A slip of paper is given to A, who marks it with either a plus or a minus sign; the probability of his writing a plus is $\frac{1}{3}$. He then passes the slip to B, who may either leave it or change the sign before passing it on to C. Next C passes the slip to D after perhaps changing the sign; finally D passes it to an honest judge after perhaps changing the sign. The judge sees a plus sign on the slip. It is known that B,  C, and D each change the sign with probability $\frac{2}{3}$. What is the probability that A originally wrote a plus?

Let x be the event that A left by the plus sign and y be the event that the final result was a plus sign. We have to fixed out $\mathrm{P}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\frac{\mathrm{P}(\mathrm{x}\cap \mathrm{y})}{\mathrm{P}\left(\mathrm{y}\right)}$ for final plus sign it must be an even no of times changes if he started with a plus sign then it changes either 0, 2 or 4 times.
$$\begin{gathered} 0\text{ times }\to {\left(\frac{1}{3}\right)}^4 \\ 4\text{ times }\to {\left(\frac{2}{3}\right)}^4 \\ 2\text{ times }{\to }^4{\mathrm{C}}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^2 \\ \text{ So }\mathrm{P}\left(\mathrm{y}\right)=\frac{1}{81}+\frac{16}{81}+\frac{24}{81}=\frac{41}{81} \end{gathered}$$
for $\mathrm{P}(\mathrm{x}\cap \mathrm{y})$ to occur there has been an even number of times.
But A did not make one of them hence there are either 0 or 2 changes. Probability of 0 changes is ${\left(\frac{1}{3}\right)}^4$ but probability of 2 changes is
$$\begin{gathered} \begin{array}{l}{ }^3{\mathrm{C}}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^2=\frac{12}{81}\\ \mathrm{P}(\mathrm{x}\cap \mathrm{y})=\frac{1}{81}+\frac{12}{81}=\frac{13}{81}\end{array} \\ \text{ So }\mathrm{P}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\frac{13/81}{41/81}=\frac{13}{41} \end{gathered}$$