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Author: panda
Date: 14-06-2006, 18:41

WC GS and CL GS both have 4 team groups, but of course CL is 6 matches.

Has it been calculated by anyone what is the minimum possible points to qualify out of a group, with favourable results (I guess for WC G it is 3 points; one team gets 9 and the other 3 get 3.) and what is the points total after which qualification is GUARANTEED (I think in CL GS, with 10 you qualify)?

Further to this, did anyone work out the maths percentages - for each points total, assuming all teams equal strength, what is the % chance of qualifying?

Author: executor
Date: 14-06-2006, 18:46
Edited by: executor
at: 14-06-2006, 18:49

Minimum number of points is {b>2{/b>: if a team wins all 3 games an the other 3 get only draws from their other 2 games. It will be something like:

1. Team A 3 3 0 0 9p
2. Team B 3 0 2 1 2p
3. Team C 3 0 2 1 2p
4. Team D 3 0 2 1 2p

Then it's up to GD and goals scored to decide.

Minimum number of points that GUARANTEE qualification is 7, because:

1. Team A 3 2 0 1 6p
2. Team B 3 2 0 1 6p
3. Team C 3 2 0 1 6p
4. Team D 3 0 0 3 0p

Author: cska
Date: 14-06-2006, 22:52
Edited by: cska
at: 14-06-2006, 23:00

@executor
Good about thinking for draws in first case. But you missed them out in second case.
Minimum points for qualification can be 7, but also 6:
1. 1 2 0 5p
2. 1 2 0 5p
3. 1 2 0 5p
4. 0 0 3 0p
Even 3p can suffice, if all matches are draws, but winners are then decided on drawing lots. But 7 really guarantee going through (or 6 with positive head-to-head, of course).
In CL, if a team gets 12p (4w and 2l - just double Executor's calculations from 3 to 6 matches in total) and a positive goal difference in head-to-head, it goes through. 13p guarantee qualification.
Remark: the rule that goal difference was replaced by head-to-head as decider kills the enthusiasm to always play attacking football. Now 5-0 is not worth more than 1-0, because it can be even known many matches before the end of season that team A is above team B on head-to-head, regardless of 10-0 or 1-0 wins.

Re: GS: points needed to qualify?
Author: executor Date: 14-06-2006, 23:30	@cska The first example is to show the minimum possible amount of points with which a team can qualify from a WC group. The second shows the minimum amount of points with which a team is {b>100%{/b> qualified. With 6p it's 95%.

Author: iwan
Date: 14-06-2006, 23:42
Edited by: iwan
at: 15-06-2006, 12:00

Yes, so many is possible!!!
Six points can't be enough by 6-6-6-0 an TWOOOO points
can be enough by 9-2-2-2 !!

Look at the next pools with extremely situations:
Pool-B at the WCT in '50.
Pool-B at the WCT in '74.
Pool-B at the WCT in '82.
Pool-C at the ECT in '04.
Pool-A at the Youth-WCT in '05
Pool-D at the Africacup in '06.

In pool-A now it's thereticly possible
Germany,Ecuador and Costarica all off them
will end at 6 points and Poland at zero, but
too that Germany ends at 9 points and
Poland,Costarica and Ecuador each at 3.

Author: bora
Date: 15-06-2006, 00:12

@panda

executor is totally right about the first part of your question. for the percentage part, if you assume all teams equal strength before the matches, they have the same chance by this assumption. that means 50 percent for every team in the group before the tournament begins. but it will be interesting study after any match that has been played and percentages change after every match. if you are interested in any group, i can show you how it is calculated. it gets easier when more matches played.

Re: GS: points needed to qualify?
Author: conscious Date: 15-06-2006, 09:58	Excuse me Iwan, but what's WCT? World ChampionaTe? Or World Championship in fooTball? This is not to attack you in any way, I'm just curious.

Re: GS: points needed to qualify?
Author: executor Date: 15-06-2006, 10:34	I think it is World Cup Tournament.

