Where next

[Caley Mad In Berks]

1 in 56 is correct for say ICT v Berwick/Spartans coming out of the draw.  But to get my other predictions correct too it must be much more.  However I'm not convinced about PerthICT's 1 in 1680.  

[Rasczak]

1 in 56 is correct for say ICT v Berwick/Spartans coming out of the draw.  But to get my other predictions correct too it must be much more.  However I'm not convinced about PerthICT's 1 in 1680.  

 

I know it seems like it must be more, however 56 possible draws are all there are. ICT are only 6/1 against getting Berwick/Spartans at all, and then evens it will be home. The odds of each team against each team drop each draw as well so the total odds are not that high.

 

Of course for it to come out exactly as predicted, ie ICT picked first, then Berwick/Spartans, then DU, Celtic etc., would be a lot higher, not sure what exactly it would be though as I can't find the formula, but for simply the ties to be as predicted is 55/1.

[DundeeICT]

1 in 56 is correct for say ICT v Berwick/Spartans coming out of the draw. But to get my other predictions correct too it must be much more. However I'm not convinced about PerthICT's 1 in 1680.

It was actually PerfICT not me PerthICT !

[Caley Mad In Berks]

 

1 in 56 is correct for say ICT v Berwick/Spartans coming out of the draw. But to get my other predictions correct too it must be much more. However I'm not convinced about PerthICT's 1 in 1680.

It was actually PerfICT not me PerthICT !

 

 

Yes, apologies PerthICT. First time I've noticed there were two such similarly named posters.

[Charles Bannerman]

Getting first home team right is 4 in 8 (1 in 2)

Getting opponent right is 1 in 7 (accumulator =1 in 14)

Next home team right is 3 in 6

Next opponent right 1 in 5

Next home team 2 in 4

Opponent right 1 in 3

Next home 1 in 2

Opponent is last ball

Multiply odds = 1 in 1680

I'm with PerfICT on this one. The probability of ICT drawing Spartans/Berwick is 1/7 so at home that halves to 1/14.

If that happens then, with six teams left the probability of DU drawing Celtic is 1/5 which becomes 1/10 at Tannadice.

If that also happens then, with four teams left the chances of Falkirk and Hibs are 1/3 which becomes 1/6.

Now only with Raith and QoS left they are bound to draw each other - probability = 1 - but that becomes 1/2 for Starks Park.

To find the probability of several events you have to multiply the individual probabilities together.

1/14 x 1/10 x 1/6 x 1/2 = 1/1680.

The overall general expression for the probability for a draw involving an even number of  "n" teams coming out in a particular way will involve something called a product operator which will give you a sequence of terms to be multiplied together.

That sequence is (2n-2)(2n-6)(2n-10)...... and so on for n/2 terms.

 

So when n = 8 as here we have (16-2)(16-6)(16-10)(16-14)

Which gives us (as above) -  14 x 10 x 6 x 2 = 1680.... or actually 1 over 1680.

[caleyboy]

you just best me to it Charles

[Tob]

That's the difference between statistical Combinations and Permutations. This is a Combination as it makes no difference whether ICT v Spartans is 1st drawn or 4th drawn etc. Order doesn't matter therefore the odds are 1 in 1680.

[Rasczak]

 

Getting first home team right is 4 in 8 (1 in 2)

Getting opponent right is 1 in 7 (accumulator =1 in 14)

Next home team right is 3 in 6

Next opponent right 1 in 5

Next home team 2 in 4

Opponent right 1 in 3

Next home 1 in 2

Opponent is last ball

Multiply odds = 1 in 1680

I'm with PerfICT on this one. The probability of ICT drawing Spartans/Berwick is 1/7 so at home that halves to 1/14.

If that happens then, with six teams left the probability of DU drawing Celtic is 1/5 which becomes 1/10 at Tannadice.

If that also happens then, with four teams left the chances of Falkirk and Hibs are 1/3 which becomes 1/6.

Now only with Raith and QoS left they are bound to draw each other - probability = 1 - but that becomes 1/2 for Starks Park.

To find the probability of several events you have to multiply the individual probabilities together.

1/14 x 1/10 x 1/6 x 1/2 = 1/1680.

The overall general expression for the probability for a draw involving an even number of  "n" teams coming out in a particular way will involve something called a product operator which will give you a sequence of terms to be multiplied together.

That sequence is (2n-2)(2n-6)(2n-10)...... and so on for n/2 terms.

 

So when n = 8 as here we have (16-2)(16-6)(16-10)(16-14)

Which gives us (as above) -  14 x 10 x 6 x 2 = 1680.... or actually 1 over 1680.

 

 

So would this be the odds of the order of the draw being exactly as Caley Mad in Berks gave, or just for those four ties being the result ? Thinking about it it would be just for the end result as you plug any of the ties as first, second, third or fourth and the odds just transfer to the others

 

That's the difference between statistical Combinations and Permutations. This is a Combination as it makes no difference whether ICT v Spartans is 1st drawn or 4th drawn etc. Order doesn't matter therefore the odds are 1 in 1680.

