I walk $\pi$ km in one direction followed by $\pi$ km in another direction. In expectation how far am I now from my starting location? Both directions are chosen uniformly at random between $0^{\circ}$ and $360^{\circ}$.

P.S. The answer may surprise you.

  • $\begingroup$ On a Euclidean plane? Or on a sphere? Or on a torus? Or… $\endgroup$ Commented Feb 4 at 7:55
  • $\begingroup$ On the plane ... $\endgroup$ Commented Feb 4 at 10:56

3 Answers 3


Start at the origin $(0,0)$. If the two directions are $t$ and $u$, the ending location is $\pi(\cos t+\cos u, \sin t+\sin u)$ and the distance from the origin is $$\pi\sqrt{(\cos t+\cos u)^2+(\sin t+\sin u)^2} = \pi \sqrt{2+2\cos(t-u)}.$$ So the expected distance from the origin is $$\int_0^{2\pi} \int_0^{2\pi} \frac{\pi \sqrt{2+2\cos(t-u)}}{(2\pi)^2} \mathrm{d}t\, \mathrm{d}u = 4.$$

  • $\begingroup$ Yep you got it! $\endgroup$ Commented Feb 2 at 15:49
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    $\begingroup$ Wow a solution without integer programming! ;) $\endgroup$ Commented Feb 3 at 8:53
  • $\begingroup$ @DmitryKamenetsky I was not expecting integer programming, and in fact a double integration is very intuitive. $\endgroup$
    – iBug
    Commented Feb 4 at 13:40
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    $\begingroup$ @Dragongeek Yes, $2\pi$ and $0$ are the extreme values of the distribution, and the expected value $4$ is between them. $\endgroup$
    – RobPratt
    Commented Feb 4 at 20:03
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    $\begingroup$ @Dragongeek Makes sense intuitively. After the first walk, you're a distance pi from the origin. From there, over half of the angles will send you even further away, so on average, you wind up a bit further away than pi. $\endgroup$ Commented Feb 5 at 17:31

If we consider the angle between the two walks to be $2\alpha$, in which $0\le\alpha<\pi/2$ (since the case for $\pi/2\le\alpha<\pi$ is equivalent), then the distance between the two points is $2\pi\sin\alpha$. Therefore, taking the integral of that in the range yields $\pi$. Finally, we divide by $\pi/2$, which is our range, to get the expected value of $2$.

  • $\begingroup$ I recall seeing this as a way to estimate $\pi$. Perhaps it is the needle-on-floor approximation? $\endgroup$
    – Sny
    Commented Feb 2 at 15:47
  • $\begingroup$ This is the nicest and easiest way to do the integration, but there must be a factor 2 missing somewhere. I think your integral should yield $2\pi$ rather than $\pi$. $\endgroup$ Commented Feb 2 at 15:52
  • $\begingroup$ I agree that 4 makes more sense $\endgroup$
    – Sny
    Commented Feb 2 at 16:00
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    $\begingroup$ $[-2\pi \cos \alpha]_0^{\pi/2}=2\pi$ $\endgroup$ Commented Feb 2 at 16:03

Let the smaller angle between the two walks be $\alpha$, where $0\le\alpha\le\pi.$ Using the law of cosines, the distance from where you started is $\sqrt{\pi^2+\pi^2-2\pi^2\cos\alpha}$. Taking the integral of this, we get $\frac{1}{\pi}\int^\pi_0 \sqrt{\pi^2+\pi^2-2\pi^2\cos\alpha}d\alpha=4.$ So the expected value of $\sqrt{\pi^2+\pi^2-2\pi^2\cos\alpha}$ is $4$

  • $\begingroup$ I got a hateful downvote, so I fixed my solution. $\endgroup$ Commented Feb 4 at 22:58

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