5th round fixture

[Danetheblade]

Here's the full calculation.

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's the really tedious bit:

P(t1) = P(s) x P(s) = 2/16 x 1/15 = 2/240 = 1/120
P(t2) = P(o) x P(o) x P(s) x P(s) = 14/16 x 13/15 x 2/14 x 1/13 = 364/43,680 = 1/120
P(t3) = P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 14/16 x 13/15 x 12/14 x 11/13 x 2/12 x 1/11 = 48,048/5,765,760 = 1/120
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) = 14/16 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 2/10 x 1/9 = 4,324,320/518,918,400 = 1/120
(the multiplied fractions are getting silly now, so I'll skip to the simplest fraction, and I think we can see a pattern emerging)
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) = 14/16 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 = 1/120
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) = 14/16 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 = 1/120
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) = 14/16 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 = 1/120
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) = 14/16 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 = 1/120

P(sd) = P(t1) + P(t2) + P(t3) + P(t4) + P(t5) + P(t6) + P(t7) + P(t8)
P(sd) = 1/120 + 1/120 + 1/120 + 1/120 + 1/120 + 1/120 + 1/120 + 1/120
P(sd) = 8/120
P(sd) = 1/15

If anyone wants to argue further that it is anything other than 1/15 then there really is no hope.

yeah... What he said.

 

[Sabella's Socks]

P(sd) = 1/15

The probability of the standard deviation* is 1/15? You've lost me...

*Perhaps better abbreviated to PTSD in this case.

 

[Grecian2000]

Stonking thread this.

Quick question to Mobile Blade though. Those bookies you've worked with, are any of them still in business????

If anyone wants to give me anything above 14/1 I'll have a swift monkey on us v every other club please (i.e 15 bets at £500 = £7500 stake). That will give me a profit of £500 for every point you're willing to give me over the 14's (i.e. £500 @ 15/1 = 7500 + 500 stake, £500 @ 16/1 = 8000 + 500 stake).

 

[Sean Thornton]

But what happens if, say, Liverpool are drawn out of the hat first? Don't our odds of playing Wednesday decrease?

At the point Liverpool are drawn either Sheffield team can be drawn out second. It has to affect the odds until the second team is known.

 

[Sean Thornton]

Stonking thread this.

Quick question to Mobile Blade though. Those bookies you've worked with, are any of them still in business????

If anyone wants to give me anything above 14/1 I'll have a swift monkey on us v every other club please (i.e 15 bets at £500 = £7500 stake). That will give me a profit of £500 for every point you're willing to give me over the 14's (i.e. £500 @ 15/1 = 7500 + 500 stake, £500 @ 16/1 = 8000 + 500 stake).

The discussion was primarily about us and them being drawn against each other. Based on the above, if that doesn't happen you'd be down £7500 no matter what points you were offered.

 

[Grecian2000]

The discussion was primarily about us and them being drawn against each other. Based on the above, if that doesn't happen you'd be down £7500 no matter what points you were offered.

Eh?
I want £500 on each of the possible 15 outcomes ( I'll have a swift monkey on us v every other club please = £500 on Us v Pigs, £500 Us v Man U, £500 Us v Man C etc. etc.). As there are ONLY 15 possible balls that we can be matched with anything over 14/1 is bunce.

 

[Sean Thornton]

Eh?
I want £500 on each of the possible 15 outcomes ( I'll have a swift monkey on us v every other club please = £500 on Us v Pigs, £500 Us v Man U, £500 Us v Man C etc. etc.). As there are ONLY 15 possible balls that we can be matched with anything over 14/1 is bunce.

Read the first line.

 

[Geordie Blade]

At the point Liverpool are drawn either Sheffield team can be drawn out second. It has to affect the odds until the second team is known.

Before a ball is drawn, there is a 1 in 15 chance that we would be drawn against Liverpool. If the first ball drawn out is Liverpool, there is still a 1 in 15 chance that we will be drawn against Liverpool. Nothing changes apart from than the chances of being home or away.

Therefore, after the first ball is drawn, our chances of drawing Wednesday are still 1 in 15. That's true, no matter who is drawn for the first ball.

 

[Sean Thornton]

Before a ball is drawn, there is a 1 in 15 chance that we would be drawn against Liverpool. If the first ball drawn out is Liverpool, there is still a 1 in 15 chance that we will be drawn against Liverpool. Nothing changes apart from than the chances of being home or away.

Therefore, after the first ball is drawn, our chances of drawing Wednesday are still 1 in 15. That's true, no matter who is drawn for the first ball.

At the moment Liverpool are drawn, either Sheffield team can be drawn against them.

 

[Grecian2000]

Read the first line.

