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Any mathematician's in the house?


de-fence

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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 by de-fence
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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 by Badger
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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

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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 by aintforever
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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.

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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 :smug:).

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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.

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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 :smug:).

 

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)

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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 by pedg
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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 :smug:).

 

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.

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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

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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.

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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 :smug:).

 

I endorse this message

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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

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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

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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

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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.

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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?

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