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The original problem was for the special case of $n=5$ and $k=3$. But I'm curious about the general solution. Let $X_i$ be the Bernoulli variable at $i$th toss and $X = X_1 \bigotimes X_2 \bigotimes ... \bigotimes X_n$. I'm considering the set $\{{A_i}\}_{i=1}^{m}$ with $m = n-k+1$ and $A_j=\{{x\in X : x_j=x_{j+1}=...=x_{j+k-1}}=1\}$. And I am thinking about solution how to solve for $|\bigcup_{i\in I\subseteq [m]}A_i|$ and use that to compute for probability. But it seems quite tedious; is there any elegant solution?

joriki
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The One
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  • Hint: $B_i$ be the event that $x_j=x_{j+1}=\dots=x_{j+k-1}=1$, and that $x_{j+k}=0$, for $1\le j\le n-k$, and let $B_{n-k+1}=A_{n-k+1}$. In words, while $A_j$ is the event that a sequence of at least $k$ ones starts at index $j$, $B_j$ is the event that a sequence of length exactly $k$ starts at index $j$. It is true that $\bigcup_{i\in I\subseteq [m]}A_i=\bigcup_{i\in I\subseteq [m]}B_i$. The advantage of using $B_i$ instead of $A_i$ is that counting the intersections of subsets of the events $B_i$ is much easier, allowed you to get a formula for $\bigcup_{i\in I\subseteq [m]}B_i$ using PIE. – Mike Earnest Feb 02 '20 at 00:28
  • This is a problem of Runs. There’s basically 3 approaches. 1) set up a linear recursion and/or markov chain— a nice one of these is a Residual Life matrix and it looks like a generalized Fibonacci matrix. 2) exact solution with inclusion-exclusion — ugly but you can truncate with bonferonni upper and lower bounds. 3) elegant — poison approximation... it takes a lot of work but you can even bound the error with Stein-Chen if you want to learn that – user8675309 Feb 02 '20 at 00:32

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