5th round fixture

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If a local team is drawn first, that’s the only chance there will be of winning that bet. So the new information is that very fact. So probability is moot other than the 1/15 chance which no one is arguing about. Any other of the 14 teams drawn first also gives new information. That the derby draw is still possible as long as neither side comes out second ball. That’s because the betting is in play after the first ball.
Nail on head. Why didn’t we think of this before? Before a ball is drawn the probability is 1/15. If a Sheffield club comes out first the probability is still 1/15. Well done.
 

I never said it doesn’t change as the draw progresses, it just only changes after the even numbered balls. You are harking back to the 2/15 chance of one of the Sheffield teams coming out as ball 2, but that isn’t all that’s left of the draw.

I can see you’ve edited, but you asked for my calculation so here you go.


Probability of Sheffield Derby = P(sd)
Probability of Sheffield Derby as Tie 1 P(t1)... etc
Probability of drawing Sheffield club = P(s)
Probability of drawing any other club = P(o)

We'll take the probability of each ball in order giving us the outcome of a Sheffield Derby until we get the two clubs out together, so P(t) = P(b1) x P(b2) x P(b3), etc. We could do all 16 balls each time, but once we have both Sheffield clubs out then we're just left with P(o) = 1 for the remaining balls.

Here we go again (with the assumption that we know ball one is other)

P(t1) = P(s) x P(s) = 0 x 2/15 = 0
P(t2) = P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 2/14 x 1/13 = 1/105
P(t3) = P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 12/14 x 11/13 x 2/12 x 1/11 = 1/105
P(t4) = P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 2/10 x 1/9 = 1/105
P(t5) = P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 8/10 x 7/9 x 2/8 x 1/7 = 1/105
P(t6) = P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 8/10 x 7/9 x 6/8 x 5/7 x 2/6 x 1/5 = 1/105
P(t7) = P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 8/10 x 7/9 x 6/8 x 5/7 x 4/6 x 3/5 x 2/4 x 1/3 = 1/105
P(t8) = P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 1 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 8/10 x 7/9 x 6/8 x 5/7 x 4/6 x 3/5 x 2/4 x 1/3 x 2/2 x 1/1 = 1/105

P(sd) = P(t1) + P(t2) + P(t3) + P(t4) + P(t5) + P(t6) + P(t7) + P(t8)
P(sd) = 0 + 1/105 + 1/105 + 1/105 + 1/105 + 1/105 + 1/105 + 1/105
P(sd) = 7/105
P(sd) = 1/15
I think we can safely assume the probability of getting the pigs in the cup are now precisely 0 out of 10.
 
It's the same principle. Several people have proved you wrong you but you insist your version of reality is correct having, in fact, simply misanalysed the issue

No gambler in their right mind would take the odds offered before the first ball is drawn, after the first team is drawn unless it’s a Sheffield team. That’s the reality.



Didn’t see this before my last comment last comment.
 
Nail on head. Why didn’t we think of this before? Before a ball is drawn the probability is 1/15. If a Sheffield club comes out first the probability is still 1/15. Well done.

My point, as you well know, was about a non local side being drawn first. Nowhere have I suggested that’s not the case. Go and sit with the other smart arse on the smug step.
 
My point, as you well know, was about a non local side being drawn first. Nowhere have I suggested that’s not the case. Go and sit with the other smart arse on the smug step.
But how could 14 balls cause the probability to change if drawn first, but the other two not? I’ve calculated that it doesn’t change for those 14 balls, can you show me an alternate answer?
 
My point, as you well know, was about a non local side being drawn first. Nowhere have I suggested that’s not the case. Go and sit with the other smart arse on the smug step.
In our defence its difficult not to be smug in this situation
 
There are 15 teams or pairs who we could possibly be drawn against, so there is a 1 in 15 chance we could be drawn against any of them. 1 in 15 chance we draw Wednesday, 1 in 30 chance it's at Bramall Lane, one in 30 chance it's at Hillsborough. Imagine someone carries a bag with all our potential opponents/pairs written down on screwed up bits of paper, in that bag are the names of our 5th round opponents; there's a 1 in 15 chance it could be any of them.
What he said - 1 in 30. Order doesn't matter. There are 15 possible opponents and two possible formats (H/A) so 30 different possible outcomes, all equally likely
 
See post 164 I was the probably first one to show you how misguided you were and 200 posts later you still haven't grasped it
 
But how could 14 balls cause the probability to change if drawn first, but the other two not? I’ve calculated that it doesn’t change for those 14 balls, can you show me an alternate answer?

As I’ve said, l think there are two additional events that could occur to change the odds. Any explanation as to why that isn’t the case without using probability and taking into account that we’re in play would be welcome and I’d be happy, and embarrassed, to fess up. In other words, can you simplify it by telling me where l’m going wrong based on the above. I’d appreciate it, seriously.
 
Right apologies that I've missed the last 7 pages of this
So are we working out if a non Sheffield team being drawn out as the 1st ball affects the probability of a Sheffield derby from after that point?
 

Right apologies that I've missed the last 7 pages of this
So are we working out if a non Sheffield team being drawn out as the 1st ball affects the probability of a Sheffield derby from after that point?


It’s Reading or Cardiff So my whole “it has to be Liverpool first” argument ( ? ) is clearly out of the window. Aah well......
 