Re: GS: points needed to qualify?
Author: Forza-AZ Date: 15-06-2006, 11:52 Edited by: Forza-AZ at: 15-06-2006, 11:53	@cska Even 3p can suffice, if all matches are draws, but winners are then decided on drawing lots. Depends on the scores if lots are needed. When for example this are the scores: A-B 2-2 A-C 1-1 A-D 0-0 B-C 0-0 B-D 0-0 C-D 0-0 Then this is the final table (all positions decided on goals scored): 1.A 3-3 (3-3) 2.B 3-3 (2-2) 3.C 3-3 (1-1) 4.D 3-3 (0-0)

Re: GS: points needed to qualify?
Author: executor Date: 15-06-2006, 12:17	This reminds me of a "beautiful" group in WC '94: {pre> 1.Mexico 3-3 4p 2.Ireland 2-2 4p 3.Italy 2-2 4p 4.Norway 1-1 4p {/pre> At that time some teams that finished 3rd in their group qualified and this was the case with this group. Luckly Ireland won against Italy 1-0 otherwise we would've seen lots. It would've been incredible if all the teams would've had GD 2-2, {i>n'est-ce pas?{/i>

Re: GS: points needed to qualify?
Author: panda Date: 15-06-2006, 15:19	OK, great, thanks a million guys. So, are the percentages also available (like 6 pts is 95%) What about for CL, where it's 6 games?

Re: GS: points needed to qualify?
Author: conscious Date: 16-06-2006, 07:18	Chances are: 0 pts 0% 1 pt 0% 2 pts 1.2% 3 pts 7.9% 4 pts 54.3% 5 pts 98.8% 6 pts 97.5% (now, that's interesting!) 7 pts 100% 9 pts 100% This is assuming that all matches produce one of the three results with equal probability, and the teams equal on points are ordered randomly.

Author: cska
Date: 16-06-2006, 09:45

@conscious
There was something strange to me that the probability to go through with 5p is greater than with 6p. So, I checked it and I will prove below that both probabilities are equal.
I will just describe how probabilities are actually calculated so that everyone can draw conclusions. I'll begin with the most elementary case - 2p and going through.
First, we set what is the number of points our team will have (i.e., we set for example 2p). Then, we calculate the probability to go through (i.e. to be 2nd out of 4) in the following way:
1. If the team has 2p, we know for sure that it has 2 draws and 1 loss. So, that fixed 3 results out of 6 matches in the group. I.e., we predetermine that team A drawed versus B and C and lost to D, for example. This means that we now have only 3 unknown matches with probabilities to be calculated. The we model out all possible rankings with Team A fixed (2pts) and other teams variable. Then we model out all favorable rankings for Team A. In this case:
1. Team D 9p
2. Team A 2p or A 2 or B 2 or B 2 or C 2 or C 2
3. Team B 2p or C 2 or A 2 or C 2 or B 2 or A 2
4. Team C 2p or B 2 or C 2 or A 2 or A 2 or B 2
You see that we always assume that Team D must have won 3 times. We have one match (D vs A) fixed (our A lost). So, the other two matches have probability of 1:9 (1:3 x 1:3) that D won two times. We also know that B must draw versus C, otherwise the above ranking is not possible. The probability is 1:3.
There are 6 permutations for the other 3 teams to be ranked with equal points on GD or head-to-head. In 2 of them team A is 2nd. This is 1:3 (or 2:6 which is the same). So, the total probability that A is 2nd, is (1:9) x (1:3) x (1:3) = 1:81 = 1.2% as Conscious listed it.

