Talking of FA Cup Draws

Odds on that DT will make a negative post before the draw (If we're in it) and another as the draw concludes.

 

That's the calculation for the probability that the first match drawn is Man City v Arsenal.

If we don't care which team is at home the calculation would be 2/64 x 1/63

If you conducted the draw an infinite number of times, on average you would draw either Man City or Arsenal first once every 32 draws. And of those draws you would pair them together one time in 63.

But the draw doesn't end when the first match is decided. It continues.

The chances are neither Man City nor Arsenal will be drawn in that first tie.

There are now 62 teams in the draw. Another match is drawn. Another chance for a Man City v Arsenal game. And if neither club is drawn there's another chance from a draw of 60 teams. And so on.

I'm not sure of the exact maths for the calculation that two specific teams will face each other and I don't have the motivation to work it out. It's statistically improbable, but it's far more likely than the maths in this thread suggests.

 

I went for simplicity more or less to explain that the AND basically multiplies the odds.

I did have a gander to see if someone had done this calculation. And they have, for 2 specific teams to be drawn together in the 3rd round (at any point int the draw)

Unfortunately it's a maths paper, that's behind a paywall. And I wasn't paying it just to quote the odds on here and look all smart and that haha.

It will have allorts of "to the power of" and divisions in it I suspect to reflect the entire draw. I do accept that my odds somewhat only cover the 1st 2 teams out of the hat.

Far easier just to accept balls out of a bag and just marvel at the simplicity of it.

This is where the paper exists. Unsure if anyone is able to get hold of this without a payment though. And then we can all get a definitive answer rather than trash our brains

https://www.jstor.org/stable/10.4169/college.math.j.47.4.282

 

You can bet your arse that reciprocal will be mentioned at least a dozen times.

If you kept repeating the draw, but started again if Man City v Arsenal (or vice versa) isn't the first fixture out, even though you'd only have to draw one ball before starting again most of the time, it'd probably take days to get the fixture.

Going through the draw and only starting again when Man City and Arsenal are not paired and I reckon you'd get the fixture within an hour.

 

The probability of Arsenal being drawn against Man City is 1 in 63. Arsenal have to play one of 63 teams, each with equal probability.

 

The probability of an all Premier team tie is the total number of possible all Premier Ties ÷ the total number of Ties = 10÷32 = 0.3125. Since a probabilty can only be between 1 and 0. In betting terms 10÷32 is 32 to 10 or 3.2 to 1.

 

This isn't right.

 

Ok.... Ill put this here. Because in Math terms, you are looking at odds or probability of some event happening WITHOUT REPLACEMENT that also are intrinsically linked to each other as either the preceding or next event. (this is important)

The probability of picking out a Prem team out in the 3rd round draw starts at - 20/64
The probability of picking out a Prem team out next is - 19/63

Therefore the probability of picking out a prem tie in the 1st fixture is - 380/4032

which equates to - 0.094 (or 9.4%) or roughly 10.5 / 1 for a Prem Tie to be picked out at the 1st attempt

Now it doesn't take much more to say, that these odds drastically change through the draw as "without replacement", the odds can be anything depending on what you have picked out beforehand. However, the fact that you are wanting to link specific events (number drawn) linked in the next ball, means you have to multiply one event by the other. It will never ever be 1/63 for any 2 teams to be drawn. the only way these odds can happen, is if you start off with say Arsenal is already drawn. (Which you will of course then wiped out any mathematical equation of the 1st team being drawn out (ie the 20/64 which I noted first above)).

lets say if you get to tie number 5 (balls number 9 & 10 to come out of the bag) and no prem teams have yet come out, the odds/probability change to as follows

The probability of picking out a Prem team out at the 9th attempt would then be - 20/56
The probability of picking out a Prem team out next is - 19/55

Therefore the probability of picking out a prem tie as the 5th game ends up - 380/3080

which equates to - 0.123 (or 12.3%) or roughly 8 / 1

 

This question is actually very complex. However, fortunately an American professor of Mathematics (Patrick Sullivan) produced a paper on exactly this theme.

The likelihood of 20 premier league teams in a draw of 64, with no replacement, with no all premier league ties being drawn at all, is around 1.2%. In other words around 99% certainty there would be at least one all premier league tie amongst the 3rd round ties.

Interestingly this actually happened in 2005......I think...

 

That's the paper I linked above.

 

@nezbfc Great research buddy, I should read all the posts before dashing to answer.