Problem source

Four students, Arthur, Bertha, Chandra and Delilah, exchange gossip. When Arthur hears a rumour, he tells it to one of the other three without saying who told it to him. He decides whom to tell by choosing at random amongst the other three, omitting the ones that he knows have already heard the rumour. When Bertha, Chandra or Delilah hear a rumour, they behave in exactly the same way (even if they have already heard it themselves). The rumour stops being passed round when it is heard by a student who knows that the other three have already heard it. Arthur starts a rumour and tells it to Chandra. By means of a tree diagram, or otherwise, show that the probability that Arthur rehears it is $3/4$. Find also the probability that Bertha hears it twice and the probability that Chandra hears it twice.

Solution source

Without loss of generality, $C$ will tell $B$ about the rumour. 
If $B$ tells $D$ then $D$ can either tell $A$ or $C$ at which point either $A$ is told or the rumour stops spreading.

\begin{center}
\begin{tikzpicture}
    % Nodes
    \node (B) {C$\rightarrow$B};
    
    % Nodes branching from B
    \node (A_top) [above right=0.5cm and 1.5cm of B] {A};
    \node (D) [below right=0.5cm and 1.5cm of B] {D};
    
    % Nodes branching from D
    \node (A_end) [above right=0.5cm and 1.5cm of D] {A};
    \node (C_end) [below right=0.5cm and 1.5cm of D] {C};

    % Connections from C to B
    % \draw[->] (C_start) -- (B);

    % Connections from B
    \draw (B) -- node[above, pos=0.5] {$1/2$} (A_top);
    \draw (B) -- node[below, pos=0.5] {$1/2$} (D);

    % Connections from D
    \draw (D) -- node[above, pos=0.5] {$1/2$} (A_end);
    \draw (D) -- node[below, pos=0.5] {$1/2$} (C_end);

\end{tikzpicture}
\end{center}

Therefore $\mathbb{P}(\text{Arthur rehears}) = 3/4$

For the chances Chandra hears it twice, still WLOG, assume she tells B:


\begin{center}
\begin{tikzpicture}
    % Nodes
    \node (B) {A$\rightarrow$C$\rightarrow$B};
    
    % Nodes branching from B
    \node (A_top) [above right=0.5cm and 1.5cm of B] {A};
    \node (D) [below right=0.5cm and 1.5cm of B] {D};
    
    % Nodes branching from D
    \node (A_end) [above right=0.5cm and 1.5cm of D] {A};
    \node (C_end) [below right=0.5cm and 1.5cm of D] {C};

    \node (D_two) [above right=0.5cm and 1.5cm of A_top] {D};

    \node (B_end_two) [above right=0.5cm and 1.5cm of D_two] {B};
    \node (C_end_two) [below right=0.5cm and 1.5cm of D_two] {C};

    \node (B_end_three) [right=1.5cm of A_end] {B};

    % Connections from C to B
    % \draw[->] (C_start) -- (B);

    % Connections from B
    \draw (B) -- node[above, pos=0.5] {$1/2$} (A_top);
    \draw (B) -- node[below, pos=0.5] {$1/2$} (D);

    % Connections from D
    \draw (D) -- node[above, pos=0.5] {$1/2$} (A_end);
    \draw (D) -- node[below, pos=0.5] {$1/2$} (C_end);

    \draw (A_top) -- node[above, pos=0.5] {$1$} (D_two);

    
    \draw (D_two) -- node[above, pos=0.5] {$1/2$} (B_end_two);
    \draw (D_two) -- node[below, pos=0.5] {$1/2$} (C_end_two);
    
    \draw (A_end) -- node[below, pos=0.5] {$1$} (B_end_three);

\end{tikzpicture}
\end{center}

So her chance of hearing it twice is $\frac12$

The person who hears it 3rd has a $\frac12$ chance of hearing it twice, but the person who hears if 4th has no chance. Therefore Bertha has a $\frac14$ chance of hearing it twice.