So, for 5p we know that it's possible only if Team A has 1 win and 2 draws, and 6p are always 2 wins and 1 loss. In these 2 cases it's much easier to model out only cases in which Team A is 3rd (it cannot be 4th with 5 or 6p, but I'll not spoil time to proved it mathematically, see my above post that positive balance /W>L or W=L and GD>0/ always means minimum 2nd out of 4/).
The only cases to be 3rd with 5p and 6p are:
1. 1 2 0 5p and 1. 2 0 1 6p
2. 1 2 0 5p and 2. 2 0 1 6p
3. 1 2 0 5p and 3. 2 0 1 6p
4. 0 0 3 0p and 4. 0 0 3 0p
If the 4th team steals any point, then automatically one of the other 3 teams loses a point and 5p (or 6p) become enough for 2nd place.
However, Conscious, observe that to analyze the variations, you can set the results of Team A (1 win and 2 draws or 2 wins and 1 loss) and only analyze the outcome of the other 3 matches.
There are 27 ways for the other 3 matches to end. (3x3x3)
And only 1 combination brings to the above listed ranking. And only 2 out of 6 ways of ordering the first 3 teams is unfavorable for our team (B or C 1st, C or B 2nd and A 3rd). This is 1:3 probability.
So, the total probability that Team A is 3rd with 5p or 6p is:
(1:27) x (1:3) = 1:81 = 1.2%.
So, the chance for Team A to go through is 100-1.2=98.8% in both cases.

Re: GS: points needed to qualify?
Author: panda Date: 16-06-2006, 11:10	I read the calculations carefully and it seems right to me. But I am not trained in stats so I can only really say thank you to you both. If you can be bothered to work out percentages for CL 6 matchdays, I would also find that very interesting.

Author: ferdi
Date: 16-06-2006, 11:49
Edited by: ferdi
at: 16-06-2006, 13:13

conscious was right, while cska made a mistake. The probability that a team advances with 6 points is 79/81 = 97.5 %.

panda wrote: {i>I read the calculations carefully and it seems right to me. But I am not trained in stats{/i>

Careful reading is not a matter of stats.

There are two ways to find the error: If you allready have an idea where the mistake might be found, then you can find it easily without careful reading. For this way being trained in stats might be helpful.

The other way is to find the error by careful reading: For every sentence in the text, ask yourself if it has been proven before, or if it is evident to you. And if it is neither evident to you nor proven before, then it bears probably the conceiled error. This way you will find the wrong sentence with no doubt. But I have to admit that this method provides undue hardship proportional to a higher power of the author's verbosity.

Author: panda
Date: 16-06-2006, 13:14

@ferdi, conscious, cska

OK, so what is the mistake cska made?

- if team A does NOT progress in 2/81 cases, and one of those 81 cases is shown in the table above (3 teams on 6 points, and our team, A, is ranked 3), what is the other case - or are we saying that there are TWO permutations that arrive at this table where A is ranked 3.

In other words, BCA and CBA (D is the zero team) - these are two results of the 81 outcomes.

But if it's 2/81 and the table looks the same for 5 pts, so we are assuming again that BCA and CBA are possible how come the probability is different 1/81 that team A does not progress?

Re: GS: points needed to qualify?
Author: ferdi Date: 16-06-2006, 13:26	panda wrote: {i>(D is the zero team){/i> Think about this sentence in both cases in question (5 pts and 6 pts). In the 6-pts case there are two different possible zero teams, in the 5-pts case there is only one possible zero team. A loses to B and beats C and D, then C and D are possible zero teams. A draws with B and C, and beats D, then only D is the possible zero team.

Re: GS: points needed to qualify?
Author: conscious Date: 16-06-2006, 14:13	Yes, if you have 5 points, you have beaten only one team, and only they can become a team with no poins in an unfavourable scenario. But if you have 6 points, you have beaten two teams, either can become a team with no poins, so there are two possible ways for an unfavourable scenario to develop. How ironic!

