Probabilty of an event reoccuring x times in a row within n trials

Question

How can I calculate the probablity of an event to reoccur 9 times in a row within 6000 trials if there is a 1/3 chance of the event to occur in a single trial.

Answer 1

This is a Bernoulli process, so you can use:

$$P(k)=\binom{n}{k}p^kq^{n-k}$$

where $n=6000$, $k=9$, $p={1\over3}$ and $q=1-p={2\over3}$:

$$P(9)=\binom{6000}{9}\left(\frac{1}{3}\right)^9\left({2\over3}\right)^{6000-9}\approx4,98*10^{-1059}$$

I saw that you wrote "an event to reoccur 9 times in a row"...in this case it is different and we have:

$$P(9)=5992p^kq^{n-k}=5992\left(\frac{1}{3}\right)^9\left({2\over3}\right)^{6000-9}\approx 3.3*10^{-1056}$$

but someone check it please.

Answer 2

I will focus on the complementary probability that in $~6000~$ consecutive (independent) Bernoulli trials, at no point is there an occurrence of $~9~$ consecutive successes. The math is somewhat complicated. However, this response provides the formulas needed in order to have a computer program compute the exact probability.

Let $~S~$ denote the collection of all possible occurrences of the $~6000~$ trials. Since each trial either fails or succeeds, you have that $~|S| = 2^{6000}.$

For $~k \in \Bbb{Z_{\geq 0}}, ~k \leq 6000,~$ let $~S_k~$ denote the subset of $~S~$ that represents having exactly $~k~$ failures. Therefore,

$$|S_k| = \binom{6000}{k}.$$

For any element $~x \in S_k,~$ where $~x~$ represents a set of $~6000~$ trials that had exactly $~k~$ failures, the probability of the element $~x~$ occurring is

$$p(k) = \left[ ~\frac{2}{3} ~\right]^k \times \left[ ~\frac{1}{3} ~\right]^{6000 - k} = \frac{2^k}{3^{6000}}.$$

Let $~g(k)~$ denote the number of elements in $~S_k~$ that represent a set of $~6000~$ trials, where

• There were exactly $~k~$ failures.
• At no point in the set of $~6000~$ trials, was there an occurrence of $~9~$ consecutive successes.

Then, the desired computation is

$$\sum_{k=0}^{6000} \left[ ~p(k) \times g(k) ~\right] = \frac{1}{3^{6000}} \times \left[ ~2^k \times g(k) ~\right].$$

So, the problem is reduced to providing a formula for $~g(k).$

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Consider the following tableau

-F----F-------F...F---F---

which represents $~k~$ failures distributed among $~6000~$ positions. These $~k~$ failures create $~(k+1)~$ islands, and the total size of the sum of these islands is $~6000 - k.~$

For $~i \in \{1,2,\cdots,x_{k+1}\},~$ let $~x_i~$ denote the size of the $~i$-th island. Now consider the following enumeration problem: compute the number of solutions to

• $x_1 + x_2 + \cdots + x_{k+1} = 6000 - k.$

• $x_1, x_2, \cdots, x_{k+1} \in \Bbb{Z_{\geq 0}}.$

• $~x_1, x_2, \cdots, x_{k+1} \leq 8.$

The number of solutions will exactly equal $~g(k).~$ That is, the distribution of the $~k~$ failures is precisely determined by the size of the variables $~x_1,x_2,\cdots,x_{k+1}.~$ Further, there will be no occurrence of $~9~$ (or more) consecutive successes if and only if each of the variables $~x_1,x_2,\cdots,x_{k+1}~$ is less than $~9.~$

So, the entire problem has been reduced to performing the above computation, as a function of $~k.~$ The first thing to do is determine the lower bound for $~k.~$ You need

$$8 \times (k+1) \geq 6000 - k \implies 9k \geq 5992 \implies k \geq 666.$$

Therefore, for $~k < 666,~$ you have that $~g(k) = 0.~$ So now, without loss of generality, $~k \geq 666.$

To compute $~g(k),~$ I will follow the model in this answer, which provides a blueprint of how to combine Inclusion-Exclusion with Stars-And-Bars to attack this generic type of problem.

First, assume that if $~a \in \Bbb{Z_{\geq 0}},~$ and $~b \in \Bbb{Z},~$ that

• $\displaystyle b < 0 \implies \binom{b}{a} = 0.$

• $\displaystyle b < a \implies \binom{b}{a} = 0.$

Now, assume that $~k~$ is some fixed element in $~\{666, 667, \cdots, 6000\},~$ and define:

• $\displaystyle T_0 = \binom{k+1}{0} \times \binom{6000 - [9 \times 0]}{k}.$
• $\displaystyle T_1 = \binom{k+1}{1} \times \binom{6000 - [9 \times 1]}{k}.$
• $\displaystyle T_2 = \binom{k+1}{2} \times \binom{6000 - [9 \times 2]}{k}.$
• Similarly, for $~r \in \{3,4,\cdots,k+1\},$
  
  $\displaystyle T_r = \binom{k+1}{r} \times \binom{6000 - [9 \times r]}{k}.$

Then, following the analysis in the linked article,

$$g(k) = \sum_{r=0}^{k+1} (-1)^{r+1} T_r.$$

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$\underline{\text{Final Summary}}$

The probability of no occurrence of $~9~$ consecutive successes can be computed as follows:

For $~k \in \{666,667,\cdots,6000\}~$:

$$p(k) = \frac{2^k}{3^{6000}}.$$

Assuming that $~k~$ is fixed:

$$T_0 = \binom{k+1}{0} \times \binom{6000 - [9 \times 0]}{k}.$$

$$T_1 = \binom{k+1}{1} \times \binom{6000 - [9 \times 1]}{k}.$$

$$T_2 = \binom{k+1}{2} \times \binom{6000 - [9 \times 2]}{k}.$$

For $~r \in \{3,4,\cdots,k+1\},$

$$T_r = \binom{k+1}{r} \times \binom{6000 - [9 \times r]}{k}.$$

$$g(k) = \sum_{r=0}^{k+1} (-1)^{r+1} T_r.$$

Then, the overall computation of the probability of no occurrences of 9 consecutive successes is

$$\sum_{k=666}^{6000} \left[ ~p(k) \times g(k) ~\right].$$