In the initial stages of this analysis, we model a team's chances of advancing in the playoffs as being an independent trial - not dependent on the outcome of the playoffs in previous years.
We will also assume, for the moment, that the probability of advancing is constant.
We don't actually expect this to be true. As players age, they get better, or worse; the makeup of the team will change, the makeup of their opponents will change. But we can justify this simplification in the following way.
Let us suppose that the probability of failing to advance in each of four years is A,B,C, and D respectively. From this, it follows that the probability of failing in all four cases is the product of the probabilities = A * B * C * D.
Now, we define P = ( A + B + C + D ) /4. In other words, P is the average probability of failure. Can we legitimately substitute P for these other four values?
Support we define four more values (a,b,c,d) such that
A = P - a
B = P - b
C = P - c
D = P - d
it follows that the sum of these four values is zero.
0
= (A+B+C+D) - (A+B+C+D)
= 4P - ( P - a + P - b + P - c + P - d )
= 4P - 4P + (a+b+c+d)
= a + b + c + d
And from this, it follows that
A*B*C*D
= (P-a)(P-b)(P-c)(P-d)
= P ^ 4 - (a+b+c+d) P^3 + (ab+ac+ad+bc+bd+cd)P^2 - (abc+abd+acd+bcd)P + abcd
From the observation above, the second term vanishes. If the largest of a,b,c,d is small, relative to P, then multiplying two or more of them togather is going to be very small, relative to P. And so the variation can be excused in rough analysis.
For example, suppose that A,B,C,D are .43,.39,.39,.39. This approximation uses the average probability (.4), and concludes that the probability of losing four in a row is 0.0256; the correct answer is 0.02550717. The error is 0.364%. We can live with that for now.
May 7, 2005 8:43 PM
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