Author: cska
Date: 16-06-2006, 14:39

@ferdi
Correct ! I missed out the fact that if a team beats two opponents, it's possible that any of them can lose it's other two matches. So, it's true that theoretically 5p give greater probability of advancing than 6p. By the way, having in mind that the total points per game are not fixed (can be 3p for win/loss or 2p in case of draw), then it is not abnormal that 5p give better chances - actually having 5p means that Team A takes 7p out of the maximum points for the other teams (3p for the win and 2p for each draw), while having 6p means only 6p taken out from maximum for the other teams (3p for each win and 0 for the loss).
In the fixed 3 matches for Team A, its opponents win totally 2p (0 1 1) and 3p (0 0 3) in the two cases discussed. In the variable matches (w/o Team A), the other 3 teams can have maximum 9p.
So, the 3 teams can distribute among themselves a total maximum of 11p or 12p, which means that no 2 teams can have simultaneously 5p in the 1st case and 6p in the 2nd.
Now about CL. Most of you may think that we can just double the points for the doubled number of matches. However, this is not the case. Again, we can have the extreme cases, when team D gets 0p and we have a tie between A, B and C. You can observe that if D takes any point from A, then A must be more successful versus B and C, which decreases their points, and if D takes points from D, this directly facilitates A. So, the extreme case is really when D has 0p. Compared to WC, now we can have more distributions.
A, B and C have 6p each from the matches versus D. A, B and C play 6 matches between each other and can distribute min 12p (6 draws and 1p for each of the two participants) and max 18p.
The whole case now is that we can leave team D out and consider a group of 3 teams. Now the question is how many points must Team A win except versus Team D in order to be 2nd out of 3 (assuming Team D is 4th with 0p). There are 6 matches between A, B and C in which they can distribute min 12p (6 draws) and max 18p (6 wins/losses).
So, the minimalistic case is that 3 teams can have all their matches drawn (4p each+6p from D=10p in the group) and the extreme case is 6p for each team (2 wins and 2 losses), which is 12p in the group. There is also a median case with 11p (3w 2d 1l for each of the 3 teams and 0w 0d 6l for D).
So, the three possibilities are: 4-0-2, 3-2-1 and 2-4-0. So, 12p is the maximum for a possible 3-tie and 13p guarantee 100% the 2nd place. On the other hand, 4p is the minimum for 2nd place (6 wins for D, and the other 3 teams draw each match beteen each other). So, 3p guarantee 100% no more then 3rd place and elimination.
By the way, one of the posters mentioned WC'94 and the group of Italy (all teams 4p each), but remember the group of Holland (3 teams 6p each and 2 wins were not enough to go through).
So, such extreme cases are not only theoretical...

Author: ferdi
Date: 16-06-2006, 16:54

For champions league: At first sight there are 12 matches, with three possible outcomes each, means 3^12 = 531441 possible outcomes. But of course there are symmetries.

If we group the 12 matches to 6 duels (match and rematch), then there are 6 possible outcomes for each match: 6-0; 4-1; 3-3; 2-2; 1-4; 0-6. This reduces the number of possible outcomes to 6^6=46656 different cases. Of course the cases 4-1; 3-3 and 1-4 have double weight!

I should be possible to write a little computer program that covers those 46656 cases, calculates the weight of each case and the table in each case, the number of points of team A, and whether A is qualified or not.

Author: cska
Date: 16-06-2006, 18:10

@ferdi
Don't group matches in "duels" (match-rematch).
You will omit the cases with odd number of points like 3-2-1. You only quote 6-0, 4-2, 3-3, etc., but fail to include a case with 1 win and 1 draw, for example.
Consider this:
1. A 3-2-1 11p
2. B 3-2-1 11p
3. C 3-2-1 11p
4. D 0-0-6 0p
It is possible, if A draws once with B and C, B draws once with C and A wins once to B, B to C and C to A.
By the way, CL is different than WC in calculations for 6 matches instead of 3, because in WC you cannot have 2p per match, but in CL you can have 2x2=4p in a pair of matches ("duel") - 1w plus 1d.

Author: ferdi
Date: 16-06-2006, 18:19
Edited by: ferdi
at: 16-06-2006, 18:26

cska wrote:

{i>You only quote 6-0, 4-2, 3-3, etc., but fail to include a case with 1 win and 1 draw, for example.{/i>

I did not write 4-2, but 4-1, and this means the case one win, one draw. There are two possibilities for this case: Home win, away draw, or home draw, away win. Similar for 3-3: Home loss, away win, or home win, away loss. And for 1-4: Home loss, away draw, or home draw, away loss.