 

Where I have dropped the ball is by working out that there are only 56 possible permutations of the draw, ie ties possibly available.

I've worked this out in the same way that if there were a league of 8 teams, and each played the others home and away, how many total games would there be?  Answer 56.

 

As there are only 56 possible games to be played, then we are predicting the correct 4 out of those 56, so now I think about it more, then it is only 1 in 56 of you getting 1 tie correct, once we go into the other ties it does come out to more. Confirming what was said in CMIB's further posts, then as there are then 30 possible ties once you have the first 2 teams, 12 possible and 2 possible, would you multiply or add these ? CMIB added them to get 100, multiplying gets 40320 which is factorial 8. So would that be the odds of getting the right draw order, ie ICT first through to QOS last ?

 

No I think I have lost it, numbers used to be my thing as well 

Edited February 9, 2015 by Rasczak

[Sneckboy]

Yeah, 40320/1 would be correct for correctly predicting the draw exactly. Trying to forget that it's a draw pairing teams for a moment, it's effectively trying to forecast the correct sequence of 8 numbered balls coming out of a bag. (8*7*6*5*4*3*2*1) = 40320.

[12th Man]

So who's right, there's 12 permutations for the semis alone with 4 teams. Even keeping 2 groups of 4 apart would be 144.

The order of the draw is also irrelevant so not all permutations can be counted its home or away.

Edited February 9, 2015 by 12th Man

[+Fraz]

Statistical analysis thread for this p*sh.

[Caley Mad In Berks]

Yes, I'm convinced now that 1 in 1680, as PerfICT said from the start, is correct.  Well done to him for getting it right so quickly, and to others who confirmed it.  Btw, I thought you taught history CB!  You weren't Leslie Frewin in an earlier life were you?

 

Sorry to those of you who found it boring, pi*h or off topic.  I didn't realise what I was starting when I originally raised the question.

[CELTIC1CALEY3]

I've looked in crystal ball

 

ICT v Berwick/Spartans

 

Dundee Utd v Celtic

 

Falkirk v Hibs

 

Raith v QOS

 

 

Will that do us?

The obvious conclusion Holmes, is that this man has an imbalanced Crystal ball. 

[Charles Bannerman]

Yeah, 40320/1 would be correct for correctly predicting the draw exactly. Trying to forget that it's a draw pairing teams for a moment, it's effectively trying to forecast the correct sequence of 8 numbered balls coming out of a bag. (8*7*6*5*4*3*2*1) = 40320.

I agree. One way of looking at this is if you take the 1/1680 for these four pairings to emerge in any order and then allow for the probability of the order which is 1/4 for the 1st one being 1st, 1/3 for the second coming next of the 3 that are left, then 1/2 for the third one and of course 1 for the last one, that gives a further 1/(2 x 3 x 4) = 1/24.

1/24 x 1/1680 = 1/43320 = 1/8! (factorial 8 which is 8 x 7 x 6...x1)

 

So Rasczak on the one hand and PerfICT and myself on the other have been stating probabilities of two different outcomes - the relevant pairings happening in any order and these pairings coming out in a specific order. The former is 24 (or 4!) times as likely as the latter.

 

It's also interesting to note that 1680 = 8!/4! and also = 14 x 10 x 6 x 2 as stated in my earlier post.

 

In response to CMIB's query about what I used to teach... it was Chemistry but relatively few people seem to know this!

I have been taken as a teacher of Maths, History, PE and on one occasion David Currie even introduced me on Saturday Sportscene as an English teacher!

[PerfICT]

Yes, I'm convinced now that 1 in 1680, as PerfICT said from the start, is correct.  Well done to him for getting it right so quickly, and to others who confirmed it.  Btw, I thought you taught history CB!  You weren't Leslie Frewin in an earlier life were you?

 

Sorry to those of you who found it boring, pi*h or off topic.  I didn't realise what I was starting when I originally raised the question.

 

I thank you kind sir, although I started to doubt my own logic for a second! Obviously spent too much time in the bookies!

[Tob]

Well it will be a lot easier in the next round. No home/away ties so it's a 1 in 3 chance of predicting the draw. God knows what the numbers would be like for predicting the outcome of the draw for Round 4 but if anyone wants to have a go...

[Charles Bannerman]

Well it will be a lot easier in the next round. No home/away ties so it's a 1 in 3 chance of predicting the draw. God knows what the numbers would be like for predicting the outcome of the draw for Round 4 but if anyone wants to have a go...

Round 4? The last 16? The numbers are enormous!

I make the probability of any particular eight game draw coming out in any order to be 1 in 43,243,200 - in other words more than three times less likely even than winning the 6 number Lotto. And the chances of that draw coming out in a particular order are 108,864 times less even than that - roughly 40 billion to one!

 

Round of 32 anyone?

