2
$\begingroup$

A coin is tossed repeatedly. A head has probability $p$ and a tail $1-p$. The outcomes of the tosses are independent. Let $E$ denote the event that the first run of $r$ successive heads occurs earlier than the first run of $s$ successive tails. And let $A$ denote the outcome of the first toss. How do you show that

$\mathbb{P}(E|A=$ head$)=p^{r-1}+(1-p^{r-1})\mathbb{P}(E|A=$ tail$)$.

And what is a similar expression for $\mathbb{P}(E|A=$ tail$)$? And what is $\mathbb{P}(E)$? Thanks in advance!

$\endgroup$
2
$\begingroup$

For $i=1,\ldots,r,\;$ define event $B_i = \text{"First tail occurs on $i^{th}$ toss"}$. Since the process ends after $r$ successive heads then, given initially that $A=H$, exactly one of these $r$ events must occur.

Note that, given $A=H$, if $B_r$ occurs then $E$ also occurs, which is to say $P(E\mid B_r\cap A=H) = 1$. Also, given $A=H$, if any of the other $B_i$ events occur then it is like we are starting over but, instead, initially given $A=T$. Therefore, conditioning on whether or not $B_r$ occurs,

\begin{eqnarray*} P(E\mid A=H) &=& P(E\mid B_r\cap A=H)P(B_r\mid A=H) + P(E\mid B_r^c\cap A=H)P(B_r^c\mid A=H) \\ && \\ &=& p^{r-1} + P(E\mid A=T)(1-p^{r-1}). \qquad\qquad\qquad\qquad\qquad(1) \\ \end{eqnarray*}

$\\$

Next, we can find $P(E\mid A=T)$ by way of $P(E^c\mid A=T)$ and that is found in a similar way to $P(E\mid A=H)$. For $i=1,\ldots,s,\;$ define event $C_i = \text{"First head occurs on $i^{th}$ toss"}$. Then,

\begin{eqnarray*} P(E\mid A=T) &=& 1 - P(E^c\mid A=T) \\ && \\ &=& 1 - [P(E^c\mid C_r\cap A=T)P(C_r\mid A=T) + P(E^c\mid C_r^c\cap A=T)P(C_r^c\mid A=T)] \\ && \\ &=& 1 - [1\cdot (1-p)^{s-1} + P(E^c\mid A=H)(1-(1-p)^{s-1})] \\ && \\ &=& 1 - [(1-p)^{s-1} + (1 - P(E\mid A=H))(1-(1-p)^{s-1})] \\ && \\ &=& (1-(1-p)^{s-1}) P(E\mid A=H). \qquad\qquad\qquad\qquad\qquad(2) \\ && \\ \end{eqnarray*}

We now proceed to find $P(E)$. Substitute $(2)$ into $(1)$:

\begin{eqnarray*} P(E\mid A=H) &=& p^{r-1} + (1-p^{r-1}) (1-(1-p)^{s-1}) P(E\vert A=H) \\ && \\ \therefore P(E\mid A=H) &\times& (p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}) \;\; = \;\; p^{r-1} \\ && \\ P(E\mid A=H) &=& \dfrac{p^{r-1}}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}} \qquad\qquad\qquad(3) \\ \end{eqnarray*}

Substitute this into $(2)$:

\begin{eqnarray*} P(E\mid A=T) &=& \dfrac{p^{r-1} - p^{r-1} (1-p)^{s-1}}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}} \qquad\qquad\qquad(4) \\ \end{eqnarray*}

$\\$

Finally, using results $(3)$ and $(4)$,

\begin{eqnarray*} P(E) &=& P(E\mid A=H)P(A=H) + P(E\mid A=T)P(A=T) \\ && \\ &=& \dfrac{p^r + p^{r-1}(1-p) - p^{r-1} (1-p)^s}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}} \\ && \\ &=& \dfrac{p^{r-1} - p^{r-1} (1-p)^s}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}}. \end{eqnarray*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.