This reduces the number of cases from 9^6 to 6^6. (But only if we still assume that each result has the same probability, i. e. no home advantage.)

Author: cska
Date: 16-06-2006, 18:44

@ferdi
Very good !!! Absolutely right !
Sorry for the mistake, I didn't notice that you were describing point distributions. Yes, from point "point of view" (nice: point as adjective and point as noun), there are 6 possible outcomes, while in general there are 9 outcomes of 2 matches (3 cases are mirror ones as you noticed).
Well, for better calculations you can just arrange teams A, B, C and D one below another and write their W-D-L's.
If you sum up all W-D-L in the bottom line you will find that W is always equal to L and is always between 0 and 12. You will find also that D is always even (because appears for 2 teams simultaneously) and is always between 0 and 12. And if you start the model with 0-6-0 for all teams (all matches are draws), then each new W in the model adds L too and drops out 2 D's.
You can start with A - you can fix each ome of its W-D-L and start considering all cases with the other 3 teams having their matches with A fixed. One combination of results will be enough, because there are mirrors.
Anyway... That's just assumption, hoping that I can do a program too.
Great Argentina and great Holland !
I hope one of them reaches the final and wins it !

Author: conscious
Date: 16-06-2006, 19:06

Double round-robin:

0 pts 0/729 = 0%
1 pt 0/4374 = 0%
2 pts 0/10935 = 0%
3 pts 0/18954 = 0%
4 pts 1/32805 = 0.003%
5 pts 60/48114 = 0.12%
6 pts 1873/55404 = 3.4%
7 pts 11508/65610 = 17.5%
8 pts 37898/69984 = 54.2%
9 pts 50166/58320 = 86.0%
10 pts 52716/54675 = 96.4%
11 pts 43712/43740 = 99.94%
12 pts 25505/25515 = 99.96%
13 pts 21870/21870 = 100%
14 pts 10935/10935 = 100%
15 pts 4374/4374 = 100%
16 pts 4374/4374 = 100%
18 pts 729/729 = 100%

Same assumptions as above. Please note that it's not entirely adequate for the CL, as in case of equal points positions are determined by head-to-head and are not really random.

Re: GS: points needed to qualify?
Author: panda Date: 16-06-2006, 19:12	Fantastic, guys! I hope bert puts this thread in the seleced topics when the time comes.

Re: GS: points needed to qualify?
Author: conscious Date: 16-06-2006, 19:16	We should probably compile a similar table for UEFA Cup groups (single round robin, 5 teams). May be I'll code a correct program for the CL (if it's not going to be too difficult...)

Re: GS: points needed to qualify?
Author: ferdi Date: 17-06-2006, 09:59	conscious wrote: {i>Please note that it's not entirely adequate for the CL, as in case of equal points positions are determined by head-to-head and are not really random.{/i> This shouldn't matter, since the situation is symmetric for all teams, i. e. for every case where A is better than B in head-to-head, there is a corresponding case where B is better than A.

Author: panda
Date: 17-06-2006, 14:55

@ferdi

I sense you a very precise person. Well, I feel like I'm getting a lot of education from you on different threads - for example, I had an intuition that "D is the zero team" was not correct, that there could be 2 possible zero teams-but the intuition was not strong enough for me to query.

Anyway, I await whoever posts the EC QG figures with interest. Actually, do we need 2 sets of figures, are some QGs a different number of teams in EC?

Author: ignjat63
Date: 17-06-2006, 21:38
Edited by: ignjat63
at: 17-06-2006, 21:48

panda

if you want figures for EC QG, then you cant suppose that all teams are of equal strength if these figures are to make sense. I looked at results in QGs for EC2004 between 1st pot teams and 2nd pot teams. If 1st pot teams were home teams then results are W-D-L 6-3-1 for home teams and when 2nd pot teams were home teams then W-D-L was 3-5-2 for home teams also.

