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A coin is tossed $m+n$ times $(m>n)$.Show that the probability of atleast $m$ consecutive heads is $\frac{n+2}{2^{m+1}}$.

I could not attempt this question,except few initial steps.Let $H$ and $T$ denote turning up of the head and tail.$\therefore P(H)=P(T)=\frac{1}{2}$

Please help me.

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    $\begingroup$ Have you looked at some small, concrete examples? What about $m = 2, n = 1$? $m$ being whatever and $n = 0$ or $1$?That's the first thing I would do. $\endgroup$
    – Arthur
    Commented Sep 29, 2015 at 10:54
  • $\begingroup$ Markov chains are a powerful tool for solving such problems. $\endgroup$
    – hardmath
    Commented Sep 29, 2015 at 14:37

1 Answer 1

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The probability the first $m$ tosses are heads is $\dfrac1{2^m}$.

The probability that the $j$th is tails and the next $m$ are heads is $\dfrac1{2^{m+1}}$, providing that $1 \le j \le n$.

If $n \le m$ then these are disjoint events, so the probability any of them occurs is $\dfrac1{2^m}+ n \times \dfrac1{2^{m+1}} = \dfrac{n+2}{2^{m+1}}.$

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    $\begingroup$ What happens if $n>m$,will they not be disjoint then?@Henry $\endgroup$
    – Brahmagupta
    Commented Sep 29, 2015 at 11:57
  • $\begingroup$ @Brahmagupta: if $n \gt m$ so $m+n \ge 2m+1$ then for example you could have $m$ heads, then $1$ tail, and then $m$ more heads. This event would be counted initially and again when $j=m+1$. $\endgroup$
    – Henry
    Commented Sep 29, 2015 at 12:21

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