Theory of 'Runs' and probability

Question

I want to know how did author compute the probability of eleven runs if all arrangements are equally probable.

My attempt to answer my own question:

This example indicates wide applicability of the probability model of placing randomly r balls into n cells. Here n cells = 16 seats and r balls = 5 persons. Such an event is completely described by its occupancy numbers $ r_1,r_2,..., r_n$ where $r_k$ stands for the number of persons in the kth seat. Every n-tuple of integers satisfying $r_1 + r_2 + ...+ r_n= r, r_k \geq 0 $ desribes the possible configuration of occupancy numbers. With indistinguishable persons, two distributions are distinguishable only if the corresponding n-tuples $(r_1, r_2, ..., r_n)$ are not identical.

The number of distinguishable distributions is $$Arrangement_{r,n} =\binom{n + r -1}{r} =\binom{n + r -1}{n-1}$$

So, in this case, the total sample space is $\binom{16 + 5 -1}{5} = \binom{20}{5} = \binom{16 + 5 -1 = 20}{ 16 - 1 =15}= 15504.$

Now, we can choose 5 seats to occupy from 11 runs in $\binom{11}{5}=462$ ways.

We can choose 5 seats from 10 runs in$ \binom{10}{5}= 252$ ways.

We can choose from 5 seats from 9 runs in $ \binom{9}{5}=126$ ways.

And lastly we can choose 5 seats from 8 runs in $ \binom{8}{5}= 56$ ways.

So, the total number of ways of selecting 11 runs are 462 + 252 + 126 + 56 = 896. The total sample space is 15504.

Hence the probability of eleven runs if all arrangements are equally probable is $\frac{896}{15504} = 0.0578 $ which matches with author's answer as well.

Is this answer computed by me and its method of computation correct?

Note: This paragraph is taken from the book " An Introduction to Probability Theory and its Application Volume 1" written by William Feller.Third edition Page number 59

Answer 1

Is this answer computed by me and its method of computation correct?

No, because it doesn't have a sample space of equally-probable outcomes. The stated problem is to find the probability that there will be $11$ runs in a random arrangement of $5$ 'O's and $11$ 'E's, these arrangements being equally probable. Thus, the sample space is the set of such arrangements, of which there are $\binom{16}{5}=8736$. The correct answer is then $${\binom{5-1}{5-1}\binom{11-1}{5}\over \binom{16}{5}}={\binom{10}{5}\over \binom{16}{5}}={504\over 8736}={3\over 52}=0.05769...$$

Obviously (since 'O'- and 'E'-runs must alternate), generally the number of 'O'-runs and the number of 'E'-runs must either be equal or they must differ by exactly $1$; thus, here $11$ runs can occur only with $5$ 'O'runs and $6$ 'E' runs (and not vice versa, because there are only $5$ 'O's altogether). Now, the number of arrangements of $a$ 'O's and $b$ 'E's, such that there are $n_1$ 'O'-runs and $n_1+1$ 'E'-runs is (by a "stars-and-bars" argument), $${{a-1}\choose{n_1-1}}{{b-1}\choose{n_1}}$$ which gives the stated result. This is consistent with Feller's Theorem 11.18, which he states as a problem at the end of the section; so the answer "0.0578..." given in the book is not quite correct.

(For what it's worth, I also confirmed this result by computing it via Python by brute force, inspecting all the length-$16$ binary sequences having $5$ 'O's and $11$ 'E's, tallying those having exactly $11$ runs.)

EDIT: I just discovered an old posting at math.s.e. that draws the same conclusions as above.