5th round fixture

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Sean Thornton said:
The moment anyone is drawn we are in play with new information. Liverpool is just an example. It could be anyone of 14 clubs.
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What information about who plays who do you learn from a random team being drawn first? We already knew a random team would be drawn first so that’s not new information.

Alternatively, what would be the new information on who plays who if SUFC were drawn first? What draws would become more likely? We’ve got to learn something for the probability to change.
 
Geordie Blade said:
Show your workings. What are the exact odds?
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I’ve said, longer than 15/1. They have to be.

There's an event that has to occur before 15/1 comes back into play. An event that could affect the price because of the chance that a derby will take place. If you’re saying that that won’t affect the 15/1 then you aren’t understanding the point. Whatever the odds actually are, they have to be longer because of the event yet to occur . Only if both local teams balls were taken out of the bag before the second team were drawn could the odds remain at 15/1. Probability before any team is drawn can’t possibly nullify that fact.
 
Blade83 said:
What information about who plays who do you learn from a random team being drawn first? We already knew a random team would be drawn first so that’s not new information.

Alternatively, what would be the new information on who plays who if SUFC were drawn first? What draws would become more likely? We’ve got to learn something for the probability to change.
Click to expand...

Anyone of 14 teams being drawn first is new information as it affects the chance of a derby In that first tie. Any random team would include both Sheffield teams. There’s the difference.

The probability changes after the first ball is drawn as we are now in play and there are two possible draws that can affect the probability of 15/1 IN THAT FIRST TIE. That’s what we learn.
 
Sean Thornton said:
I’ve said, longer than 15/1. They have to be.

There's an event that has to occur before 15/1 comes back into play. An event that could affect the price because of the chance that a derby will take place. If you’re saying that that won’t affect the 15/1 then you aren’t understanding the point. Whatever the odds actually are, they have to be longer because of the event yet to occur . Only if both local teams balls were taken out of the bag before the second team were drawn could the odds remain at 15/1. Probability before any team is drawn can’t possibly nullify that fact.
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See my post 164 for how you calculate the probability. Which to save anyone looking back is 1 in 15
 
Blade83 said:
But

But there’s also now no risk of either club drawing Liverpool after ball number 2, which makes the odds of a derby in the remaining games higher.
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How many times do leave to reiterate that my comments are in regard to the first tie?
 
anglian blade said:
See my post 164 for how you calculate the probability. Which to save anyone looking back is 1 in 15
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Where does that take in the event of one of the other fourteen teams being drawn out first and creating a further two possible events which could alter that probability? In the first tie!

The first tie is 15/1 on a derby. IF any other team is drawn out first, there are two possible events that could occur to down that bet. Those events necessarily change the odds and have to be taken into account before ball two is drawn. I’m saying at that point, the odds lengthen until ball two is drawn. Arguing that probability before any event occurs can’t possibly alter whoever is drawn first makes no sense. The odds change.
 
my explanation is perfectly clear. The chances of a derby being scuppered increases in the first tie , but reduces in all the remaining draws.

Once you have completed the first draw the odds change but not until then
 
diode_blade said:
Ferk me...shoutybox was shut down for less than this....
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nathxnblade said:
I think this turned into the most boring thread I’ve ever read
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ZoneManWilf said:
Ps everyone saying this thread is boring is wrong
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Well, you know what? I’ve been shite at maths my whole life. And I don’t bet, partly because I’m chronically skint and partly because I don’t understand it.

But, as an object lesson in how to exchange completely divergent, but densely, patiently argued and evidenced opinions, without lapsing into the fractious, ban-provoking shit we’ve seen on here recently, this thread has fucking nailed it.

I don‘t really care who’s right. It’s been an education, and a pleasure.

Thank you, all.
 
ShockingBadBuy said:
Well, you know what? I’ve been shite at maths my whole life. And I don’t bet, partly because I’m chronically skint and partly because I don’t understand it.

But, as an object lesson in how to exchange completely divergent, but densely, patiently argued and evidenced opinions, without lapsing into the fractious, ban-provoking shit we’ve seen on here recently, this thread has fucking nailed it.

I don‘t really care who’s right. It’s been an education, and a pleasure.

Thank you, all.
Click to expand...
Shut up you tit, you’re wrong.
 
Sean Thornton said:
I’ve said, longer than 15/1. They have to be.

There's an event that has to occur before 15/1 comes back into play. An event that could affect the price because of the chance that a derby will take place. If you’re saying that that won’t affect the 15/1 then you aren’t understanding the point. Whatever the odds actually are, they have to be longer because of the event yet to occur . Only if both local teams balls were taken out of the bag before the second team were drawn could the odds remain at 15/1. Probability before any team is drawn can’t possibly nullify that fact.
Click to expand...

You’re correct that someone isn’t understanding the point.

If you’re so sure that the odds are longer than 1 in 15 (which, by the way, is not the same as 15/1) work them out and show us.
 

Punk Blade said:
So do I need to put 13/15 of keeping my bet alive against the probability of winning the bet on the remainder of the draw?
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14/1 x 13/15?

As l said, you’d be better off waiting and getting the full 14/1 , or not :) , if the bet wins through.....,
 
Geordie Blade said:
You’re correct that someone isn’t understanding the point.

If you’re so sure that the odds are longer than 1 in 15 (which, by the way, is not the same as 15/1) work them out and show us.
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Quoting probability from pre first ball on changing events must mean the probability changes.

point taken.

Whatever the odds are, calculating them is irrelevant, feel free to mock the fact l haven’t worked them out, they have to be longer as they can be affected by two possible eventualities that could reduce 1/15 written in stone before any eventuality occurs. If you really believe that two possible events that could affect the current odds and that they remain the same as before the event started, I’ll leave it with you.