But the chances of us being drawn against them are exactly the same as being drawn against any of the other 14 clubs (obviously taking replay's out of the equation).
Therefore, the odds are 14/1, any odds greater than this (as suggested earlier in the thread) applied to every possible outcome are guaranteed to provide the bettor with a profit.
If the true odds are 14/1 setting the overround to 100%, then offering 13/1 would set it to 107.8% (the bookie wins) and 15/1 to 92.4% (the punter wins).
If you can find ANY event with an overround of < 100% there is certain level of staking that guarantees a profit.

 

[Sean Thornton]

But the chances of us being drawn against them are exactly the same as being drawn against any of the other 14 clubs (obviously taking replay's out of the equation).
Therefore, the odds are 14/1, any odds greater than this (as suggested earlier in the thread) applied to every possible outcome are guaranteed to provide the bettor with a profit.
If the true odds are 14/1 setting the overround to 100%, then offering 13/1 would set it to 107.8% (the bookie wins) and 15/1 to 92.4% (the punter wins).
If you can find ANY event with an overround of < 100% there is certain level of staking that guarantees a profit.

I wasn't disagreeing with your calculations just pointing out it was about us drawing them/vice versa.

My point, from the start, has been that if Liverpool (or anyone of 14 clubs) are drawn first, the odds on an all Sheffield tie change because either one of them can be drawn next. Therefore the chances of that game occurring lengthen until the second team is drawn. Then it resets.

 

[MobileBlade]

P(sd) = P(t1) + P(t2) + P(t3) + P(t4) + P(t5) + P(t6) + P(t7) + P(t8)
P(sd) = 1/120 + 1/120 + 1/120 + 1/120 + 1/120 + 1/120 + 1/120 + 1/120

The bit you got wrong is this bit. P(t2) is reliant upon P(t1) therefore it has to be multiplied to it not added. They are not separate events. Therefore it should be 1/120 x 1x120 etc. Thanks for working it all out though. That was the bit I couldn't get my head around so your answer is 1/960. If P(t1) has taken place and ball 4 or 8 comes out then it is 1 in 15. Until then it is 1/120 however the odds for one of the 8 coming in is 1/960.

 

[Punk Blade]

The bit you got wrong is this bit. P(t2) is reliant upon P(t1) therefore it has to be multiplied to it not added. They are not separate events. Therefore it should be 1/120 x 1x120 etc. Thanks for working it all out though. That was the bit I couldn't get my head around so your answer is 1/960. If P(t1) has taken place and ball 4 or 8 comes out then it is 1 in 15. Until then it is 1/120 however the odds for one of the 8 coming in is 1/960.

You haven’t calculated 1/120 x 1/120, etc. You have calculated 1/(8x120). You mean to calculate (1/120)^8, which gives 1/42,998,169,600,000,000.

You've misunderstood the correct calculation. P(t1) is the ENTIRE draw with a derby as the first tie drawn. P(t2) is the ENTIRE draw with a derby as the second tie drawn. They ARE separate events. There are eight possible ties where a derby could be drawn, and I’ve calculated the probability of each (not surprisingly it’s the same for each) and added them up. The first tie not being a derby is calculated first in each instance (for the seven instances where the derby is after the first tie drawn) as 14/16 x 13/15 (the total non-Sheffield teams remaining over the total number of balls remaining).

If the probability of us drawing Wednesday is 1/960, and given the probabilities of us drawing any other ball must be equal, you have a draw where there are apparently 960 possible outcomes for SUFC: one is Wednesday, 14 are the other balls and 945 are unaccounted for.

If you had calculated what you meant and the probability of us drawing Wednesday is 1/42,998,169,600,000,000 (this is hilarious) then you have 42,998,169,599,999,985 outcomes where we don’t have a fifth round tie.

You’ve had an absolute mare here.

 

[Geordie Blade]

At the moment Liverpool are drawn, either Sheffield team can be drawn against them.

That's correct. And it doesn't change anything.

The chance of either Sheffield team being drawn second (given that Liverpool have been drawn first) will be 2 in 15. So, you can 'bank' the 2/15 chance of us not drawing the pigs (the * bit below). On the other hand, the second team out might not be a Sheffield club. There is a 13/15 chance of that. If that's the case, with 14 teams now left in the draw, our chances of drawing the pigs would now be 1 in 13 (or 12/13 that we don't draw them).

So, the chances of not drawing the pigs would be: [(2/15)]* + [(13/15) x (12/13)] = 14/15. Leaving a 1 in 15 chance of drawing them!

 

[MobileBlade]

You haven’t calculated 1/120 x 1/120, etc. You have calculated 1/(8x120). You mean to calculate (1/120)^8, which gives 1/42,998,169,600,000,000.

