5th round fixture

[Elsecar_Blade]

They can't come out anywhere in the draw though. They have to come out successively and as stated in my previous reply 1st and 2nd or 3rd and 4th or 5th and 6th etc. For them to come out 3rd and 4th they cannot come out 1st or 2nd so you have the odds of that not happening as well. ucandomagic did the odds.

But United can come out anywhere. The only variable you're only interested in whether Wednesday come out in the opposite spot. It doesn't matter a jot where United come out, only whether Wednesday come out opposite to the odds. Therefore, there's 1 possible spot opposite United and 14 other spots that aren't which makes it a 1 in 15 shot or 14/1.

Yes, the odds change as you move through the draw because basically the pot is shrinking. If you're asking what are the odds NOW of a Sheffield derby in Round 5, it's 14/1. But if Liverpool come out first, you then need neither Sheffield side second out against them and then, wherever United are drawn after that, United to be opposite Wednesday. If I could be bothered, I'd work out how the odds of a Sheffield derby change as we move through the draw ... but I can tell you that right now it's 14/1.

 

[Bruce Wayne]

If 14 are drawn out and neither Sheffield team has yet been drawn then the likelihood of a Sheffield derby is 100%.

Why are Hallam FC still in?

 

[MobileBlade]

Found this on line in a paper published by the Royal Statistical Society.

"The BBC website recently published a feature using the FA Cup draw to highlight basic calculations in probability. As an example, in a round with 32 teams, the probability of any two teams being drawn to play each other is 1/

But United can come out anywhere. The only variable you're only interested in whether Wednesday come out in the opposite spot. It doesn't matter a jot where United come out, only whether Wednesday come out opposite to the odds. Therefore, there's 1 possible spot opposite United and 14 other spots that aren't which makes it a 1 in 15 shot or 14/1.

No. There has to be two consecutive balls, one for United, One for Wednesday. They have to come out in the right position. United can't come out 2nd and Wednesday 3rd for instance. If United came out first then yes it would be 1 in 15. But there is only a 1 in 16 chance of United coming out first. Thherefore the chance of United or Wed 1st and Wed 2nd is 1 in 120. If one of them doesn't come out 1st, they can't come out 2nd and so it becomes 1 in 14 of coming out 3rd and 1 in 13 of coming out 4th . However for this to happen you also have the 14 out of 16 chance them not coming out first and the 13 out of 15 chance of them not coming out 2nd.

So now the formula becomes 1/120 + (14/16 x 13/15 x 2/14 x 1/13). You carry that on all the way through as fractions obviously not divisions.

 

[MobileBlade]

That BBC link doesnt seem to work so perhaps a simple example with show the reasoning.

Suppose both Sheffield teams get to the semi-final with say Man City and Liverpool. There are 6 possible combinations of matches SU v SW, SU v MC, SU v L, SW v MC, SW v L , L v MC. There are 2 semi finals So the chances are 1 in 6 times two matches equals 1 in 3

Relating this to the round with 16 teams, there are 240 possible combinations and 8 matches so the odds are 8/240 or 1 in 15

You are right about the first formula but wrong on 2nd. Number of matches doesn't come into it. Its just the combinations. So the number of combinations is therefore 2 in 240 or 1 in 120.

 

[Elsecar_Blade]

No. There has to be two consecutive balls, one for United, One for Wednesday. They have to come out in the right position. United can't come out 2nd and Wednesday 3rd for instance. If United came out first then yes it would be 1 in 15. But there is only a 1 in 16 chance of United coming out first. Thherefore the chance of United or Wed 1st and Wed 2nd is 1 in 120. If one of them doesn't come out 1st, they can't come out 2nd and so it becomes 1 in 14 of coming out 3rd and 1 in 13 of coming out 4th . However for this to happen you also have the 14 out of 16 chance them not coming out first and the 13 out of 15 chance of them not coming out 2nd.