As I’ve said, l think there are two additional events that could occur to change the odds. Any explanation as to why that isn’t the case without using probability and taking into account that we’re in play would be welcome and I’d be happy, and embarrassed, to fess up. In other words, can you simplify it by telling me where l’m going wrong based on the above. I’d appreciate it, seriously.
The two events being that either Sheffield team could be drawn as ball two? They were already included in the original probability before a ball was drawn, so that calculation still stands. They were always possible, so drawing one ball hasn’t changed that. Pairing off two teams is the first significant event that affects the outcome.

The thing is, as much as you want an explanation without the maths, maths is how we get to the right answer. I can see that it could seem counter-intuitive that drawing one ball doesn’t make a difference, but is a hunch or a gut feeling, calculating it is the way to prove or disprove it.
 
It’s Reading or Cardiff So my whole “it has to be Liverpool first” argument ( ? ) is clearly out of the window. Aah well......
Well I'm now upset cause with a bit of filtering and tinkering, my spreadsheet could've solved this and saved the last 7 pages of debate :(;)
 
The two events being that either Sheffield team could be drawn as ball two? They were already included in the original probability before a ball was drawn, so that calculation still stands. They were always possible, so drawing one ball hasn’t changed that. Pairing off two teams is the first significant event that affects the outcome.

The thing is, as much as you want an explanation without the maths, maths is how we get to the right answer. I can see that it could seem counter-intuitive that drawing one ball doesn’t make a difference, but is a hunch or a gut feeling, calculating it is the way to prove or disprove it.

Thanks for that. I get it. Can l ask you about the actual odds, which is where I’ve been coming from since the beginning? To clarify.

What are the actual odds for an all local draw with sixteen teams.

What are the actual odds for a Sheffield team to draw any other team

What are the actual odds for any team to be drawn first?

What are the actual odds for either local side to be drawn second?

What are the actual odds for either local side to be drawn second?

l’m not being a twat, l’d like your input on the odds.
 
So probably of an all Sheffield tie in this year's FA Cup now the Pigs have just got Man City is zero, right?
😂
 
What are the actual odds for an all local draw with sixteen teams.

1 in 15.

What are the actual odds for a Sheffield team to draw any other team

14 in 15.

What are the actual odds for any team to be drawn first?

For any given team, it is 1 in 16.

Of course, the chance of ‘a team’ being drawn first is 100%.

What are the actual odds for either local side to be drawn second?

Before a ball is drawn, 1 in 16.

After the first ball is drawn:
  • If the first ball is a non-Sheffield team, 1 in 15 for a given team, or 2 in 15 for either Sheffield team.
  • If the first ball is a Sheffield team, it’s 1 in 15 for the other team.
 
This thread has kept me thoroughly entertained while waiting news about Berge and Retsos. Thanks to all involved.

I am firmly in the one in fifteen camp. However, to answer Sean Thornton in a different manner. Yes, we do have some new information if Liverpool are drawn first, but it doesn't make any difference to the answer.

If Liverpool are drawn first, there is a 2 in 15 chance of this scuppering a Blades vs Wendy tie after the second ball is drawn (as there is a 1 in 15 chance of either the Blades or Wendy being the second ball out). Therefore, there is a 13 in 15 chance that Liverpool being drawn first won't destroy the chance of a Sheffield derby elsewhere in the draw.

If that doesn't happen, there are now only 14 balls left in the bag (if you'll pardon the imagery). That means the odds of an all Sheffield draw becomes a 1 in 13 for the remaining games.

The probability of an all Sheffield derby after Liverpool is drawn first but before the second ball is drawn is dependent on both of these events: 13 in 15 multiplied by 1 in 13. This will give you 13 in 195 chances of the Derby coming up elsewhere in the draw... Which simplifies to a 1 in 15 probability.

I hope this helps. If not, I can safely say there is a 1 in a million chance of an all Sheffield derby in the next round given the sty will turn into an abattoir when City visit.
 
This thread has kept me thoroughly entertained while waiting news about Berge and Retsos. Thanks to all involved.

I am firmly in the one in fifteen camp. However, to answer Sean Thornton in a different manner. Yes, we do have some new information if Liverpool are drawn first, but it doesn't make any difference to the answer.

If Liverpool are drawn first, there is a 2 in 15 chance of this scuppering a Blades vs Wendy tie after the second ball is drawn (as there is a 1 in 15 chance of either the Blades or Wendy being the second ball out). Therefore, there is a 13 in 15 chance that Liverpool being drawn first won't destroy the chance of a Sheffield derby elsewhere in the draw.

If that doesn't happen, there are now only 14 balls left in the bag (if you'll pardon the imagery). That means the odds of an all Sheffield draw becomes a 1 in 13 for the remaining games.

The probability of an all Sheffield derby after Liverpool is drawn first but before the second ball is drawn is dependent on both of these events: 13 in 15 multiplied by 1 in 13. This will give you 13 in 195 chances of the Derby coming up elsewhere in the draw... Which simplifies to a 1 in 15 probability.

I hope this helps. If not, I can safely say there is a 1 in a million chance of an all Sheffield derby in the next round given the sty will turn into an abattoir when City visit.


Thank you. I shall try to take it in :(
 

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