You've misunderstood the correct calculation. P(t1) is the ENTIRE draw with a derby as the first tie drawn. P(t2) is the ENTIRE draw with a derby as the second tie drawn. They ARE separate events. There are eight possible ties where a derby could be drawn, and I’ve calculated the probability of each (not surprisingly it’s the same for each) and added them up. The first tie not being a derby is calculated first in each instance (for the seven instances where the derby is after the first tie drawn) as 14/16 x 13/15 (the total non-Sheffield teams remaining over the total number of balls remaining).

If the probability of us drawing Wednesday is 1/960, and given the probabilities of us drawing any other ball must be equal, you have a draw where there are apparently 960 possible outcomes for SUFC: one is Wednesday, 14 are the other balls and 945 are unaccounted for.

If you had calculated what you meant and the probability of us drawing Wednesday is 1/42,998,169,600,000,000 (this is hilarious) then you have 42,998,169,599,999,985 outcomes where we don’t have a fifth round tie.

You’ve had an absolute mare here.

No. You are getting mixed up now. You have calculated the chances of two balls from 16 being picked on either occasions. The 2nd round is dependant upon the first round because you cannot be drawn on either ball in first round for the second round to happen. Therefore the odds of two balls, whether that be any of the 120 combinations, is 1/960 over 8 rounds of balls before the draw commences. This then reduces through each round until you reach the last two when it will, obviously, be 1/1. You had 8 separate events at 1/120 which is not possible at the 2nd round but is the situation before the draw commences.

 

[Punk Blade]

No. You are getting mixed up now. You have calculated the chances of two balls from 16 being picked on either occasions. The 2nd round is dependant upon the first round because you cannot be drawn on either ball in first round for the second round to happen. Therefore the odds of two balls, whether that be any of the 120 combinations, is 1/960 over 8 rounds of balls before the draw commences. This then reduces through each round until you reach the last two when it will, obviously, be 1/1. You had 8 separate events at 1/120 which is not possible at the 2nd round but is the situation before the draw commences.

Just for clarity can you answer me two questions:

a) what do you believe is the probability of us drawing Wednesday?
b) do you agree that the probability of us drawing Wednesday must be equal to the probability of us drawing each of the other 14 balls?

 

[Bristol Blade]

MobileBlade are you by any chance related to TheJosh from this famous thread?

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[Dec]

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.

Just to strip this back... you're saying that the probability of two teams in a semi final being drawn versus each other are 1 in 120??? Think about that a bit.

 

[Sean Thornton]

That's correct. And it doesn't change anything.

The chance of either Sheffield team being drawn second (given that Liverpool have been drawn first) will be 2 in 15. So, you can 'bank' the 2/15 chance of us not drawing the pigs (the * bit below). On the other hand, the second team out might not be a Sheffield club. There is a 13/15 chance of that. If that's the case, with 14 teams now left in the draw, our chances of drawing the pigs would now be 1 in 13 (or 12/13 that we don't draw them).

So, the chances of not drawing the pigs would be: [(2/15)]* + [(13/15) x (12/13)] = 14/15. Leaving a 1 in 15 chance of drawing them!

To clarify, and if I’m being thick here I’ll gladly hold my hand Up but rather than quote 1/120 or whatever can you explain why before any team is drawn the odds of a Sheffield derby are 1/15 and this doesn’t change when Liverpool are drawn first. ( l don’t know what you mean by “bank” the2/15). Those odds must change because of the chance of either Sheffield club being drawn to play Liverpool. I’m not asking about subsequent draws as such. I doubt a bookmaker offering 1/15 at that point would get many takers.

An explanation without numbers would help me

 

[diode_blade]

Ferk me...shoutybox was shut down for less than this....

Where's Rachel Riley and Carol Vorderman wearing there mankini's when ya need them....

 

[TJB.]

P(t1) = P(s) x P(s) = 2/16 x 1/15 = 2/240 = 1/120
P(t2) = P(o) x P(o) x P(s) x P(s) = 14/16 x 13/15 x 2/14 x 1/13 = 364/43,680 = 1/120
P(t3) = P(o) x P(o) x P(o) x P(o) x P(s) x P(s) = 14/16 x 13/15 x 12/14 x 11/13 x 2/12 x 1/11 = 48,048/5,765,760 = 1/120
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) = 14/16 x 13/15 x 12/14 x 11/13 x 10/12 x 9/11 x 2/10 x 1/9 = 4,324,320/518,918,400 = 1/120

What does the 'All Weapons' cheat from GTA have to do with all of this?

 

[Sean Thornton]

Ferk me...shoutybox was shut down for less than this....

Where's Rachel Riley and Carol Vorderman wearing there mankini's when ya need them....

