Bernoulli trials are sequences of 0 and 1. What is the average length, the number of trials, that you need to perform until you reach n ones in sequence?

I tried to make it with a 2D recursion enter image description here

Here first index of $a$ is a counter of trials whereas second index is the level (the row). I have combined diagonals in groups to show that the whole group transits at level 0 at next column with whenever a failure occurs (with prob q).

The initial condition $$a_{00} = 1$$

The other values in the first row, which corresponds to after-failure condition for every number of trials performed so far,

$$a_{col+1,0} = a_{col,0}\cdot q + a_{col,1} \cdot q + \cdots + a_{col,col}\cdot q = q \cdot \sum_0^{col}{a_{col,i}}$$

For instance $a_{2,0}$ corresponds to paths 00 and 10, through $a_{11}$ and $a_{10}$ correspondingly.

The higher rank rows correspond to sequences of 1. If you get to the row n this means that you had n ones in sequence. Therefore, coefficients decay geometrically as we progress downwards along the column:

$$a_{trials,level} = p\cdot a_{trials+1,level+1}$$

We have got 3 equations for the induction:

$$ \begin{cases}\begin{array}{l} a_{0,0} &= 1 \\ a_{m+1,0} &= q \cdot \sum_0^{col}{a_{m,i}} \\ a_{m+1,n+1} &= p\cdot a_{m,n} \end{array}\end{cases}$$

I am familiar with generating functions and can define $X_m(x) = x_0 + x_1 \cdot x + x_2 \cdot x^2 + \cdots$ and transform last equation into

$$a_{m+1,0} + a_{m+1,0} \cdot x + a_{m+1,2} \cdot x^2 + \cdots - p\cdot x \cdot a_{m,0} - p\cdot x^2 \cdot a_{m, 1} - \cdots = a_{m+1,0} + x(a_{m+1,1} - p \cdot a_{m, 0}) + x^2(a_{m+1,2} - p \cdot a_{m, 1}) + \cdots = a_{m+1,0} = X_{m+1} - p\cdot x\cdot X_m$$

Resulting $$a_{m+1,0} = X_{m+1} - p\cdot x\cdot X_m$$ is also a recursion of the kind $X_{m+1} = a\cdot X_m + b$, where $b = a_{m+1,0}$ and $a = p\cdot x$ which has a solution

$$Y(y) = {x_0+y(x_0-b)\over (1-y)(1-ax)} = {X_0 + y(X_0-a_{m+1,0}) \over (1-y)(1-pxy)} = y_0 + y_1\cdot y + y_2 \cdot y^2\ldots.$$

You see, the things are becoming rather complex.

I also ignored the fact that the table must have a limited number of rows because we must stop at the row and compute its average, $E[level] = \sum_{col=0}^\infty{(level + col) \cdot a_{level, col}}$. That is our purpose. Currently, with infinite number of rows, we reset to the first row whenever 0 is Bernoulli despite desired level may be already achieved at the time. This distorts the true $a_{m,n}$ probabilities.

I expect that there is an easier method, along the lines of average tree size, which computes average, bypassing the probability distribution.


1 Answer 1


You are correct that the method used in the solution to the average tree size question is simpler. This is found in Ross' text $\it Introduction \ to \ Probability \ Models,$ 8th Edition. I will follow the approach given there.

Let $T_n$ be the number of trials needed to get $n$ consecutive successes, where the probability of success is $p.$ Let its expected value be $M_n.$ Conditioning on $T_{n-1},$

$$M_n = E[T_n] = E \left[ E \left[ T_n|T_{n-1} \right] \right]$$

Now consider that if you have had $n-1$ successes, you either reach your goal on the next trial or you restart: $$E \left[ E \left[ T_n|T_{n-1} \right] \right] =(T_{n-1}+1)(p)+(T_{n-1}+1+E[T_n])(1-p)$$

This simplifies to

$$ E \left[ E \left[ T_n|T_{n-1} \right] \right] =T_{n-1}+1+(1-p)E[T_n].$$

Taking the expectation of both sides gives $$M_n = M_{n-1} + 1 + (1-p)M_n,$$ which gives the recursive $$M_n = \dfrac{1}{p} + \dfrac{M_{n-1}}{p} $$

But the time $T_1$ until the first success is geometric and so we know $$M_1 = {{1} \over {p}}$$

So $$M_2 = \dfrac{1}{p} + \dfrac{1}{p^2}$$

and in general

$$M_n = \dfrac{1}{p} + \dfrac{1}{p^2} + \cdots + \dfrac{1}{p^n}$$


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