probability that point is within triangle formed by three lines of bearing This is a basic probability problem, and I can get the solution. But for a while I was using a wrong approach. My question for the forum is why my initial approach was wrong.
The problem: You're on a ship sailing off a coast in a 2-dimensional world, and to fix your position you take three lines of bearing. Each of the three readings is independent and has independent errors. Errors are symmetrical for each bearing, i.e. you're equally likely to get a bearing which is too high as a bearing which is too low. The three observed lines of bearing will form a triangle. What is the probability of your ship being inside the triangle.
The solution: The probability is 25%. To see this, consider each line of bearing as having equal probability of being too high or too low. You therefore have eight error possibilities: +++, ++-, +-+, etc. If you draw out all eight, you'll see that in two of them the ship is in the triangle formed. Hence, 25%.
My original, incorrect approach: I drew out three observed lines of bearing, and considered the possibility that the ship was in the triangle formed. However, the three lines divide the plane into only seven areas. In this approach, I considered the probability that the ship was "above" or "below" a given line of bearing as 50%. (As opposed to the correct approach, where the probability that an observed line was "above" or "below" the correct position of the ship was 50%.) But, like I said, the ship has only seven (not eight) possible combinations of "above" and "below" for the three lines of bearing -- one of those eight combinations is a geometrical impossibility. So this line of reasoning was not fruitful, and I couldn't get anywhere until I thought the correct way.
My question for the forum: Why was my original approach wrong? I'm thinking it has something to do with considering the position of the ship as the random variable rather than the errors of each bearing line as random variables. Or else, in drawing three bearing lines, I'm already excluding some probabilities (i.e. making a choice)?  
I know I'm being fast and loose with my terminology, but I'm hoping that if you think of three bearing lines on a plane, my meanings are clear.
 A: Indeed, the questions


*

*"before taking the bearings, what is the probability that the ship will end up inside the triangle?" and

*"given the bearings, what is the probability that the ship is inside the triangle?"


are not the same.  The answer to the first is 1/4, assuming the errors are symmetric; the answer to the second depends on the size and shape of the triangle, and probably on the error distribution too.
An extreme example of this occurs if the three lines of bearing just happen to intersect at a point, in which case the probability of the ship being exactly at that point is obviously zero.
A: From what I understand, I don't think your approach is wrong. It's just that you seem to be forgetting that by assuming a particular triangle you are implicitly restricting to one of the two possible cases. The (crappy) diagram below tries to show it. The zones "above" and "below" each bearing line is signaled by bold and thin (respectively) colored lines. The two possible triangles correspond to the cases: 
1. Below superior, above middle, below inferior (thin green, bold red, thin blue) -+-
2. Above superior, below middle, above inferior (bold green, thin red, bold blue) +-+
Both of this have 1/8 probabilty, and both must must be sum (leading to the 25%)
It's true that your approach can lead to a false reasoning. One could say: "Both cases are symetric, equally probable; hence the global probability that the ship falls in the triangle, must be equal to that same probability given that (conditioned to) we are in case 1". This is true. The one continues "Restricting to this case: one knows that the probability of falling in each side of the line is 1/2, hence the probability of falling in the triangle is 1/8". So? The fallacy is that when we condition on the event "we are in case 1" we are altering the probabilities, it's no longer true that all the events (falling on one or other side of the line, indepentently, it's 1/2.

A: This is not a "basic probability problem" but a statistical problem. Note that if your measurements were exact two bearings (in different directions) would suffice to determine the exact position of the ship. Now you have three measurements, albeit with errors.  You may assume that the error for a single measurement is normally distributed with a certain $\sigma$ and that the errors for different bearings are independent. But note that a given  error $\delta$ in a bearing produces a much larger insecurity about the position if the landmark is far away, etc. It all amounts to this: You tell a statistician the three bearings and the estimated distances to the corresponding landmarks. Then ideally he will compute for you the smallest domain such that with, say, 95% probability, your ship is in that domain. If your three bearings more or less intersect in one point then this domain will be centered near that point, but otherwise you can't say much without actually doing the calculations. 