At least no ones been called a (P where P=(c) P1(u) P2 (n) P3 (t) ) or having a 1/15 chance of being a P(ig)

 

[MobileBlade]

MobileBlade are you by any chance related to TheJosh from this famous thread?

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Ha ha. If you saw you'd say definitely not!

 

[MobileBlade]

Just for clarity can you answer me two questions:

a) what do you believe is the probability of us drawing Wednesday?
b) do you agree that the probability of us drawing Wednesday must be equal to the probability of us drawing each of the other 14 balls?

I'm going back to my original answer which is before the draw is made it is 1 in 120 simply because that is number of combinations of balls that are available. The moment the first ball is drawn that all changes. The chances of them coming out in any of the rounds before the draw is made 1 in 940 that is because there are 8 attempts to get 1 set of balls. That is the same for any combination of two balls. If however you asking the probability of United drawing Wednesday AND Man City drawing Man Utd for instance then by your argument it would still be 1 in 15 but obviously you the chances of this are much less.

Ps can we pack in now? Think we've bored everyone silly enough!!

 

[diode_blade]

At least no ones been called a (P where P=(c) P1(u) P2 (n) P3 (t) ) or having a 1/15 chance of being a P(ig)

No one is using the right formula
E=MC2

Enda=MO x Tufty squared

 

[Sean Thornton]

No one is using the right formula
E=MC2

Enda=MO x Tufty squared

“Somebody I never met
But in a way I know
Didn't think that you could get
So much from a picture show
Man dies first reel
People ask what's the deal?
This ain't how it's supposed to be
Don't like no aborigine”

 

[MCRBlade]

I think this turned into the most boring thread I’ve ever read

 

[Geordie Blade]

To clarify, and if I’m being thick here I’ll gladly hold my hand Up but rather than quote 1/120 or whatever can you explain why before any team is drawn the odds of a Sheffield derby are 1/15 and this doesn’t change when Liverpool are drawn first. ( l don’t know what you mean by “bank” the2/15). Those odds must change because of the chance of either Sheffield club being drawn to play Liverpool. I’m not asking about subsequent draws as such. I doubt a bookmaker offering 1/15 at that point would get many takers.

An explanation without numbers would help me

It's hard to do it without numbers, but I promise it's true!

Liverpool's ball has to come out at some point. I understand why you might think that Liverpool's ball coming out first might mean that the chances of us playing Wednesday go down because the 2nd ball could be us or the pigs. However, by the same logic, it should also apply if Liverpool's ball comes out, say, 11th (where the '12th' team could be us or the pigs) or if Liverpool's ball comes out 16th (where the '15th' team could have been us or the pigs).

Essentially, Liverpool will always play against one of the other fifteen teams no matter where they're drawn out. So, them being drawn out first gives us no information that we didn't already know. We knew they'd get drawn, and they did! Everything else is still subject to exactly the same chance as before (*although as previously stated, the chances of a home/away draw would have changed).

 

[Sean Thornton]

It's hard to do it without numbers, but I promise it's true!

Liverpool's ball has to come out at some point. I understand why you might think that Liverpool's ball coming out first might mean that the chances of us playing Wednesday go down because the 2nd ball could be us or the pigs. However, by the same logic, it should also apply if Liverpool's ball comes out, say, 11th (where the '12th' team could be us or the pigs) or if Liverpool's ball comes out 16th (where the '15th' team could have been us or the pigs).

Essentially, Liverpool will always play against one of the other fifteen teams no matter where they're drawn out. So, them being drawn out first gives us no information that we didn't already know. We knew they'd get drawn, and they did! Everything else is still subject to exactly the same chance as before (*although as previously stated, the chances of a home/away draw would have changed).

Liverpool was an example. It could be any of the other 14 teams.

 

[Sean Thornton]

It's hard to do it without numbers, but I promise it's true!

Liverpool's ball has to come out at some point. I understand why you might think that Liverpool's ball coming out first might mean that the chances of us playing Wednesday go down because the 2nd ball could be us or the pigs. However, by the same logic, it should also apply if Liverpool's ball comes out, say, 11th (where the '12th' team could be us or the pigs) or if Liverpool's ball comes out 16th (where the '15th' team could have been us or the pigs).

Essentially, Liverpool will always play against one of the other fifteen teams no matter where they're drawn out. So, them being drawn out first gives us no information that we didn't already know. We knew they'd get drawn, and they did! Everything else is still subject to exactly the same chance as before (*although as previously stated, the chances of a home/away draw would have changed).

Anyone of the 14 being drawn out first DOES give us information we weren’t aware of before that first drawn ball. It tells us that as either Sheffield team could be drawn second the likelihood of a derby is lessened until the second ball is drawn.