Do multiple observers increase the probability of detecting an event? I will describe my question with an example:
A machine manufactures toys and places them on a conveyor belt. A toys expert overlooking the conveyor belt observes the toys. If a toy is faulty he can successfully detect it with probability $P_{d}$. It is possible to have more experts overlooking the process.
If I have one expert and he detects a faulty toy, then I am $P_d$ sure that the toy is faulty (right?). Intuitively I understand that if more experts detect the toy as faulty then probability of the toy being faulty should increase (right?). What happens if 2 experts report the toy as fault and the other 2 report is as not faulty? What is the probability of the toy being faulty then?  How can you work these questions out formally with probability theory?
 A: Say we have a specific toy on the conveyor belt and only one toy expert. Let $Z$ be a 0/1-variable, with a 1 indicating that the toy is faulty. Let $D$ be another binary variable, with a 1 indicating that our toy expert detects the faulty toy. Using your notation,
$ P_d  = P(D=1 | Z=1). $
In words, conditional on the toy being faulty $P_d$ is the probability of the expert detecting it. This is not the same as $P(Z=1 | D=1)$ which in words can be described as the probability of a faulty toy conditional on the expert classifying the toy as faulty. The two probabilities are related through Bayes's rule and are equal if and only if the probability of a faulty toy equals the probability of the expert classifying it as faulty. This can be seen from Bayes' rule which states that for two events, $A,B$, such that $P(B) \neq 0$,
$ P(A | B) = \frac{P(B |A)P(A)}{P(B)} $.
If you have more experts you'll have to assume something about the relationship between their judgements. Are their judgements independent? Independent conditional on the state of the toy? Let's assume the latter for now. This might not be realistic as one would expect them to react on the same things. However, otherwise you'll have to specify some other kind of dependence between their judgements. If we assume this conditional independence and let $D_1, D_2$ be the decisions of experts 1 and 2 we for instance get, using Bayes's rule again,
$ P(Z=1 | D_1 = 1, D_2 = 1) = \frac{P(D_1 = 1, D_2 = 1 | Z = 1)P(Z=1)}{P(D_1 = 1, D_2 = 2)} $.
Using the conditional independence we can write $P(D_1 = 1, D_2 = 1 | Z = 1) = P(D_1 = 1 | Z = 1)P(D_2 = 1 | Z=1)$. If we also assume something about the probability $P(D_1 = 1, D_2 =1)$ we get a result. To answer your original question we'll have to compare this probability with 
$P(Z=1 | D_1 = 1) = \frac{P(D_1 = 1| Z=1)P(Z=1)}{P(D_1=1)}$
If we assume that $P(D_1 = 1, D_2 =1) = P(D_1 = 1) P( D_2 =1)$ we see that the chance of a faulty toy is strictly larger when we have two experts that agree (as opposed to the case with only one expert) whenever 
$\frac{P(D_2 =1 | Z=1 )}{P(D_2 =1)} > 1$.
This last condition is definitely a fair assumption as expert number 2 would otherwise be very poor at his/her job.
A: You have to carefully distinguish between conditional probabilities and joint probabilities.  
Let the "expert" be represented by the variable $E$ that takes the values $\{b,g\}$ (for "bad" and "good" status of the toy). Let the toy be represented by the variable $T$, having the same support.  
The way you describe it (if the toy is faulty, then the expert will detect it...") we have
$$P_d = \Pr(E=b \mid T=b)$$
i.e. it is a conditional probability. So, no, in general you are not "$P_d$ sure that the toy is faulty", because $P_d$ presupposes that it is, indeed, faulty. Υou have to take into account the possibility that the expert will make a mistake - declare a toy as faulty while in reality it is not.  
It appears that you want to know "given that the expert declared the toy as faulty, what is the probability that the toy is indeed faulty?" This is written as
$$ \Pr(T=b \mid E=b ) =?$$
By standard rules
$$\Pr(E=b \mid T=b) \cdot \Pr(T=b) = \Pr(T=b \mid E=b ) \cdot \Pr(E=b)$$
$$\Rightarrow \Pr(T=b \mid E=b) = \frac {\Pr(T=b)}{\Pr(E=b)}\cdot P_d$$
Now ${\Pr(E=b)}$, or an estimate of it, is the total number of times the expert declared a toy as faulty divided by the total number of toys that passed in front of him.
Obtaining an estimate for $\Pr(T=b)$ requires sampling also from toys that passed the expert. In general, you need to obtain an estimate for the joint probability distribution, from which all other probabilities, conditional and marginal(unconditional) will flow, i.e. to determine the four probabilities  
1) $\Pr(T=b, E=b)$
2) $\Pr(T=b, E=g)$
3) $\Pr(T=g, E=b)$
4) $\Pr(T=g, E=g)$
This requires sampling.
