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Consider repeated independent trials of two outcomes S (success) or F (failure) with probabilities $p$ and $q$, respectively. Determine the distribution of the number of trials required for the first occurrence of totally 5 S and a consecutive 5S, which is a special case of totally $n$ S containing at least one $r$-consecutive S..

The total number of S is 5. And it must be a 5-consecutive. The trail pattern must be in the form of F...FSSSSS, i.e., a bunch of F followed by 5 S. Denote the desired random variable as $N$. Then Pr($N=k$) = $q^{k-5}p^5,k=5,6,...$. But it turns out that $\sum^\infty_{k=5}q^{k-5}p^5=p^5\sum^\infty_{k=5}q^{k-5}=p^5\sum^\infty_{k=0}q^k=p^5/(1-q)=p^4<1$.

This is weird. What is wrong here? Is the problem definition wrong or the calculation wrong? Actually I am considering the distribution of the number of trials required for the first occurrence of totally $n$ S containing at least one $r$-consecutive S. I just look at the special case of $n=5$ and $r=5$. I tried to solve this general problem in a recurrence way and the base case is $n=r$.

Junk Warrior
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    In the sample space, there are possibilities to get other patterns, such as FFSSFFSS... – YJT Sep 02 '20 at 21:49
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    $p^4$ is clearly the probability that the first S, whenever it happens, will immediately be followed by four more Ss. So the calculation is correct – Henry Sep 02 '20 at 21:51
  • The sum you have is not normalized. You need to divide by $\sum_{k=0}^{\infty} q^k$. To get the expectation of the number of failures, you would need: $$\frac{ \sum_{k=0}^{\infty} k q^k}{\sum_{k=0}^{\infty} q^k}$$. – Robby the Belgian Sep 02 '20 at 21:54
  • @Henry According to you, the two requirements (totally 5 S and must be a 5-consecutive) just automatically require 4S to follow the first S. Thus the problem is just equivalent to the number of trials to get the first S. Is this what you mean? This is still strange. Is it consistent with the general case of totally n S containing at least one r-consecutive? After all, my goal is to use n=r as the base case for recursion. – Junk Warrior Sep 02 '20 at 22:05
  • @JunkWarrior You are the one who seems to be setting the requirement "The total number of S is 5. And it must be a 5-consecutive." I have said that your calculation that the probability of that happening is $p^4$ is clearly correct. – Henry Sep 02 '20 at 22:39
  • @Henry Please help me with my true concern at https://math.stackexchange.com/questions/3797044/distribution-of-the-number-of-trials-required-for-the-first-occurrence-of-the-ev – Junk Warrior Sep 02 '20 at 23:56

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The trail pattern must be in the form of F...FSSSSS, i.e., a bunch of F followed by 5 S.

No, e.g. you could have FSFFFSSFSSFSSFFSSSSS.

Also, note that they're asking two different questions here: (1) distribution of trials to get 5 successes total, and (2) distribution of trials to get 5 successes in a row.

Henry
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Daniel McLaury
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  • It cannot be FSFFFSSFSSFSSFFSSSSS. If so, there are 12 S. The total number of S must be 5. The general question is the first time to get totally n S containing at least one r-consecutive. The two requirements (total number and the r-consecutive) must be satisfied at the same time. – Junk Warrior Sep 02 '20 at 21:55
  • @JunkWarrior: From context, we can infer that they are asking two separate questions: first, calculate the distribution of the number of trials you will need to get five S's total. Second, separately, calculate the distribution of the number of trials you will need to get five consecutive S's. – Daniel McLaury Sep 03 '20 at 03:05