uknowsit Posted March 23, 2011 Share Posted March 23, 2011 Hi all, just need some help with my analysis. I'm focusing on 2-2 draws using english league 1 and league 2. If i estimate that the chance of a 2-2 draw occurring in either of these leagues is equal to 5% (20.0). how can i calculate that during any given weekend (of 24 matches) at least 3 will end 2-2 - and also what are the chances of the nth game (say 10th for example) being a 2-2 draw! I've done basic binomial and discrete probability stuff for A level but that was a LONG time ago now. Any experts out there care to help? (I'm aware there are websites that can produce the answer - i'm just trying to understand how they come to it) :ok Quote Link to comment Share on other sites More sharing options...
Rushian Posted March 23, 2011 Share Posted March 23, 2011 Re: Maths help - Possibly using Binomial Assuming the matches are independent (i.e. one ending in a 2-2 draw does not increase or decrease the chances of another ending in a 2-2- draw), then the probability of having exactly k matches ending in a 2-2 draw are given by: P(k) = p^k * (1-p)^(n-k) * n! / [(n-k)! * k!] where: n = total number of matches k = exact number of 2-2 draws p = chance of a 2-2 draw and n! = n * (n-1) * (n-2) * ... * 2 * 1 So if you wanted exactly 3 out of 24 matches ending in 2-2 the probability would be: P(k=3) = 0.05^3 * 0.95^21 * 24! / (21! * 3!) = 0.0862 = 8.62% In excel this is given by "=BINOMDIST(3;24;0.05;0)" If you wanted at least 3 matches, it's the same as subtracting the chance of having less than 3 matches (i.e. at most 2) from 100%. This is given in Excel using: "=100% - BINOMDIST(2;24;0.05;1)" which is equal to 11.6%. Hope this helps :ok Quote Link to comment Share on other sites More sharing options...
uknowsit Posted March 23, 2011 Author Share Posted March 23, 2011 Re: Maths help - Possibly using Binomial Thanks Rushian, I ran the data through http://stattrek.com/Tables/Binomial.aspx and it showed me 11.6% chance - or at least i think it did. Where have i got confused there?? Edited to say for 3 or more 2-2 draws occuring Quote Link to comment Share on other sites More sharing options...
Rushian Posted March 23, 2011 Share Posted March 23, 2011 Re: Maths help - Possibly using Binomial Thanks Rushian' date=' I ran the date through http://stattrek.com/Tables/Binomial.aspx and it showed me 11.6% chance - or at least i think it did. Where have i got confused there?? My fault. "1-BINOMDIST(3;24;0.05;1)" gives you the probability of more than 3 (i.e. 4 or more). You wanted the probability of "3 or more" which means that the right formula is: "=1-BINOMDIST(2;24;0.05;1)" which is 11.6%. This agrees with your webpage. I'll amend my previous post accordingly so that there is no future confusion. Quote Link to comment Share on other sites More sharing options...
uknowsit Posted March 23, 2011 Author Share Posted March 23, 2011 Re: Maths help - Possibly using Binomial Thanks very much for your assistance and clarifaction Quote Link to comment Share on other sites More sharing options...
Grex Posted March 23, 2011 Share Posted March 23, 2011 Re: Maths help - Possibly using Binomial I think PL should start two new forums.... "Sophisticated System & Strategies" for people who conribute to threads like this, and...... "Systems & Strategies for Ejits".... like me :loon Quote Link to comment Share on other sites More sharing options...
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