Not the best analogy but if you bet on a team to win a match and they go a goal behind, their odds lengthen. You appear to be saying because the probability was say, evens, that doesn’t change the odds when they are a goal down, because their is still time to play.

Blinded by Theory v reality.
 
Good draw to get us through to the quarters. Would have liked Pompey away but hey ho. Nice to not get knocked out in the 3rd round 😁 also, Bournemouth are shit
 
Punk Blade said:
This is your error. Your instinct is that the probability will change after one ball is drawn, so you are clinging onto an abstract notion without being able to articulate it or back it up with any evidence.

I calculated it, and you’re wrong.
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The probability before the first ball doesn’t change after the first ball is drawn then even though there are two distinct possibilities that can affect things. Okay.

The probability that the odds can’t/won’t change at all must mean that there’s no chance of us or them being the second ball drawn. Clearly not the case. If we come out second, jobs done. That, or them being drawn second ball is clearly an option. And affects the odds. That’s the real world. Events In play can affect odds. You’re saying that’s not the case.
 
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Sean Thornton said:
Quoting probability from pre first ball on changing events must mean the probability changes.

point taken.

Whatever the odds are, calculating them is irrelevant, feel free to mock the fact l haven’t worked them out, they have to be longer as they can be affected by two possible eventualities that could reduce 1/15 written in stone before any eventuality occurs. If you really believe that two possible events that could affect the current odds and that they remain the same as before the event started, I’ll leave it with you.

Not the best analogy but if you bet on a team to win a match and they go a goal behind, their odds lengthen. You appear to be saying because the probability was say, evens, that doesn’t change the odds when they are a goal down, because their is still time to play.

Blinded by Theory v reality.
Click to expand...

I’m genuinely not sure if you’re on the wind up, but I’ll reply.

Several people have shown why the odds will be 1 in 15. I’m not mocking you for not working it out, but I promise that, if you try, you will get to the same answer 1 in 15.

Your analogy isn’t quite the same. I agree that a goal in a game would change the odds. It is new information.

I know that it is hard to follow, but the first team out of the hat in a cup draw is not new information.

In your analogy, we didn’t know (at the start of the game) whether that team would score. Therefore, a goal being scored would be new information.

In the cup draw example, we already know that Team A (or whoever) are going to get drawn at some point. Therefore, them coming out of the hat first, is not new information. It only becomes information when the second ball is out and we know who they’ve been paired with.

You might say that them picked first is new information, but that’s not true. Them being first would have exactly the same impact as it would if we were somehow told that they were going to get picked second. Or third. Or fourth. Of fifth... etc.

Another way of putting it is that, before the draw, we definitely knew that Liverpool* would be playing against someone (we just don’t know who). After the first ball is out, we still know that Liverpool are going to play someone, but we are still non the wiser (nor any less certain) who. That’s why it doesn’t change the odds, whether that’s in theory or in reality.

*Or whoever.
 
Sean Thornton said:
Quoting probability from pre first ball on changing events must mean the probability changes.

point taken.

Whatever the odds are, calculating them is irrelevant, feel free to mock the fact l haven’t worked them out, they have to be longer as they can be affected by two possible eventualities that could reduce 1/15 written in stone before any eventuality occurs. If you really believe that two possible events that could affect the current odds and that they remain the same as before the event started, I’ll leave it with you.

Not the best analogy but if you bet on a team to win a match and they go a goal behind, their odds lengthen. You appear to be saying because the probability was say, evens, that doesn’t change the odds when they are a goal down, because their is still time to play.

Blinded by Theory v reality.
Click to expand...
Punk Blade said:
This is your error. Your instinct is that the probability will change after one ball is drawn, so you are clinging onto an abstract notion without being able to articulate it or back it up with any evidence.

I calculated it, and you’re wrong.
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Sean Thornton seems to be one of those people who believe if heads have come up three times in a row in a toss then tails must be favourite for the fourth toss.
 
Sean Thornton said:
The probability before the first ball doesn’t change after the first ball is drawn then even though there are two distinct possibilities that can affect things. Okay.

The probability that the odds can’t/won’t change at all must mean that there’s no chance of us or them being the second ball drawn. Clearly not the case. If we come out second, jobs done. That, or them being drawn second ball is clearly an option. And affects the odds. That’s the real world. Events In play can affect odds. You’re saying that’s not the case.
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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
 
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
 
Geordie Blade said:
I’m genuinely not sure if you’re on the wind up, but I’ll reply.

Several people have shown why the odds will be 1 in 15. I’m not mocking you for not working it out, but I promise that, if you try, you will get to the same answer 1 in 15.

Your analogy isn’t quite the same. I agree that a goal in a game would change the odds. It is new information.

I know that it is hard to follow, but the first team out of the hat in a cup draw is not new information.

In your analogy, we didn’t know (at the start of the game) whether that team would score. Therefore, a goal being scored would be new information.

In the cup draw example, we already know that Team A (or whoever) are going to get drawn at some point. Therefore, them coming out of the hat first, is not new information. It only becomes information when the second ball is out and we know who they’ve been paired with.

You might say that them picked first is new information, but that’s not true. Them being first would have exactly the same impact as it would if we were somehow told that they were going to get picked second. Or third. Or fourth. Of fifth... etc.

Another way of putting it is that, before the draw, we definitely knew that Liverpool* would be playing against someone (we just don’t know who). After the first ball is out, we still know that Liverpool are going to play someone, but we are still non the wiser (nor any less certain) who. That’s why it doesn’t change the odds, whether that’s in theory or in reality.

*Or whoever.
Click to expand...

Last comment .....

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.
 

Punk Blade said:
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
Click to expand...

Yes, dawned on me about your calculation, sorry.

Just posted my last comment. It’s been, interesting :)
 
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