Edited February 9, 2015 by Charles Bannerman

[Tob]

There are 32 teams in Round 4 Charles. Detention!

[Charles Bannerman]

There are 32 teams in Round 4 Charles. Detention!

All these Highland League clubs getting in without having to go through the Qualifying Cup is confusing me! In old money the round of 32 (to me the best day in Scottish football) was the third round where the top teams came in for the first time. I don't know what the hell round is what these days but will try to work out something for 32 just for the hell of it when I get home later tonight. The number will be massive!

[Tob]

Nice to be giving you some homework for a change. I do know the arithmetic to calculate the answer so I will be checking your working very carefully :)

 

12.576 Sextillion to 1 I made it!!!

Edited February 9, 2015 by Tob

[Charles Bannerman]

Nice to be giving you some homework for a change. I do know the arithmetic to calculate the answer so I will be checking your working very carefully :)

 

12.576 Sextillion to 1 I made it!!!

I'm not much into the various "zillions" but we do have the same answer because I get (to five significant figures) 1.2576 x (10 to the 22) or 12576000000000000000000 different permutations.

Put in different terms, that's about 1800000000000 times the population of the earth, or alternatively, roughly the number of "air" molecules in an "empty" beer can.

But that's only the odds against that particular set of matches. If you want them to come out in a particular order, you have to multiply by a further 16! (factorial 16) which is 2.09 x (10 to the 13) or 20900000000000. That gives 2.63 x (10 to the 35)  or 38 million million millon millon times the population of the earth, or quarter of a million times as many molecules as are contained in the earth's entire stock of sea water!

 

Did I ever teach you?

 

And have we gone off topic?

[12th Man]

I cant remember what the topic was or what the replies were but I remember one poster simply typing a while ago.

 

Fun at parties.

[Sneckboy]

I found an interesting article pertinent to the odds discussion.

Apparently, in 2008, a punter posted a message on a comments forum prior to the Champions League quarter-final draw - correctly predicting the outcome!!!!

 

A post left on a newspaper's internet forum at 10.28am this morning, 90 minutes before the draw took place in Nyon, Switzerland, correctly predicted the entire quarter-final draw.   

The poster claimed bookmakers were no longer willing to accept bets on the draw. When all four ties predicted in the post came to fruition it caused fevered internet speculation that some sort of manipulation might have taken place.

But a Uefa spokesman said: "It is just a lucky guess."

 

This of course caused media outlets to speculate on what the odds were of 'guessing' the outcome. And as there were clearly no resident statisticians to-hand, various newspapers and websites were publishing all manner of guesses as to what the correct odds would be!!!

As it happens, the odds for this punter's prediction were deemed to be 1 in 105. (He had the 4 ties correct, but not in the order they were drawn or with the 'correct' first home team)

 

1 in 105 explanation

the first team drawn can be paired with 7 other teams and

• for each of these 7 choices the third team drawn can be paired with 5 other teams
• and for each of these choices the 5th team drawn can be paired with 3 other teams
• (at this point the last two teams must be paired together).
• So that makes 7 times 5 times 3 combinations which is 105.
  So the probability of correctly guessing all four ties is 1 in 105

This is where much of the confusion arises. There are 3 different odds for predicting a quarter-final, it would appear.

 

Predicting the draw exactly as it comes out, ball by ball = 1 in 43,320

Predicting the 4 ties (in any order) but with the correct home/away combination = 1 in 1,680

Predicting the correct 4 pairings (without regard to home/away) = 1 in 105

 

e.g. (1 in 105)

Celtic will play Dundee United

Raith will play Inverness

Berwick/Spartans will play Hibs

Falkirk will play QOS

 

The odds of getting the Celtic-Dundee Utd pairing the 'right way round' would require a multiplication of 2.

(1 in 210)

Dundee United v Celtic

Raith will play Inverness

Berwick/Spartans will play Hibs

Falkirk will play QOS

 

The odds of getting the Raith-Inverness pairing the 'right way round' would require a further multiplication of 2.

(1 in 420)

Dundee United v Celtic

Inverness v Raith

Berwick/Spartans will play Hibs

Falkirk will play QOS

 

The odds of getting the Berwick/Spartans-Hibs pairing the 'right way round' would require a further multiplication of 2.

(1 in 840)

Dundee United v Celtic

Inverness v Raith

Hibs v. Berwick/Spartans

Falkirk will play QOS

 

The odds of getting the Falkirk-QOS pairing the 'right way round' would require a further multiplication of 2.

(1 in 1,680)

Dundee United v Celtic

Inverness v Raith

Hibs v Berwick/Spartans

QOS v Falkirk

 

Which brings us back to the 1,680 figure!!

[Caley Mad In Berks]

Wow, what did  I start when i had a peak in my crystal ball?    At least it got the Dundee Utd v Celtic  tie spot on, probably the most important one of the lot for us, ensuring that one major rival is knocked out!

[Sneckboy]

This is the best thread on the forum! I have no idea how Fraz hasn't embraced it!