So now the formula becomes 1/120 + (14/16 x 13/15 x 2/14 x 1/13). You carry that on all the way through as fractions obviously not divisions.

How many other teams are called Sheffield Wednesday? 1
How many other teams are not called Sheffield Wednesday? 14

Right now, it's 14-1.

The chances of United vs. Wednesday coming out as the first tie? 1/8 x 1/15 = 1/120 ... or 119-1. As you have said. You could work out all the other chances of it being the second tie, third tie etc and add them all up - and they'll add together to make the decimal equivalent of 1/15 ... or 14-1.

If you're specifically talking about United vs. Wednesday at Bramall Lane as the first tie? 1/16 x 1/15 = 1/240 ... or 239-1.

 

[ScottishBlade93]

who cares about odds of drawing them let’s just see who we get and take it from there? I fancy us against anyone at the lane.

 

[MobileBlade]

How many other teams are called Sheffield Wednesday? 1
How many other teams are not called Sheffield Wednesday? 14

Right now, it's 14-1.

The chances of United vs. Wednesday coming out as the first tie? 1/8 x 1/15 = 1/120 ... or 119-1. As you have said. You could work out all the other chances of it being the second tie, third tie etc and add them all up - and they'll add together to make the decimal equivalent of 1/15 ... or 14-1.

If you're specifically talking about United vs. Wednesday at Bramall Lane as the first tie? 1/16 x 1/15 = 1/240 ... or 239-1.

The question was the chances of there being an all Sheffield game. 1 in 15 is only if one of them is drawn first. To calculate the chance you have to consider them coming out 3rd and 4th, 5th and 6th or 7th and 8th against them coming out in any other order such as 6th and 15th.

 

[anglian blade]

You are right about the first formula but wrong on 2nd. Number of matches doesn't come into it. Its just the combinations. So the number of combinations is therefore 2 in 240 or 1 in 120.

Unfortunately you are wrong. If you dont believe it then you can argue with the Royal Statistical Society and the American Statistical Association https://www.statslife.org.uk/sports...your-team-getting-four-home-fixtures-in-a-row

 

[MobileBlade]

Unfortunately you are wrong. If you dont believe it then you can argue with the Royal Statistical Society and the American Statistical Association https://www.statslife.org.uk/sports...your-team-getting-four-home-fixtures-in-a-row

Thats not quite to do with this question. That talks about chances of ANY 2 clubs being drawn against each other. In this case we are talking about two specific clubs being drawn against each other. Not the same.

 

[Corazon]

The question was the chances of there being an all Sheffield game. 1 in 15 is only if one of them is drawn first. To calculate the chance you have to consider them coming out 3rd and 4th, 5th and 6th or 7th and 8th against them coming out in any other order such as 6th and 15th.

But that's exactly what Wombwell said. If you do all those calculations and add them up, the answer will be 0.666 recurring - the decimal equivalent of a 1 in 15 chance, or 14/1.

 

[MobileBlade]

But that's exactly what Wombwell said. If you do all those calculations and add them up, the answer will be 0.666 recurring - the decimal equivalent of a 1 in 15 chance, or 14/1.

You can't add up then, I'll say no more.

 

[anglian blade]

Thats not quite to do with this question. That talks about chances of ANY 2 clubs being drawn against each other. In this case we are talking about two specific clubs being drawn against each other. Not the same.

I think you misunderstand what they mean by ANY. The chances of us playing wednesday are exactly the same as for us playing any of the other 14, 1 in 15. If you mean what are the chances of any two unnamed clubs playing each other that of course will be 100%,

 

[Corazon]

You can't add up then, I'll say no more.

Oops, missing zero. Should have been 0.0666 recurring obviously

 

[anglian blade]

It's exactly the same odds whoever is drawn first. Each team will definitely be drawn at some point whether it's first ball or sixteenth ball. It's who they are playing that affects the odds.

