Let S be a set of events and let p_x be the probability of an event x occurring in one period. The probability of a run of n x's is p_xⁿ(1−p_x) for n≥1. (Runs of length zero are excluded because their inclusion would constitute double counting. A run of length zero for x is simultaneously the beginning of a run of a non-x.)

It must be verified that the set of probabilities p_xⁿ(1−p_x) constitute a proper set of probabilities for the set of run lengths; i.e., that they sum to unity.

Σ_n=1^∞p_xⁿ(1−p_x) = p_x(1-p_x)Σ_n=1^∞p_x^n-1 = p_x(1-p_x)(1/(1-p_x)) = p_x
 

Therefore the probabilities p_xⁿ(1−p_x) must be divided by p_x to get a proper set of probilities; i.e., the proper probability of a run of n x's is p_x^n-1(1−p_x).

The expected run length is then

E{n} = Σ_n=1^∞np_x^n-1(1−p_x) = (1−p_x)Σ_n=1^∞np_x^n-1
 

The sum Σ_n=1^∞np_x^n-1 is equal to 1/(1−p_x)² and therefore

E{n} = (1−p_x)/(1−p_x)² = 1/(1−p_x)
 

The expected length of a run is then

Σ_x∈S p_x/(1−p_x)
 

This formula tends to give dominance to the largest values of p_x.

For example, if S={a, b, c} and p_a=p_b=p_c=1/3 then E{n_a}=E{n_b}=E{n_c}=1/(1-1/3) = 1/(2/3) = 3/2 and E{n} = 3(1/3)(3/2) = 3/2.

If p_a=1/4, p_b=1/2 and p_c=1/4 then E{n_a}=E{n_c=1/(1-1/4)=4/3 and E{n_b}=1/(1-1/2) = 2. Thus E{n} = (1/4)(4/3) + (1/2)(2) + (1/4)(4/3) = (1/3) + 1 + (1/3) = 5/3.

(To be continued.)