Monty Hall problem with 7 doors According to the Monty Hall Problem, I select a door, say door number $1$. Monty then opens $3$ of the remaining $6$ doors that do not contain the prize. The question is should I switch.
Note that all the doors are said to have equal chance of containing the prize.
My attempt:
$$P(1) = \frac{1}{7}$$
Once Monty opens three doors, we know that none of them contain the prize, so, we update our beliefs $$P(i|j, k, m) = \frac{1}{4}$$
Thus any of the closed doors has probability $\frac{1}{4}$ of containing the prize, so switching is pointless.
What am I doing wrong here?
 A: Under your formulation of the problem, there are seven possible positions for the prize, each equally likely.  Under only one of those positions was the prize behind the door you selected (the first door).  Thus far, prior to revealing any doors, you should agree that the probability of winning is $1/7$.
Now, the opening of the other doors does not change the fact that in only one of these seven scenarios, you would win.  For example, if the prize was behind door #3, the host opens any other three doors excluding #1 and #3, and you would lose by staying with #1.  We can enumerate the outcomes after the non-winning doors are revealed.  Let $P$ represent the location of the prize, $*$ represent an opened door (which has nothing behind it), and $E$ represent an empty but unopened door.  Then the following list might represent the instructions to the host for opening the required doors--note that he never opens door #1 nor a door labeled $P$:
$$(P, *, *, *, E, E, E) \\
(E, P, *, *, *, E, E) \\
(E, *, P, *, *, E, E) \\
(E, *, E, P, *, E, *) \\
(E, E, *, *, P, *, E) \\
(E, *, E, *, *, P, E) \\
(E, *, *, E, E, *, P) \\
$$
So for example, if the prize is behind door #3, and you always choose door #1, the host would open doors #2, 4, and 5.  It becomes clear that both before and after opening the doors, each of the outcomes is equally likely.  If you stay with door #1, you win only $1$ out of $7$ times.  If you switch to any other unopened door, you would lose with certainty $1$ out of $7$ times, and in the other $6$ cases where the door you originally chose did not have the prize, you now have a $1/3$ probability of selecting the correct door; thus the probability of winning by switching is $0(1/7) + (1/3)(6/7) = 2/7$.
