Solving another game theoretical problem In the wardens
calculation it is assumed that if $A$ were to be pardoned, then with equal probability the warden
would tell $A$ that either $B$ or $C$ would be executed. However, The warden can assign probabilities $\gamma$ and $1-\gamma$ to these events as shown in Table 1.
Thus following events hold:
\begin{align}
A &= \text{"$A$ is pardoned"} \\
B &= \text{"$B$ is pardoned"} \\
C &= \text{"$C$ is pardoned"} \\
W &= \text{"Warden tell $A$ that $B$ will be executed".} \\
\end{align}
I would like to calculate $\mathbb{P}(A|\mathcal{W})$ as a function of $\gamma$. For what values of $\gamma$ is $\mathbb{P}(A|\mathcal{W})$ less than, equal to, or greater than $\frac{1}{4}$ 
I struggle somehow to approach the start.
$$
 \begin{array}{r|cccc}
 & \texttt{Prisoner pardoned} & \texttt{Warden tells A} & \texttt{Probability}\\
 \hline
 & A &  \texttt{B dies}  & \texttt{with probability} \ \gamma \\
 & A &  \texttt{C dies} & \texttt{with probability} \ 1-\gamma\\
 & B &  \texttt{C dies} &   \\
    & C &  \texttt{B dies} &   \\
\end{array}
$$
 A: We have events:
\begin{align}
A &= \text{"$A$ is pardoned"} \\
B &= \text{"$B$ is pardoned"} \\
C &= \text{"$C$ is pardoned"} \\
W &= \text{"Warden tell $A$ that $B$ will be executed".} \\
\end{align}
By Bayes Theorem we have:
\begin{align}
P(A\mid W) &= \dfrac{P(W\mid A)P(A)}{P(W\mid A)P(A) + P(W\mid B)P(B) + P(W\mid C)P(C)} \\
& \\
&= \dfrac{\gamma\cdot\dfrac{1}{3}}{\gamma\cdot\dfrac{1}{3} + 0\cdot\dfrac{1}{3} + 1\cdot\dfrac{1}{3}} \\
& \\
&= \dfrac{\gamma}{\gamma+1}.
\end{align}
Notes:


*

*Setting $\gamma=\dfrac{1}{2}$ you get the original example with resulting probability $\dfrac{1}{3}$.

*Setting $\gamma=0$ we get a resulting probability of $0$. This is because if $W$ occurs it's because $C$ is pardoned and the warden is forced to say $B$ - and if $C$ is pardoned, $A$ cannot be.

*Setting $\gamma=1$ we get a resulting probability of $\dfrac{1}{2}$. This is because the warden will say $B$ only if $B$ is not pardoned. Effectively, the question has changed from "which of $B$ and $C$ will be executed?" to "will $B$ be executed?", with "B" meaning yes and "C" meaning no. If answered yes, then $A$ and $C$ are equally likely to be pardoned.
