How do I add multiple probabilistic results of a single experiment? Let's say I've conducted an experiment that yields either a positive or negative result with a 50% probability of each. Three people attempt to determine the result of the experiment. They all only see part of the result, so can only give me probabilistic answers. The first tells me that it was a success with a 60% probability, the second with a 75% probability, and the third with only a 30% probability. Based on this information, how do I calculate the overall probability that the experiment had a positive result?
Edit:
I don't think it's an average because if two people told me they thought it was 90% correct, I think my overall probability would be greater than 90%, specifically, I think it would be 1-(1-0.9)*(1-0.9) = .99. The problem is that doesn't seem to work for all cases, for example when one says 50% and the other says 51%, I think the result should be 51%, but I get: 1-(1-0.5)*(1-0.51)=.755, which cannot be right.
Edit:
Here's another example which may be easier to explain. I ask a guy on the street if town is to my left or my right. He says he'll tell me, but me may lie depending on the roll of a die. He takes a 10 sided die out and says, "OK if it's any of some 7 numbers I'm thinking of, I'll tell you the truth." He rolls it and tells me it's to the left. Then he says he'll roll it again, but this time choose 6 numbers. He rolls it again and again tell me it's to the left. Finally, we do it a third time, but now he chooses 8 numbers and tells me it's to the right. What is the probability that town is to my left?
 A: Without any additional information I think your best bet is to do a calculation of the mean or median.
You provided this equation as being possible but being wrong and it is indeed very wrong which you can see by how it gets ridiculous so quickly. With only one more person guessing there is a 50% chance the probability quickly gets higher and higher.
$1-(1-0.5)*(1-0.51)*(1-0.5)=0.8775$
If you want something more rigorous based on these three people (like if these three people are real) then I would suggest that you get them to come to some sort of consensus and get more people. Better yet, have them bet on it in a prediction market.
A: Your die throwing witness is a precise problem, so let's answer that.  
It's easy to change the numbers, so let's just use the ones you provide.  We analyze the two cases:
Case I: the true direction is Left. (probability $.5$)
Then you get the first answer of left with probability $.7$
You get the second answer of left with probability $.6$
You get the third answer of right with probability $.2$
Hence case I arises with probability $.5\times .7\times .6\times .2=.042$
Case II: the true direction is Right. (probability $.5$)
Then you get the first answer of left with probability $.3$
You get the second answer of left with probability $.4$
You get the third answer of right with probability $.8$
Hence case II arises with probability $.5\times .3\times .4\times .8=.048$
Given that you got the given answers, we know we are in one of these two cases.  Then the probability that you are in fact in case I is $$\frac {.042}{.042+.048}=.4\overline6$$