To give the maths behind that:

If Liverpool are drawn first the chances of the 2nd ball not being a Sheffield one is 13/15. The chances in the rest of the draw is now 1 in 13, (only 14 teams left ) . Multiply the two and the result is 1 in 15

 

[MobileBlade]

To give the maths behind that:

If Liverpool are drawn first the chances of the 2nd ball not being a Sheffield one is 13/15. The chances in the rest of the draw is now 1 in 13, (only 14 teams left ) . Multiply the two and the result is 1 in 15

There are 16 balls in a bag. You needs balls 4 and 8 to come out in either order consecutively either 1st and 2nd or 3rd and 4th or 5th and 6th or 7th and 8th (and so on to 15th and 16th). You are saying there is a 1 and 15 chance of that happening?

 

[Barney Rubble]

There are 16 balls in a bag. You needs balls 4 and 8 to come out in either order consecutively either 1st and 2nd or 3rd and 4th or 5th and 6th or 7th and 8th (and so on to 15th and 16th). You are saying there is a 1 and 15 chance of that happening?

At this moment in time the odds on us being drawn out 1st and 2nd are 120/1. There are equal chances of being drawn together on the 7 subsequent draws . So a total of 8 draws, each with a chance of 1 jn 120. So 120 should be divided by 8, not multiplied and 120/8 =15, so a 14/1 chance of us being drawn together.

 

[MobileBlade]

At this moment in time the odds on us being drawn out 1st and 2nd are 120/1. There are equal chances of being drawn together on the 7 subsequent draws . So a total of 8 draws, each with a chance of 1 jn 120. So 120 should be divided by 8, not multiplied and 120/8 =15, so a 14/1 chance of us being drawn together.

You are forgetting for them to come out 3rd and 4th then one of them can't be drawn 1st or 2nd. Similarly for them to come out 5th and 6th one of them can't come out in the first 4.

If the odds were that short, there would be millions more winners each time the lottery was drawn. Instead you have more chance of being struck by lightening.

 

[MobileBlade]

Actually for those doubting this article has far more numbers involved but the principle is the same:

http://www.webmath.com/cgi-bin/lottery.cgi?count=15&low=1&high=20

 

[AWetFlannel]

Not sure that the mechanics of the draw change the probability.

Once teams start to be drawn it obviously all changes, but before the draw begins we have a 1 in 15 chance of being drawn against the pigs.

 

[Sabella's Socks]

There are 16 balls in a bag. You needs balls 4 and 8 to come out in either order consecutively either 1st and 2nd or 3rd and 4th or 5th and 6th or 7th and 8th (and so on to 15th and 16th). You are saying there is a 1 and 15 chance of that happening?

United have to play someone in the next round.
There are 15 teams we can play
1 of those teams is W*******y
Therefore the probability of playing W*******y is 1 in 15

There are all sorts of Home/Away, this ball/that ball combinations, but they just cancel each other out.

It's a 1 in 15 chance that our opponents will be W*******y.

 

[MobileBlade]

United have to play someone in the next round.
There are 15 teams we can play
1 of those teams is W*******y
Therefore the probability of playing W*******y is 1 in 15

There are all sorts of Home/Away, this ball/that ball combinations, but they just cancel each other out.

It's a 1 in 15 chance that our opponents will be W*******y.

As I've said about 100 times, only if United or Wednesday are the first ball out of the bag.

 

[Sabella's Socks]

As I've said about 100 times, only if United or Wednesday are the first ball out of the bag.

One of those factors that cancels out.

We have to play someone. There are fifteen teams we can play. One of them is W*******y. Therefore it's 1 in 15.

(Out of interest, do you think our chances of being drawn against Man United are more or less than our chances of being drawn against W*******y, or are they the same.)

 

[MobileBlade]

One of those factors that cancels out.

We have to play someone. There are fifteen teams we can play. One of them is W*******y. Therefore it's 1 in 15.

