Why this probability is true $\Pr[M_3\mid D_1]=1/2$ in Monty Hall Problem. I want to understand the three conditional probabilities of the Bayes theorem of the Monty Hall Problem.
Suppose that $M_i$ for $i=\{1,2,3\}$ is the event where we have a goat in the $i$th door and suppose that $D_j$ for $j=\{1,2,3\}$ is the event where a car is behind the $j$th door.
Now, we have the following probabilities:

$\Pr[M_3\mid D_1]=1/2$
$\Pr[M_3\mid D_2]=1$
$\Pr[M_3\mid D_3]=0$

Now, for $\Pr[M_3\mid D_2]=1$ because we are given car in behind door no. $2$, therefore there must be a goat in door 3 since we know car isn't in door $3$. For $\Pr[M_3\mid D_3]=0$ because we are given a car in door no. $3$, and probability to find goat in door $3$ is $0$ since the problem ensure that only one thing behind each door. Now, I don't understand why $\Pr[M_3\mid D_1]=1/2$ for me it should be $1$. Why? because we are given the car in door no. 1, and so we are very sure that goat would be in any door except $1$, so the probability should be equal 1.
Could someone argue with me why the probability of this statement is true $\Pr[M_3\mid D_1]=1/2$?
 A: Let $C_n$ represent the event that the car is behind door $n$
So $P(C_n)=1/3$
Without loss of generality we can assume that you originally guess Door #1
Now let $O_n$ represent the event the Monty opens door $n$. 
The critical thing to keep in mind is that he won't open door #1 and he won't open a door with a car behind it
$$\begin{eqnarray*}
P(O_1)&=&0 \\
\\P(O_2|C_1)&=&1/2
\\P(O_2|C_2)&=&0
\\P(O_2|C_3)&=&1\\
\\P(O_3|C_1)&=&1/2
\\P(O_3|C_2)&=&1
\\P(O_3|C_3)&=&0
\end{eqnarray*}$$
**** EDIT *******
I think this is really where the answer to your question lies:
What you wrote as $P(M_3|D_1)$ is what I wrote as $P(O_3|C_1)$ , so your $M_3$ is not the event that there is a goat behind door #3 but rather the event that Monty opened door #3 (revealing a goat, of course.)
$P(O_3|C_1)=1/2$ because we are always looking at the case in which your initial guess is for Door #1, in this particular situation, the car is actually behind door #1, so Monty is free to open either doors 2 or 3.
****  END EDIT *******
Notice that $P(O_2)=p(O_3)=1/2$
Let $P(O_nC_m)$ represent the probability that Monty opens the door $n$ and the car is behind door $m$
Using the formula
$P(O_nC_m)=P(O_n|C_m)P(C_m)=1/3P(O_n|C_m)$
$$\begin{eqnarray*}
P(O_1C_m)&=&0
\\
\\P(O_2C_1)&=&1/6
\\P(O_2C_2)&=&0
\\P(O_2C_3)&=&1/3\\
\\P(O_3C_1)&=&1/6
\\P(O_3C_2)&=&1/3
\\P(O_3C_3)&=&0
\end{eqnarray*}$$
now we can calculate the conditional probabilities $P(C_m|O_n)$ using the formula $$P(C_m|O_n)= \frac{P(O_nC_m)}{P(O_n)} = 2  P(O_nC_m)$$
$$\begin{eqnarray*}
P(C_m|O_1)&=&0 \\
\\P(C_1|O_2)&=&1/3
\\P(C_2|O_2)&=&0
\\P(C_3|O_2)&=&2/3\\
\\P(C_1|O_3)&=&1/3
\\P(C_2|O_3)&=&2/3
\\P(C_3|O_3)&=&0
\end{eqnarray*}$$
So Monty definitely won't open door #1
If he opens door #2 the car is twice as likely to be behind door #3 as it is behind door #1
If he opens door #3 the car is twice as likely to be behind door #2 as it is behind door #1
**** EDIT #2 ****
As mentioned by Hurkyl, $P(O_2|C_1)=1/2$ is actually an assumption of even-handedness on Monty's part. The final result that you gain advantage from switching your choice of door works out to be independent of Monty's door-opening strategy. 
If we were to let $$P(O_2|C_1)=x; x\in [0,1]$$
we would also require $$P(O_3|C_1)=1-x$$
then we get ...
$$ P(O_2)=\frac{1+x}{3} \,\,\text{ and }\;\; P(O_3)=\frac{2-x}{3} $$
and finally ...
$$ P(C_1|O_2)=\frac{x}{1+x} \,\,\text{ and }\;\; P(C_1|O_3)=\frac{1-x}{2-x} $$
note that both of these probabilities fall in the interval $[0, \frac12]$ , so it is always to your advantage to switch your choice of door . 
In the most extreme cases, where $x=0$ or $x=1$, situations can arise in which it makes no difference whether you change your choice of door. 
*** END EDIT #2 *****
A: You are correct in that $P(M_3 | D_1) = 1$.
You are mistaken in that you believe wikipedia claims $P(M_3 | D_1) = 1$.
There are four basic families of events here:


*

*$M_i$ - there is a goat behind door $i$

*$D_i$ - there is a car behind door $i$

*$H_i$ - Mr. Hall opens door $i$

*$C_i$ - the contestant picks door $i$


What wikipedia claims is that
$$ P(H_1 \mid C_1, D_1) = 0 $$
$$ P(H_2 \mid C_1, D_1) = \frac{1}{2} $$
$$ P(H_3 \mid C_1, D_1) = \frac{1}{2} $$
That is, if the contestant picks door #1, and that really is the door with the car, then Mr. Hall picks one of the other doors to open with equal probability.
(note this is an assumption — one generally assumes the problem is symmetric, but in 'real world' problems, one must account for the fact that whatever is in Mr. Hall's place may have a preference for one door over another, and might open, say, the middle door rather than either end door when given a choice)
Basically, you have misidentified the information you're given. When Mr. Hall opens door #3, you only recognized that you learned there was a goat behind it — you've failed to acknowledge that you also learned that Mr. Hall did not open door #2. Properly handling that piece of information is of crucial importance to correctly understanding the probabilities involved.
