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| mathematics | |
| Author: kurt
Date: 27-07-2005, 11:57 | with 6 teams of the big five in the seeded pot round 3 and 3 unseeded teams of the big five in the unseeded pot what is the chance that the clubs will meat each other at least one ? i think that there will be no match between them, uefa has magic hands |
| Re: mathematics | |
| Author: isidromv
Date: 27-07-2005, 12:30 Edited by: isidromv at: 27-07-2005, 12:37 | Including the country protection, except for Liverpool, the probability is 73.6%. I think there will be one match between them: Liverpool - Everton. Assuming this pair, the remaining probability is 50.5%. |
| Re: mathematics | |
| Author: bert.kassies
Date: 27-07-2005, 14:43 | From 2004 CL3 draw procedure: "In the event of a clash between two clubs from the same football association, the club drawn second will be put on hold. In their place, another ball from the corresponding group pot (seeded or not) will be drawn. The ball put on hold will then be moved to the next match and placed as the visiting team. Another ball will then be drawn from the other group pot (seeded or not) and placed as the home team of this game. If the clash occurs with the last two clubs drawn, the club drawn second will be switched with the corresponding seeded or non-seeded team of Match 1. " |
| Re: mathematics | |
| Author: Lunaris
Date: 27-07-2005, 15:08 | bert i guess kurt was asking how many big-five clashes there would be for example england-spain (man u - betis) and so on |
| Re: mathematics | |
| Author: bert.kassies
Date: 27-07-2005, 17:16 | isidromv, please help me. I have no problem to calculate the chance without country protection: 1 - 13/16 * 12/15 * 11/14 * 10/13 * 9/12 * 8/11 = 1 - (10*9*8)/(16*15*14) = 78.6 % But do you assume a certain order of events for the case with country protection? |
| Re: mathematics | |
| Author: isidromv
Date: 27-07-2005, 17:21 | Bert, I just assume 15 possible rivails of Villarreal. Manchester and Inter, instead of 16. 1 - 13/15 * 12/14 * 11/13 * 10/13 * 9/12 * 8/11 = 1 - (10*9*8)/(15*14*13) = 0.736 |
| Re: mathematics | |
| Author: bert.kassies
Date: 27-07-2005, 17:32 | But why don't you consider another possible order of events? First (Liverpool, Monaco, Werder) and then (Villarreal, Manchester, Inter): 1 - 13/16 * 12/15 * 11/14 * 10/12 * 9/11 * 8/10 = 0.721 Given the draw procedure there must be one answer to the question. |
| Re: mathematics | |
| Author: isidromv
Date: 27-07-2005, 18:05 | Yes Bert, you are right. I thought the order doesn't matter, but it does. |
| Re: mathematics | |
| Author: kurt
Date: 27-07-2005, 20:31 | i would think that the chance would be more then 100 %,and you are right i mean a clash between the big five |
| Re: mathematics | |
| Author: walter-wade
Date: 29-07-2005, 11:23 | 1-((13/15)^4+(13/16)^2)=0.627561 |
| Re: mathematics | |
| Author: isidromv
Date: 29-07-2005, 12:29 Edited by: isidromv at: 29-07-2005, 12:29 | This is not correct, you can not due (13/15)^4*(13/16)^2 because the individual events are not independent. In other words: Manchester versus a non-big-5 team, and Villarreal versus a non-big-5 team are not independent events, so the probability of boths is not 13/15*13/15, it is 13/15*12/14. |
| Re: mathematics | |
| Author: byl
Date: 30-07-2005, 07:25 Edited by: byl at: 30-07-2005, 08:10 | imo chances are 0.716... with country protection for Liverpool and chances are 0.736... without country protection for Liverpool explanation: in the event that liverpool doesn't have country protection for the calculation, assume all seeded teams get drawn first then assume that first betis, then udinese and then everton gets placed somewhere randomly. (this assumption changes nothing to the possibility or probability of big five (G5) confrontations, the important word is RANDOMLY!!!) betis can draw 15 teams (not villareal) 5/15 are G5, 10/15 are not G5 suppose that betis doesn't draw a G5 country (P=10/15) udinese can then draw 14 teams (not inter and not the team that betis drew) 5/14 are G5, 9/14 not so the chances that udinese draws a non G5 team in the event that betis draws a non G5 team are 10/15*9/14 suppose that betis and udinese both drew non G5 teams (P=10/15*9*14=90/210) everton can draw 13 teams (not ManU and not the 2 teams that Villareal and Udinese drew) 5/13 are G5, 8/13 not so chances that everton, udinese and villereal draw all not G5-teams are 90/210*8/13 = 720/2730 = 0.264... In all other cases at least 1 G5-confrontation occurs, so the chance that there is a G5 confrontation = 1-0.264...=0.736 ************************************* in the event that liverpool does have country protection things are a bit more complicated because then it is important in which order betis, udinese and everton will be drawn. anyway there are 3 possibilities (everton gets drawn 1st, 2nd or 3rd from the 3 teams), each with equal probability (p=1/3) for each of these possibilities you do the same calculation as above this leads to the following probability that no G5-confrontation occurs p=1/3*(10/14*9/14*8/13+10/15*9/13*8/13+10/15*9/14*8/12)= 0.284.. In all other cases at least 1 G5-confrontation occurs, so the chance that there is a G5 confrontation = 1-0.284...=0.716... ********************************* Hope the above is correct, not very likely given my points on probability calculus way back when i was a student  |
| Re: mathematics | |
| Author: bert.kassies
Date: 30-07-2005, 12:33 | byl, congratulations. Your analysis appears correct to me. I don't know which ranking system was applied when you were a student, but in the Dutch system (1-10) I would apply mark 10  |