(Out of interest, do you think our chances of being drawn against Man United are more or less than our chances of being drawn against W*******y, or are they the same.)

They are the same but they are not 1 in 15. They are 1 in 120 for it to be us and them first two out of hat.

 

[Sabella's Socks]

They are the same but they are not 1 in 15. They are 1 in 120 for it to be us and them first two out of hat.

Ok. I think we are probably at cross purposes. Possibly.

The probability of a Sheffield derby being the first fixture out of the bag is 1 in 120.

The probability of a Sheffield derby being any of the fixtures is 1 in 15.

 

[Barney Rubble]

You are forgetting for them to come out 3rd and 4th then one of them can't be drawn 1st or 2nd. Similarly for them to come out 5th and 6th one of them can't come out in the first 4.

If the odds were that short, there would be millions more winners each time the lottery was drawn. Instead you have more chance of being struck by lightening.

I think that the chance of one of the teams being drawn before the other is counterbalanced by the increased likelihood of us being drawn together if they've not already been drawn out. Yes there are billions of total combinations of a draw of 16 teams, but in 1 in 15 of all those combinations we would be drawn together.. I think 5,420,511,625,866.6 times out of 1,307, 674,368,000 possible permutations produces this outcome, but that couild be complete bollocks.

 

[MobileBlade]

Ok. I think we are probably at cross purposes. Possibly.

The probability of a Sheffield derby being the first fixture out of the bag is 1 in 120.

The probability of a Sheffield derby being any of the fixtures is 1 in 15.

Nope because the two balls have to come out consecutively at a specific point. They cannot come out 2nd and 3rd for instance.

Try a different way. Your local club has a lottery. There are 16 numbers to pick from. You can pick two numbers. To win both your numbers have to come out. How many entries are there and what are your chances of winning?

 

[Punk Blade]

They are the same but they are not 1 in 15. They are 1 in 120 for it to be us and them first two out of hat.

We have to draw somebody. The probabilities of us drawing each other ball in the draw must therefore sum to 1.

If the probability of us drawing each ball is 1/15, with there being 15 other balls, 1/15 x 15 is 1.

If the probability of us drawing each ball is 1/120 then 15 balls x 1/120 is 15/120. What are the other 105 potential outcomes?

 

[Punk Blade]

Nope because the two balls have to come out consecutively at a specific point. They cannot come out 2nd and 3rd for instance.

Try a different way. Your local club has a lottery. There are 16 numbers to pick from. You can pick two numbers. To win both your numbers have to come out. How many entries are there and what are your chances of winning?

If it’s like the cup draw where ALL 16 balls get drawn, then everyone’s entering, everyone will definitely win and the club will go bankrupt.

 

[MobileBlade]

We have to draw somebody. The probabilities of us drawing each other ball in the draw must therefore sum to 1.

If the probability of us drawing each ball is 1/15, with there being 15 other balls, 1/15 x 15 is 1.

If the probability of us drawing each ball is 1/120 then 15 balls x 1/120 is 15/120. What are the other 105 potential outcomes?

That is explained much further up. You have 16 balls in a hat. Means you have a 2/16 chance of you or opponent you want coming out first and a 1/15 chance coming out 2nd. 1/15 x 2/16 = 2/240 or 1/120. If you want to add it up its 15+14+13+12 etc for combinations.

 

[Blade83]

Or to calculate the probability using the permutations logic, we need to work out the number of permutations that include us against Weds divided by the total number of possible draws without that restriction, so:

- number of combinations that put us together: There are 8 draws that will put us together 1&2, 3&4 etc and each can be reversed (United 1st or United 2nd) and we need to times this by 14! (14*13*12 etc), which is the number of possible orders for the other 14 clubs in the draw once we know one pair is us vs Weds;

- divided by the number of combinations of all possible draws, which is just 16!

This gives (0.067 or 1 in 15).