforever

[de-fence]

Nah that can't be right because its statistically certain that they will both get drawn at some point. If you said what are the odds of them both getting drawn and it had to be the first (for example) tie then it would be 1/64 x 1/63. But as it can occur at any point during the draw that 1/64 doesn't apply which leaves you with just 1/63. That's the way I saw it anyway.

 

I think what my mate was tryna say is that the odds of Man City getting drawn to Man Utd are 1/63. But also that the odds of Man Utd getting drawn against Man City are 1/63. Therefore as there's two possibilities you can multiply them to get 2/63. For some reason that just doesn't fit quite right with me.... though I couldn't tell you why.

[Baj]

I make it 1 in 2016, but what do I know...

[jawillwill]

If you're asking what are the odds of City drawing United once City had already been picked out of the hat, then the answer is 1/63.

 

If, on the other hand, you're asking what the odds of them being the first 2 teams out the hat (before any balls have been drawn at all), then the answer is 1/4032 (i.e. 1/(64 x 63)).

 

HTH (source: A* GCSE back in the day ).

[Baj]

Btw, I got that by taking the total of different combinations of games as being 64^2 - 64, then dividing those odds by 2 as they would appear twice in those total possible games (home and away each)

[Baj]

ah, so my maths gives them as the odds of being at a particular point in the draw as well, ok fair enough

[Baj]

ah yes, so if you take my amount and divide by 32 (total number of games in the draw) you get 1 in 63, ok, not far off, heheh

[aintforever]

I think what my mate was tryna say is that the odds of Man City getting drawn to Man Utd are 1/63. But also that the odds of Man Utd getting drawn against Man City are 1/63. Therefore as there's two possibilities you can multiply them to get 2/63. For some reason that just doesn't fit quite right with me.... though I couldn't tell you why.

 

That's because they are the same event, unless you are talking home or away.

[Whitey Grandad]

Man City have got to play somebody, and there are 63 possibilities for the other team, each with an equal probability of being chosen. (If not the draw would be biased).

[Badger]

If you're asking what are the odds of City drawing United once City had already been picked out of the hat, then the answer is 1/63.

 

If, on the other hand, you're asking what the odds of them being the first 2 teams out the hat (before any balls have been drawn at all), then the answer is 1/4032 (i.e. 1/(64 x 63)).HTH (source: A* GCSE back in the day ).

 

That was the starting point for my theory - although initially misread the answer on calculator ! Eyesight obviously better in the old days,( 'O' level, predating GCSE)

[View From The Top]

1/4032. That's either City v Utd or Utd v City, before a ball has been drawn.

[pedg]

Before the draw starts the odds that 2 specific teams will meet in 1/63. Either Team A is drawn first and there is a 1/63 of team B being draw or team B is drawn first in which case there is a 1/63 chance of Team A being draw. As each of these scenarios (Team A drawn first or Team B drawn first) has a 1/2 chance of happening the odds are Team A at home to Team B 1/126, Team B at home to Team A 1/126, Team A and Team B meeting irrespective of who is at home 1/63.

 

In the case of the FA cup draw obviously as teams that are neither Team A or Team B are drawn together the chances of Team A and Team B meeting increased as the draw progresses to the obvious final scenario where they are the only 2 teams left at which point it is certain.

 

As the question says "During the draw" rather than "first out" the odds before a ball is drawn need not take into account the number of ties.

Edited 7 December, 2011 by pedg

[de-fence]

If you're asking what are the odds of City drawing United once City had already been picked out of the hat, then the answer is 1/63.

 

If, on the other hand, you're asking what the odds of them being the first 2 teams out the hat (before any balls have been drawn at all), then the answer is 1/4032 (i.e. 1/(64 x 63)).

 

HTH (source: A* GCSE back in the day ).

 

Yea I get that. But the question is for the two to be drawn against each other at any point during the draw with either side being home. I think it gets more complicated than just 1/63 but couldn't tell you why. Hoping someone could give me the comprehensive answer I'm after. Sorry if this is boring people btw, very tedious I know just like most stats.

[Whitey Grandad]

1/4032. That's either City v Utd or Utd v City, before a ball has been drawn.

Isn't that just for the first tie drawn? If so, shouldn't it be half of that?

[de-fence]

Before the draw starts the odds that 2 specific teams will meet in 1/63. Either Team A is drawn first and there is a 1/63 of team B being draw or team B is drawn first in which case there is a 1/63 chance of Team A being draw. As each of these scenarios (Team A drawn first or Team B drawn first) has a 1/2 chance of happening the odds are Team A at home to Team B 1/126, Team B at home to Team A 1/126, Team A and Team B meeting irrespective of who is at home 1/63.

 

In the case of the FA cup draw obviously as teams that are neither Team A or Team B are drawn together the chances of Team A and Team B meeting increased as the draw progresses to the obvious final scenario where they are the only 2 teams left at which point it is certain.

 

As the question says "During the draw" rather than "first out" the odds before a ball is drawn need not take into account the number of ties.

 

I think you're cracked it my friend

[pedg]

Another way to think about it is to forget about the sequential nature of the draw and think of 32 pairs of holes at the bottom of a box and that you tip in 64 balls and shake the box till one ball goes in each hole. Thinking about it that way it should be clearer, its the fact that when determining the odd's you do not need to know about how the random arranging of teams is done, only that you end up with them all paired. They could be done all at once or sequentially it does not matter as you end up with the same result.

[1976_Child]

one add feking one is more than one.

 

,gfeing blag.h.h.hhhhhh

[hamster]

43.