On the other hand, 1st pot teams against 5th pot teams won all at home 10-0-0 and all but one away 0-1-9.

Re: GS: points needed to qualify?
Author: conscious Date: 17-06-2006, 22:59	@ ferdi Thank you, that's how some thought saves a lot of work.

Author: panda
Date: 18-06-2006, 15:13

@ignjat63

No- I am not worried about unequal strength teams, even though that is obviously the case in EC QG. I am hoping for these stats as a sort of 'rule of thumb' for a team with some chance of qualifying, so they say 'yes, so many points is usually enough.'

To get into the idea of how strong the other teams in the group are is to go down the road of prediction- and say, e.g. San Marino match is a guaranteed win for the big country. I'm just interested in the stats for stats sake.

Author: conscious
Date: 18-06-2006, 17:48

UEFA Cup style group: 5 teams, 3 of them qualify

0 pts: 0/729 = 0%
1 pt: 0/2916 = 0%
2 pts: 6/4374 = 0.14%
3 pts: 166/5832 = 3.1%
4 pts: 2116.6/9477 = 22.3%
5 pts: 6636.4/8748 = 75.9%
6 pts: 6827.4/7290 = 93.7%
7 pts: 8742/8748 = 99.93%
8 pts: 4374/4374 = 100%
9 pts: 2916/2916 = 100%
10 pts: 2916/2916 = 100%
12 pts: 729/729 = 100%

@panda
6-team group with 30 matches won't fall to brute force. If you want a rule of thumb, I suggest you look through the results of past tournaments and gather some stats.

Author: cska
Date: 19-06-2006, 09:02

@ all posters here
Maybe some readers were wondered what are these stats and probabilities about and why should we care about them.
Well, look at the groups at WC - so many possibilities that teams can go through with 5p and even 4p...
However, I must make a BIG remark. These scenarios with probabilities would have been true, if we assume that all matches of Team A are fixed (i.e. results known in advance) and all the other 3 matches are variable (i.e. results are unknown and probabilities are calculated). However, at WC this is not the case. Normally, Team A can calculate probabilities after each round depending on the scenario development. For example, now the probability that Sweden goes through with 5p is 100%, because we know 4 results and there is 1 more match for Sweden. If Sweden had played all 3 matches in advance and got 5p, then the above calculated probabilities would applied.

Re: GS: points needed to qualify?
Author: executor Date: 19-06-2006, 09:44	Fans like to think that the results of their team are settled (I like this word better than fixed ). For example if your national team was in the WC, before the start of the competition you would've calculate: "We can get 3p from team A and 1p from both B and C. That would be 5p. What chances are we qualify?".

Re: GS: points needed to qualify?
Author: panda Date: 19-06-2006, 16:17	@cska Yes, of course I accept your 'big remark' - in a 4 team group it's not hard to make the calculations after each game. However, the interest in stats is like a commentary on individual group cases. At the moment it is 1-0 Switzerland-Togo. if it stays like that, and Fr can also beat Togo, with a S Korea - Swiss draw there WOULD be that unusual case, 3 teams on 5 points.

Author: Ricardo
Date: 20-06-2006, 08:34

So what are the chances now of all teams(considering 9 possibilities from the last 2 matches, and on GD it's equal chance)):
A: Germany& Ecuador 100%
B: England 100%
Sweden 89%
T&T 11%
C: Argentina& Netherlands 100%
D: Portugal 100%
Mexico 89%
Angola 11%
E: Italy 78%
Czech 44%
Ghana 55%
USA 22%
F: Brasil 100%
Australia 61%
Croatia 28%
Japan 11%
G: Switzerland 85%
S-Korea 85%
France 30%
H: Spain 100%
Ukraine 61%
Tunesia 28%
Saudi Arabia 11%

Re: GS: points needed to qualify?
Author: panda Date: 20-06-2006, 12:50	Group E figures are interesting. Group G shows - big pressure on France. I presume these are 'pure' stats figures, i.e. without referring to the strengths of the teams involved.

