Probability of Never Having a Losing RecordDate: 04/07/2001 at 20:15:40 From: Omer Bulut Subject: Probability Suppose a football team plays 8 games. The chance of winning any particular game is 50%. Determine the probability of completing the season without ever having more losses than wins during the season. It is really hard for me to do this question. Thank you, Omer Date: 04/08/2001 at 12:38:53 From: Doctor Anthony Subject: Re: Probability This is by no means a trivial problem. You can model it as the number of paths on a grid of lattice points from the origin to (m,n) such that you never cross but may touch the line y = x. In the case when m = n we have what are known as Catalan numbers. If you are interested in seeing the proof of the formulae used below you should write again. In the work that follows we are referring to ACCEPTABLE paths, that is, paths in which at all stages y > x, though we can accept y = x. The required number of acceptable paths is given by: (n-m+1) -------.C(m+n,m) n+1 Using this model with 8 football matches, we have m+n = 8 and there is equal probability at each stage (or game) that either m or n will increase by 1. Number of routes from the origin to (0,8) = (9/9).C(9,0) = 1 " " " (1,7) = (7/8).C(8,1) = 7 " " " (2,6) = (5/7).C(8,2) = 20 " " " (3,5) = (3/6).C(8,3) = 28 " " " (4,4) = (1/5).C(8,4) = 14 ------------ Total = 70 The total number of possible sequences of Win/Lose is 2^8 = 256 and so the required probability = 70/256 = 35/128 (=0.27344) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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