[1976_Child]

You don't so much need a numberist as a basic refresher course in grammar and syntax.

 

"mathematicians" : two or more numberists,

 

"mathematician's [..x..] : [..x..] which belongs to the numberist.

 

Honestly.

Edited 8 December, 2011 by 1976_Child

[Clapham Saint]

If you're asking what are the odds of City drawing United once City had already been picked out of the hat, then the answer is 1/63.

 

If, on the other hand, you're asking what the odds of them being the first 2 teams out the hat (before any balls have been drawn at all), then the answer is 1/4032 (i.e. 1/(64 x 63)).

 

HTH (source: A* GCSE back in the day ).

 

I endorse this message

[Frank's cousin]

A more laborious way, but maybe simpler to understand is to think of it as a 64 team league - which means that each team will play 63 home games and 63 away. over 126 weeks assuming 3pm saturdays ;-)

 

so if you pick any given saturday (effectively mirroring a random draw) , the odds that City will be at home is and playing Man U would be 1/126, ( ie it only happens in one week out the 126 saturadys) The odds that city will be at away at Man U will also be 1/126, meaning that odds of either would be 2 x 1/126 = 1/63

[badgerx16]

This all presupposes that the draw isn't rigged.

[Clapham Saint]

If we want to get all MLG about it...

 

There are 5.026 x10^57 different combinations in which the draw could fall.

 

8.734 x10^55 of these result in a Manchester Derby.

 

(8.734 x10^55)/(5.026 x10^57) = 1/63

[Whitey Grandad]

If we want to get all MLG about it...

 

There are 5.026 x10^57 different combinations in which the draw could fall.

 

8.734 x10^55 of these result in a Manchester Derby.

 

(8.734 x10^55)/(5.026 x10^57) = 1/63

 

Some of us did that in our heads

[Lighthouse]

Just to clarify this once and for all

 

Chance of City vs United = 1/63

Chance of City at home to Utd = 1/126 = 1/2 x 1/63

Chance of City vs Utd being first match drawn = 1/2016 = 2(1/64 x 1/63)

Chance of City at home to Utd being first match drawn = 1/4032 = 1/64 x 1/63

[norwaysaint]

[h=2]Any mathematician's in the house?[/h]

 

You might want to ask for a bit of help with your native language too.

[keithd]

i see it as being 50-50. They either got drawn against each other or they didnt

 

Move on....

 

Next....

[de-fence]

You might want to ask for a bit of help with your native language too.

Congratulations...

[teamsaint]

what are the odds of the Euro 2012 draw being fixed.........

 

Is an easier question to answer.

[pedg]

An easier statistic is what are the odds that the manchester clubs will be drawn together in the knock out stages of the champions league.

[Window Cleaner]

Now I always thought that questions of probablity can always be calculated by using factorials.

 

In this case it's factorial 64 divided by factorial 64-2 isn't it?

 

This equates simple to a one in 64x63 chance ie 4032/1

[Dibden Purlieu Saint]

1/8. Fact.

[Stubby]

if its home or away then 1/63 has to be right as Man City had 63 possible opponents.

 

This

[Whitey Grandad]

Now I always thought that questions of probablity can always be calculated by using factorials.

 

In this case it's factorial 64 divided by factorial 64-2 isn't it?

 

This equates simple to a one in 64x63 chance ie 4032/1

It's a combination (I think). There are 64 opportunities for your event (32 home, 32 away) so it's 4032/64 which = 63

[Doctoroncall]

Now I always thought that questions of probablity can always be calculated by using factorials.

 

In this case it's factorial 64 divided by factorial 64-2 isn't it?

 

This equates simple to a one in 64x63 chance ie 4032/1

 

That's the odds of those two being drawn first and second.

[Jones91]

So what are the chances of them playing each other in the last 32 of the Europa cup?

[Badger]

So what are the chances of them playing each other in the last 32 of the Europa cup?

 

considerably better than meeting in the Champions League.

[sandwichsaint]

A more laborious way, but maybe simpler to understand is to think of it as a 64 team league - which means that each team will play 63 home games and 63 away. over 126 weeks assuming 3pm saturdays ;-)

 

so if you pick any given saturday (effectively mirroring a random draw) , the odds that City will be at home is and playing Man U would be 1/126, ( ie it only happens in one week out the 126 saturadys) The odds that city will be at away at Man U will also be 1/126, meaning that odds of either would be 2 x 1/126 = 1/63

 

I'm not sure our squad is quite up to this. How many transfer windows are you proposing in a 126 week season?

[Smirking_Saint]

Thank f*ck its not a J number eh ??

[sebastian firefly]

what are the odds of some one picking the two "hot balls " out of the bag

[Patrick Bateman]

Just a thought, but the odds of team A being drawn are 1/64.

 

The odds of team B being drawn are 1/63.

 

So is the probability of A v B , is 1/ 64 X 63 ( that is 1 in 4032* ).

 

Probably not that simple

 

* ammended for error - misread calculator - FAIL

 

This is right, it's 1/4032

0.02%

[Patrick Bateman]

Although, that assumes that ALL the balls are in the bag! It's 1/4032 to be the first tie;

 

1/64 x 1/63 = 1/4032

 

If there were only 14 balls left, it would have been;

 

1/14 x 1/13 = 1/182

[Whitey Grandad]

Although, that assumes that ALL the balls are in the bag! It's 1/4032 to be the first tie;

 

1/64 x 1/63 = 1/4032

 

If there were only 14 balls left, it would have been;

 

1/14 x 1/13 = 1/182

But only if they had not already been drawn. After that the odds are pretty sh!t.