Author: badgerboy
Date: 20-06-2006, 14:48

Hmmm.

A fascinating thread from a mathematical viewpoint but since football teams aren't dice or coins...

Ignjat63 pointed out the inequality of teams in European Championship qualifiers and I guess this is true (to a greater or lesser extent) in all competitions.

So Ricardo's figures give France only a 30% chance to progress with Switzerland and South Korea on 85%. In reality France's chances of qualifying are still very high and the chances of BOTH Switzerland and South Korea progressing very low.

Ricardo's figures also disregard goal difference. I'm not too hot on the precise stats but -once you do take GD into account then Switzerland's chances of progressing have to be greater than South Korea's as it is already known that a draw for Switzerland takes them through whatever the result between France and Togo while a draw for Korea may not in the event of a French win. I guess that makes the true "probability" figures - Switzerland 89%, S.Korea 83%, France 27.5%.

Re: GS: points needed to qualify?
Author: Ricardo Date: 20-06-2006, 16:50	THe point is that France knows that a 2-0 victory will get them through, no matter what. They won't just defend a 1-0 victory when they know it's not enough. knowing is half the game

Re: GS: points needed to qualify?
Author: MalcolmW Date: 20-06-2006, 18:14	Italy 78%, USA 22%, Ghana 55%, Czech Republic 44%. So who has the other 1% chance????

Re: GS: points needed to qualify?
Author: bora Date: 20-06-2006, 18:32	i think ghana has 55.55% (56%) chance. but in these calculations there can be rounding errors.

Author: cska
Date: 21-06-2006, 11:59

@MalcolmW and Bora
You may observe that:
Italy 78% + Ghana 55% + Czechia 44% + USA 22% = 199%, not 100%.
The teams are not grouped two by two to add up to 100% as you might think.
They just show individual chances of each team to go through.
I.e., the 22% (=100-78), in which Italy fails are included in some part of the %'s that two of the other teams go through.
If people here are really interested in stats and maths, they may read about Venn's diagrams, which present visually what % of an event is explained by certain factor, but can be also used to represent what part of the 78% of Italy coincide with what part of the other three team's %'s.
Anyway, I'm not very keen right now to calculate all these figures, but it's not too hard to calculate the % for Italy and Ghana going through, fo example.
If you still don't understand why the percents add up to 199%, then look at group A before the last round. After two matches, Germany and Equador both had 100%, which is 200% totally.
Anyway, for Italy's group, all %'s are in fact fractions of 9: 78% is in fact 7/9 = 77,7777%, 55% is probably a mistaken rounding of 5/9= 55,5555%. Note: 1/9 is 11,1111%.
By the way, as two teams qualify, then %'s really should add up to 200% to cover for that.

Re: GS: points needed to qualify?
Author: MalcolmW Date: 21-06-2006, 13:54	@cska actually I am aware (and always have been) of the rounding issue. My question was therefore facetious. I realise that the conversion of calculated fractions into percentages can cause such difficulties. If Italy's 77.777r7% is converted to 78% then Ghana's 55.555r5% should show as 56%.

Re: GS: points needed to qualify?
Author: Ricardo Date: 21-06-2006, 15:42	It shouldn't add up to 100.000% anyway. There is always a chance on a war breaking out, and, more likely nowadays, some fraud coming to daylight inflicting a team or referee.

Author: cska
Date: 23-06-2006, 09:56

@MalcolmW
Absolutely right. 55,5555% should be 56%, but I suppose this was a negligible technical error of our fellow poster here.
@Ricardo
Yes, if you add outer events in the analysis, it is no longer valid in ints current shape and the %'s would not add up to 100%. However, we only assume the three possible outcomes of a normal football game - and no wars, referee scandals and other events (like in Italy now, for example).
I am now wondering if the Bulgarian association will protest the decision of UEFA to put Auxerre as replacement of Palermo.
(anyway, it's off-topis, but this outer event also impacts probabilities and uncertainties)