World series lengths competition, binomial distribution. Listed in the following table is the length distribution of World Series competion for the 58 series from 1950 to 2008 (there was no series in 1994).
WORLD SERIES LENGTHS
(note, the total = 58)
of games and # of years
4 games, 12 years
5 games, 10 years
6 games, 12 years
7 games, 24 years
total 58 games.            
Assuming that each World Series game is an independent event and that the probability of either team's winning any particular contest is 0.5, find the probability of each series length.  How well does the model fit the data?
(Compute the "expected" frequencies, that is, multiply the probability of a given-length series times 58).
Attempt: Using the binomial distribution, p(k success = x) = n_C_k * (p)^k* (1 - p)^(n - k)
Assuming probability of either team's winning is 0.5 
we have
For 4 games: we have n = 4, p = 0.5 , 1-p = 1- 0.5 = 0.5.
Thus P(for four games) = 4_C_4*(0.5)^4(0.5)^(4-4) = 0.0625
Thus times 58 we have 3.625
I don't know how to do this. Please can someone please help me?
Thank you.
 A: Call the teams A and B. To find the probability a series lasts $k$ games, we find the probability that the series lasts $k$ games and Team A wins, and double the result.
The probability team A wins $4$ in a row is $\frac{1}{2^4}$. So the probability that the series lasts $4$ games is $\frac{1}{2^3}$.
The series lasts $5$ games and A wins happens if A wins $3$ of the first $4$ games, and then wins the fifth game. The probability is $\binom{4}{3}\cdot\frac{1}{2^4}\cdot \frac{1}{2}$. For the probability the series lasts $5$ games is therefore, after some simplification, $\frac{1}{2^2}$. 
The series lasts $6$ games and A wins if A wins $3$ of the first $5$ games, and then wins the sixth. This has probability $\binom{5}{3}\cdot \frac{1}{2^5}\cdot \frac{1}{2}$. Double. The probability the series last $6$ games is $\frac{5}{16}$.
The series lasts $7$ games and A wins has probability $\binom{6}{3}\frac{1}{2^6}\cdot\frac{1}{2}$. Thus the probability the series lasts $7$ games is $\frac{5}{16}$. 
Remark: The fact that series lasting $6$ games has the same probability as the series lasting $7$ is obvious without calculation. Suppose that $5$ games have been played and the series is not over. Then some team has won $3$ and the other has won $2$. If the team that is leading wins the sixth game, the series lasts $6$. If it loses the sixth game, then the series lasts $7$. Under our assumptions the team that is leading after $5$ is equally likely to win the sixth game as to lose it.  
A casual glance seems to indicate that the theoretical proportions, under the assumptions we made, look like a poor fit to the reality. A goodness of fit test will confirm that. But the assumptions were unreasonable to begin with. 
