Monty Hall Problem - Host doesn't know confusion I am given the variation of the Monty Hall problem, where the host does not know where the car/goats are. There are three doors labelled A, B and C.  Let probability of choosing A, and the host opening the door B afterwards, be denoted by $P(B|A)$. Then, assuming A is the winning door, the probability of success is given by:
$$P(win) = \frac{P(B|A) + P(C|A)
}{1 − P(A|B) − P(A|C)
} $$
And substituting the corresponding probabilites yields the result 0.5, which I belive is the correct result of this variation of Monty Hall. Now comes the part that I am confused about: assume a probability $p$, denoting the probability that the host will reveal the car after the guest has chosen a goat. Under this condition, I choose a door and the host opens a different door revealing it to be empty. Show that the above formula reduces down to:
$$P(win) = \frac{1}{3-2p} $$
I have read the numerous other posts on this variant of the Monty Hall problem. However, I just cannot seem to obtain that second formula. 
Any help or hints would be appreciated.
Edit:
Quote of the problem:
First part:
"After you have been presented the boxes and have made your choice, the host randomly (she/he
doesn’t know where the prize is!) opens one of the other boxes that you didn’t chose and it happens to
be empty. Let for example the pair (A, B) denote the case where you choose A and the host opens the
box B afterwards; the probability of this occurring is given by $P(B|A)$."
Second part:
"Assume there is a probability p that the host reveals the keys if you have chosen an empty box.
Under this condition, you have chosen a box and the host opens a different box, revealing it to be empty.
Show that the probability becomes:
$$P(win) = \frac{1}{3-2p} $$
 A: Based on the wording of the problem, the assumptions seem to be:


*

*There is one box containing a prize and two empty boxes. You initially have no information about which box is which.

*You initially choose a box.

*If the chosen box contains the prize, the host opens one of the other boxes (with equal probability to choose either box).

*If the chosen box is empty, the host opens the prize box with probability $p$ and opens the other empty box with probability $1-p.$

*After the host opens a box, you are allowed to choose any of the three boxes, even the open box.

*In the end, after the host opens a box, you choose the box that has the greatest probability of containing the prize.

*You win if the box you choose in the end contains the prize.
Assumption 5 is the least confident assumption, since the usual wording assumes that you have a choice of your first chosen box or the other unopened box. If we change assumption 5 so that you cannot choose the open box, you lose every time the host opens the prize box. However, as the specific question is the probability of winning given that the host opens an empty box, it is safe to assume that (as in the standard Monty Hall problem) you will not choose the open box.
If $p = 1,$ the host only opens an empty box when you have already chosen the prize box, so once you see that the host has opened an empty box, you know you have already chosen the prize box. Conditioned on the fact that the host opened an empty box, you keep your original choice and win with probability $1.$
But if $p = 0,$ this is the standard Monty Hall problem.
You win with probability $\frac23.$ But
$$ \frac 1{3 - 2(0)} = \frac13. $$
So the proposed expression, $\dfrac 1{3 - 2p},$ is not the probability of winning if the host opens an empty box.

We can work the problem as follows:
Let $Q$ be the event that your first choice is the prize box.
Let $R$ be the event that the host opens the prize box.
Then
\begin{align}
P(Q \cap R) &= 0, \\
P(Q \cap R^C) &= \frac13, \\
P(Q^C \cap R) &= \frac23p, \\
P(Q^C \cap R^C) &= \frac23(1-p).
\end{align}
Then
\begin{align}
P(R^C) &= P(Q \cap R^C) + P(Q^C \cap R^C) = \frac13 + \frac23(1-p) = 1 - \frac23p, \\
P(Q \mid R^C) &= \frac{P(Q \cap R^C)}{P(R^C)}
= \frac{1/3}{1 - (2/3)p} = \frac1{3 - 2p}.
\end{align}
So we see that $\dfrac 1{3 - 2p}$ actually is the probability that your first choice is the prize box, given that the host opens an empty box.
The probability of winning, however, can be better.
An optimal strategy is, if the host opens an empty box, choose your original box if $P(Q \mid R^C) \geq \frac12,$
but choose the other unopened box if $P(Q \mid R^C) < \frac12.$
The probability that the other unopened box is the prize box, given that the host opens an empty box, is
$$ 1 - \frac 1{3 - 2p} = \frac{2 - 2p}{3 - 2p}. $$
The probability of winning if the host opens an empty box is therefore
$$
P(\mathrm{win}\mid R^C) = \begin{cases}
\dfrac 1{3 - 2p} & \text{if}\ p \geq \frac12, \\
\dfrac{2 - 2p}{3 - 2p} & \text{if}\ p < \frac12.
\end{cases}
$$

It is also interesting to ask for the unconditioned probability of winning the game.
If $p = 0$ then the problem is the standard Monty Hall problem, and
$$ P(\mathrm{win}) = P(\mathrm{win} \mid R^C) = \frac13. $$
But if $p > 0$ the answer depends on Assumption 5.
Note that in this case,
\begin{align}
P(R) &= P(Q \cap R) + P(Q^C \cap R) = \frac23p, \\
P(Q \mid R) &= \frac{P(Q \cap R)}{P(R)} = \frac{0}{(2/3)p} = 0, \\
\end{align}
so under assumption 5, if the host opens the prize box we choose it and win. Then
$$P(\mathrm{win}) = P(R \cap \mathrm{win}) + P(R^C \cap \mathrm{win})
 = P(R) + P(R^C \cap \mathrm{win}).$$
We find that
\begin{align}
P(R^C \cap \mathrm{win}) &= P(\mathrm{win}\mid R^C)P(R^C) \\
&= \left(1 - \frac23p\right) P(\mathrm{win}\mid R^C) \\
&= \frac{3 - 2p}3 P(\mathrm{win}\mid R^C) \\
&= \begin{cases}
\dfrac 13 & \text{if}\ p \geq \frac12, \\
\dfrac{2 - 2p}3 & \text{if}\ p < \frac12.
\end{cases}
\end{align}
So the overall probability of winning is
$$
P(\mathrm{win}) = \frac23p + P(R^C \cap \mathrm{win})
= \begin{cases}
\dfrac{1+2p}3 & \text{if}\ p \geq \frac12, \\
\dfrac23 & \text{if}\ p < \frac12.
\end{cases}
$$
On the other hand, if we are not allowed to choose the open box, then we lose every time that the host opens the prize box.
Then
$$P(\mathrm{win}) = P(R \cap \mathrm{win}) + P(R^C \cap \mathrm{win})
 = 0 + P(R^C \cap \mathrm{win}).$$
So the overall probability of winning is
$$
P(\mathrm{win}) = \begin{cases}
\dfrac 13 & \text{if}\ p \geq \frac12, \\
\dfrac{2 - 2p}3 & \text{if}\ p < \frac12.
\end{cases}
$$
