My simplified 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?

With regard to the second journey, there is a tiny chance (infinitely small) that she will end up exactly at the start point; and there is an equally miniscule possibility that she’ll be 20km from home. But I want the average distance (i.e. the expected distance, given that the angle of journey 2 is random). I have worked out that after the second journey, she has a 35% 33% chance of being closer to home than she was after the first journey.


5 Responses to “My simplified taxi problem”

  1. Thomas Viner on October 18th, 2006 18:34

    How did you get 35%?

    I have a mental image that looks like the MasterCard logo, but with the circles slightly closer so that the perimeter reaches the centre of the other circle. Each circle has radius 10km and is where you might be after each journey.

    Before the 2nd journey we are 10km from the start position, to still be 10km away after the 2nd journey we must land on one of the two points where the circles intersect each other. Make the top intersect point the summit of a triangle that has the two centres of the circles as its other two corners. All sides of this triangle have length 10km so it is equilateral. All angles in it are 60 degrees. This makes the answer (60*2)/360 = 1/3.

    Is that right? Maybe you agree and happen to be feeling out of character today – very lax – and rounded 33.3…% to 35%? If so might you re-file this post under “personality number two”. Number one being the spelling nazi who would abhor such imprecision. 🙂

  2. Dan on October 18th, 2006 18:55


    You’re right. I didn’t use that logic, but did some basic trig., and got confused between my tan and cos.

    cos x = 5/10, so the angle is 60 degrees, so there is indeed a 1/3 probability.

    Still don’t have the answer to my question, though…


  3. Thomas Viner on October 19th, 2006 07:13

    ok here you go:

    sqrt(n * (10^2))

  4. Thomas Viner on October 21st, 2006 00:00

    Do you agree with that answer?

    Or do I need to show my working?

  5. Francis on November 2nd, 2006 11:36

    You guys are funny, asked a problem ending “what distance…” you respond with a percentage!!!

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