Combining probabilities/information from different sources Lets say I have three independent sources and each of them make predictions for the weather tomorrow. The first one says that the probability of rain tomorrow is 0, then the second one says that the probability is 1, and finally the last one says that the probability is 50%. I would like to know the total probability given that information.
If apply the multiplication theorem for independent events I get 0, which doesn't seem correct. Why is not possible to multiply all three if all sources are independent? Is there some Bayesian way to update the prior as I get new information?
Note: This is not homework, is something that I was thinking about.
 A: In the framework of Transferable Belief Model (TBM), it is possible to combine different predictions using for instance the "conjunctive rule of combination". In order to apply this rule, you need to transform the probabilities of the predictions into basic belief assignments. This can be achieved with the so-called Least-Committed-Principle. In R:
library(ibelief)
#probabilities
p1 <- c(0.99, 0.01) # bad results for 0 and 1
p2 <- c(0.01, 0.99)
p3 <- c(0.5, 0.5)

# basic belief assignment, 
# each row represents a subset of (rain, not rain)
# each column represents one prediction
Mat <- LCPrincple(rbind(p1,p2,p3))

# combine beliefs
m <- DST(Mat, 1)

# resulting probability distribution (pignistic probability)
mtobetp(m)
# returns 0.5 and 0.5

For the second example of three independent predictions of 0.75, this approach returns a higher value:
p4 <- c(0.75, 0.25)
Mat <- LCPrincple(rbind(p4,p4,p4))
m <- DST(Mat, 1)
mtobetp(m)
#returns 0.9375 0.0625

This is not very far from the Bayesian approach shown in Arthur B's answer.
A: I think it's worthwhile to look at the weighting scheme based on inverse errors mentioned in one of the answers. If the sources are truly independent and we constrain the weights to sum to one, the weights are given by $$ w_1 = {{\sigma_2^2 \sigma_3^2} \over {\sigma_1^2 \sigma_2^2 + \sigma_1^2 \sigma_3^2 + \sigma_2^2 \sigma_3^2}},\ w_2 = {{\sigma_1^2 \sigma_3^2} \over {\sigma_1^2 \sigma_2^2 + \sigma_1^2 \sigma_3^2 + \sigma_2^2 \sigma_3^2}},\ w_3 ={{\sigma_1^2 \sigma_2^2} \over {\sigma_1^2 \sigma_2^2 + \sigma_1^2 \sigma_3^2 + \sigma_2^2 \sigma_3^2}}. $$
If, as the OP states, the forecasts are equally reliable, then all weights will simplify to $\frac{1}{3}$ and the combined forecast for the given example will be 50%. 
Note that the values of $\sigma_i$ do not need to be known if their relative proportions are known. So if $\sigma_1^2 : \sigma_2^2 : \sigma_3^2 = 1:2:4,$ then the forecast in the example would be $$f = { {{8} \over {14}}*(0) + {{4} \over {14}}*(1) + {{2} \over {14}}*(0.5) } = 0.3571 $$
A: There are a lot of complicated answers given to this question, but what about the Inverse Variance Weighted Mean: https://en.wikipedia.org/wiki/Inverse-variance_weighting

Instead of n repeated measurements with one instrument, if the
  experimenter makes n of the same quantity with n different instruments
  with varying quality of measurements...
Each random variable is weighted in inverse proportion to its
  variance.

The inverse-variance weighted average seems very straightforward to calculate and as a bonus has the least variance among all weighted averages.
A: There are two way to think of the problem. 
One is to say that the sources observe a noisy version of the latent variable "it will rain / it will not rain".
For instance, we could say that each source draws its estimates from a $Beta(a+b,a)$ distribution if it will rain, and a $Beta(a,a+b)$ distribution if it will not. 
In this case, the $a$ parameter drops out and the three forecast, $x$, $y$, and $z$ would be combined as
$$p = \frac{1}{1+\left(\frac{1}{x}-1\right)^b\left(\frac{1}{y}-1\right)^b\left(\frac{1}{z}-1\right)^b}$$
$b$ is a parameter controlling how under ($b>1$) or over ($b<1$) confident the sources are. If we assume that the sources estimates are unbiased, then $b = 1$ and the estimate simplifies as 
$$\frac{p}{1-p} = \frac{x}{1-x} \frac{y}{1-y} \frac{z}{1-z}$$
Which is just saying: the odds of rain is the product of the odds given by each source. Note that it is not well defined if a source gives an estimate of exactly $1$ and another gives an estimate of exactly $0$, but under our model, this never happens, the sources are never that confident. Of course we could patch the model to allow for this to happen. 
This model works better if you're thinking of three people telling you whether or not it rained yesterday. In practice, we know that there is an irreducible random component in the weather, and so it might be better to assume that nature first picks a probability of rain, which is noisily observed by the sources, and then flips a biased coin to decide whether or not it is going to rain.
In that case, the combined estimate would look much more like an average between the different estimates.
A: Their numbers for rain likelihood is only half the story, as we'd have to temper their predictions with the probability that they are accurate when making guesses.
Because something like rain is mutually exclusive(it's either raining or isn't, in this setup), they cannot all simultaneously be correct with 75% probability as Karsten suggested (I think, hard to tell with the confusion I hear about what it means to find "combined probability").
Taking into consideration their individual abilities to predict the weather, we could take a stab (a la Thomas Bayes, as in a generally blind shot in the dark) at what the chance of rain is tomorrow.  
Station 1 is correct in their predictions 60% of the time, the second 30% of the time, and the last station a poor 10% of the time.     
E[rain]=PxX+PyY+Pz*Z is the form we're looking at here:
(.6)(0)+(.3)(1)+(.1)(.5) = E[rain] = 35% chance of rain with made up prediction accuracies.  
A: For combining reliability, my go-to formula is r1xr2xr3÷(r1xr2xr3+(1-r1)x(1-r2)x(1-r3). So for the 3 sources of reliability 75% all saying the same thing, i would have .75^3 ÷ (.75^3 + .25^3) => 96% reliability of the combined response
