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Consider the classic Monty Hall Problem, but with four doors, labelled A, B, C, and D.

I want to calculate the probability of winning with the strategy "choose, switch, stick". I'm struggling to get from maths I think is right to the agreed upon answer, so was hoping someone could help.

I start off by picking door A. Door B is opened, so my probability distribution is as follows:

A(1/4), B(0), C(3/8), D(3/8)

This makes sense as I learnt nothing about door A, and door C and D share the remaining probability.

I then switch arbitrarily to C (C and D are indifferent). At this point, Monty will either open door A or door D.

If Monty opens door A, I propose the following distribution:

A(0), B(0), C(1/2), D(1/2)

On the other hand, if Monty opens door D:

A(4/7), B(0), C(3/7), D(0)

To calculate the probability of my strategy succeeding, I then need to combine the probability of winning depending on which door was opened.

From my understanding, this should be 3/8: http://mathworld.wolfram.com/MontyHallProblem.html. However, I haven't reached this answer regardless of how I try to combine the two scenarios.

When Monty opens the second door after you've switched to C, the conditional probability that the prize is behind door C (viz. $ frac38 $) does not change, for the same reason that it doesn't change for the door you've picked in the original Monty Hall problem. Thus, the probability that the prize is behind the remaining unopened door is $ frac58 $, regardless of whether that door is door A or door D.

Addendum: In light of Misha Lavrov's answer I should point out that in my above comment I have made the standard assumption that when Monty chooses a door to open he always picks one equiprobably from among the unselected that conceal booby prizes, and independently of any other occurrence

We need to assume something about Monty's strategy for opening doors. Let's assume that he opens a randomly chosen door of all doors that you haven't picked without the prize behind it, as is traditional.

Case 1: Monty opens B, then D.

Suppose the prize is behind door A. The probability that Monty opens door B when you pick door A, then opens door D when you switch to door C, is $frac13 cdot 1 = frac13$.

Suppose the prize is behind door C. The probability that Monty opens door B when you pick door A, then opens door D when you switch to door C, is $frac12 cdot frac12 = frac14$.

Initially, doors A and C were equally likely, but Monty's actions are $frac43$ times likelier if the prize is behind door A. So the probability is $frac47$ for door A and $frac37$ for door C (the odds are $frac43 : 1$).

Case 2: Monty opens B, then A.

Suppose the prize is behind door C. The probability that Monty opens door B when you pick door A, then opens door A when you switch to door C, is $frac12 cdot frac12 = frac14$ (as before).

Suppose the prize is behind door D. The probability that Monty opens door B when you pick door A, then opens door A when you switch to door C, is $frac12 cdot 1 = frac12$.

Initially, doors C and D were equally likely, but Monty's actions are $2$ times likelier if the prize is behind door D. So the probability is $frac13$ for door C and $frac23$ for door D (the odds are $1 : 2$).

Overall probability

But the above calculations are irrelevant for the problem you want to solve.

To figure out how good your chances are if you plan to pick a door, switch after the first opened door, and then stick with that choice no matter what, you don't need to do casework based on the second opened door, since it won't affect your decision. You've computed probabilities of $frac14, frac38, frac38$ for doors A, C, and D, so the final answer if you switch to door C is $frac38$.