Q.4.76 - A First Course in Probability

Question

Solve the Banach match problem (Example $8e$) when the left-hand matchbox originally contained $N1$matches and the right-hand box contained $N2$ matches.

Step-by-Step Solution

Answer

$\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)=\left(\begin{array}{c}{N}_1+N_2-k\\ N_1\end{array}\right){\left(\frac{1}{2}\right)}^{N_1+N_2-k+1}+\left(\begin{array}{c}{N}_1+N_2-k\\ N_2\end{array}\right){\left(\frac{1}{2}\right)}^{N_1+N_2-k+1}$

Step 1:Given information

Given in the question that, we need to solve the Banach match problem (Example 8e) when the left-hand matchbox originally contained $N1$ matches and the right-hand box contained $N2$ matches 

Step 2:Explanation

Let $E$ denote the event that the mathematician first discovers that the right-hand matchbox is empty and there are$K$ matches in the left-hand box at the time. Now this event will occur If and only if the ${\left(N_1+1\right)}^{\text{th }}$ choice of the right - hand matchbox is made at the $N_1+1+N_2-k$ trial. Following negative Binomial random variable distribution

$\left(\right.$with $\left.\mathrm{p}=\frac{1}{2};\mathrm{r}={\mathrm{N}}_1+1; \mathrm{n}={\mathrm{N}}_1+{\mathrm{N}}_2-k+1\right)$

$\mathrm{P}\left(\mathrm{E}\right)=\left(\begin{array}{c}{\mathrm{N}}_1+N_2-\mathrm{k}\\ {\mathrm{N}}_1\end{array}\right){\left(\frac{1}{2}\right)}^{{\mathrm{N}}_1+{\mathrm{N}}_2-k+1}$

As there is an equal probability that it is the left-hand box that is first discovered to be empty and there are $\mathrm{K}$ matches in the right-hand box at that time, the expected result is

$\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)=\left(\begin{array}{c}{N}_1+N_2-k\\ N_1\end{array}\right){\left(\frac{1}{2}\right)}^{N_1+N_2-k+1}+\left(\begin{array}{c}{N}_1+N_2-k\\ N_2\end{array}\right){\left(\frac{1}{2}\right)}^{N_1+N_2-k+1}$

Step 3:Final answer

the desired result is:$\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)=\left(\begin{array}{c}{N}_1+N_2-k\\ N_1\end{array}\right){\left(\frac{1}{2}\right)}^{N_1+N_2-k+1}+\left(\begin{array}{c}{N}_1+N_2-k\\ N_2\end{array}\right){\left(\frac{1}{2}\right)}^{N_1+N_2-k+1}$