little problem(s) 2022.09

[ijabbott]

Using a bit of educated guesswork for #2 ...

• Probability of S successes in a run of S trials = \(P^S\)
• ∴ Probability of less than S successes in a run of S trials = \(1-P^S\)
• Number of (overlapping) runs of S in N = \(\begin{cases} N-S+1 & S<=N \\
  0 & S>N \end{cases}\)
• ∴ Probability of no run of S consecutive successes in a run of N trials = \(\begin{cases} \frac{1-P^S}{N-S+1} & S<=N \\
  1 & S>N \end{cases}\)
• ∴ Probability of at least one run of S consecutive successes in a run of N trials = \(\begin{cases} 1-\frac{1-P^S}{N-S+1} & S<=N \\
  0 & S>N \end{cases}\)
  

To determine the probability of a run of exactly S successes in a run of N trials, determine the probability of a run of at least S+1 consecutive successes in a run of N trials and subtract from the probability of a run of at least S consecutive successes in a run of N trials.