I have two identical dice. When I throw either one of them, the
probability of it showing a 6 is \(p\) and the probability of it not showing a 6 is \(q\), where \(p+q=1\). As an experiment to determine \(p\), I throw the dice simultaneously until at least one die shows a 6. If both dice show a six on this throw, I stop. If just one die shows a six, I throw the other die until it shows a 6 and then stop.
Show that the probability that I stop after \(r\) throws
is \(pq^{r-1}(2-q^{r-1}-q^r)\), and find an expression for the
expected number of throws.
[{\bf Note:} You may use the result $\ds \sum_{r=0}^\infty rx^r =
x(1-x)^{-2}\(.]
In a large number of such experiments, the mean number of throws was \)m\(. Find an estimate for \)p\( in terms of \)m$.
Solution:
\(\,\) \begin{align*}
\mathbb{P}(\text{stop after r}) &= \mathbb{P}(\text{both stop at r}) + 2\mathbb{P}(\text{first stops before r second stops at r})\\
&= (q^2)^{r-1} p^2 + 2\cdot q^{r-1} p\cdot(1-q^{r-1}) \\
&= q^{r-1}p\left (2-2q^{r-1}+pq^{r-1} \right) \\
&= q^{r-1}p\left (2-q^{r-1}(1+p+q-p) \right) \\
&= q^{r-1}p\left (2-q^{r-1}-q^r\right) \\
\end{align*}
\begin{align*}
\E[\text{throws}] &= \sum_{r=1}^{\infty} r \mathbb{P}(\text{stop after r}) \\
&= \sum_{r=1}^{\infty} r q^{r-1}p\left (2-q^{r-1}-q^r\right) \\
&= \sum_{r=1}^{\infty} 2r q^{r-1}p-\sum_{r=1}^{\infty}r pq^{2r-2}-\sum_{r=1}^{\infty}r q^{2r-1}p \\
&=2p \sum_{r=1}^{\infty} r q^{r-1}-pq^{-2}\sum_{r=1}^{\infty}r q^{2r}-pq^{-1}\sum_{r=1}^{\infty}r q^{2r} \\
&= 2p(1-q)^{-2} - pq^{-2}q^2(1-q^2)^{-2}-pq^{-1}q^2(1-q^2)^{-2} \\
&= 2pp^{-2} -p(1+q)(1-q^2)^{-2} \\
&= 2p^{-1}-p(1+q)(1+q)^{-2}p^{-2} \\
&= 2p^{-1}-p^{-1}(1+q)^{-1} \\
&= \frac{2(1+q)-1}{p(1+q)} \\
&= \frac{1+2q}{p(1+q)} \\
&= \frac{3-2p}{p(2-p)}
\end{align*}
\(\,\) \begin{align*}
&& m &= \frac{3-2p}{p(2-p)} \\
\Rightarrow && 2mp-mp^2 &= 3-2p \\
\Rightarrow && 0 &= mp^2-(2m+2)p + 3 \\
\Rightarrow && p &= \frac{2m+2 \pm \sqrt{(2m+2)^2-12m}}{2m} \\
&&&= \frac{m+1- \sqrt{m^2-m + 1}}{m} \\
\end{align*}
If we are looking for an approximation, we could say \(p^2 \approx 0\) and \(p \approx \frac{3}{2(m+1)}\)
A hostile naval power possesses a large, unknown number \(N\) of
submarines. Interception of radio signals yields a small number \(n\)
of their identification numbers \(X_i\) (\(i=1,2,...,n\)), which are taken
to be independent and uniformly distributed over the continuous range
from \(0\) to \(N\). Show that \(Z_1\) and \(Z_2\), defined by
$$
Z_1 = {n+1\over n} {\max}\{X_1,X_2,...,X_n\}
\hspace{0.3in} {\rm and} \hspace{0.3in}
Z_2 = {2\over n} \sum_{i=1}^n X_i \;,
$$
both have means equal to \(N\).
Calculate the variance of \(Z_1\) and of \(Z_2\). Which estimator
do you prefer, and why?