Taxi Solution
Dan has a problem on his site he calls the "taxi" problem:
"Let’s assume that taxi journeys are all of an equal distance (10km, say), and that they go straight in a random direction. At the end of one journey, the driver picks up another passenger at that very drop-off spot.
After her first journey, the taxi driver is ten kilometres from her start point. But what is her expected distance from her original start point after a second journey? And is there a generic formula that tells us how far she is from his origin after n journeys?"
So my answer uses the law of cosines which if Dan had learned his trig instead of wasting time with probability he’d know is
1) The traveller starts at the first corner (A) of the triangle and heads off 10km towards point C, let’s say that’s length b.
2) Then makes a turn of some degrees and heads off another 10k towards point B, that’s length a. IF they turn more than 180 degrees here then just get the modulus as we’re dealing with triangles remember. They are now "c" distance from home where
c = sqrt (a2 + b2 – 2ab cos C)
plugging in the values you get c = sqrt(200 – 200 cosC)
So if you know the angle they turned then you can figure out the distance they are from home aka point A. This completes the round.
Now work out the angle at B (the angle at B). You’ll need this for the next bit.
For the next trip:
They ended up at point B which becomes the point "C1" in a new triangle and step 1) from above is now complete. We know the length of this side (b1) is the results from the equation we just worked out.
2) They make a turn of some E degrees from the way they’re facing, this is a new angle. To get the angle at C1 in the new triangle you need to subtract the old angle "B" from 180 and then add E.
If the result is greater than 180, then subtract from 360 to get the new C1 angle. Make sense?
That’s the only tricky part and you can keep drawing new triangles where one side is the latest distance from home and the other is 10k with a single angle known. Above is a crappy diagram but I’m too tired and this is too silly a problem to warrant a better one.
If Mr. F. Shanahan had read the question properly, he would have realised that the answer should have been a number, in kilometres. The expected distance should be agnostic of the angles they take, which he’d know if he’d turned up to his probability lectures instead of getting heads down in his trig.
I have worked out that the expected distance from home after one journey is 10km (not rocket science), and after two journeys, it’s 40/pi, but I’m not sure why.
The facetiousness of my comment is in response to his comment in my own blogworld:
http://www.osirra.com/post/1/707
Further analysis can be found here, btw:
http://www.osirra.com/post/1/714