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# Expected value?

### #1

Posted 31 January 2013 - 06:29 AM

### #2

Posted 31 January 2013 - 10:05 AM

*Vidi vici veni.*

### #3

Posted 06 February 2013 - 01:30 AM

Spoiler for This should answer it

I keep coming up with the same answer using the exact same math, but when I simulate it on the computer, it takes greater that 2.25 steps for r=1.5m (takes about 3.5 steps) and greater than 9 steps for r=3m (takes about 11.4 steps).

### #4

Posted 12 February 2013 - 06:12 PM

I can't find a neat analytic solution for the problem of finding the expected number of steps until the ant leaves the circle, though it could be done by simulation or numerical approximation. This puzzle, however, asks a different question...Mathemagician, on 30 Jan 2013 - 21:37, said:

A long time ago, I posted a problem about an ant in a circle with a diameter of 3 meters. The ant would go 1 meter in a random direction, rest, and go 1 meter in another random direction and repeat this process until it got out of the circle. What is the expected number of 1-meter steps the ant takes before escaping. It was never fully solved and I just recently got back to it and thought of this question: assume f(x) is the expected distance from the center after x steps. Is the answer to the original question merely when does f(x) become greater than 3?

### #5

Posted 16 February 2013 - 10:59 PM

I believe this problem is about a drunk man, not an ant. He takes 1 meter step, falls, gets up makes another step in random direction, and so on.

Spoiler for This should answer it

I don't see how this trip averages out to zero and returns to the same point. Clearly, it is not true for the first step. And the probatillity of getting back to the point of origin on the second step is slight.

I am not sure wheter the title I used "Two Dimensional Random Walk" is the conventional name for this problem. Must check the literature, (which I never do.)

**Edited by Prime, 16 February 2013 - 11:03 PM.**

Past prime, actually.

### #6

Posted 17 February 2013 - 02:22 AM

Oops, forgot another square root. Here is the correction/addition to the previous post.

Normally, I try to avoid using integrals and calculus in general. I don't like that branch of mathematics and don't believe in it.

So it is entirely possible I messed up the above formulas. Besides, I have not actually figured out the Integral I used for this solution. But never mind all that. I checked few sources on the internet regarding two-dimensional random walk problem. I did not feel like submerging into the complex math presented there. Imo, those sources used insufficient justification and explanation for the formulas they use. What's interesting, in my brief examination, I did not see my **concept of the average increase in distance from any given point with each step **in the treatments of Random Walk in Two Dimensions. (As shown on the diagram in my previous post.)

Are there any professional mathematicians here to shed light on this?

Past prime, actually.

### #7

Posted 20 February 2013 - 01:23 AM

To reiterate the point **Bushindo** has made already (post#4.)

There is a difference in **average distance from origin after n steps **and **how many steps are required on average to travel that far**. The average distance includes variations where the drunkard has traveled further and then returned. On the other hand, to travel further than 1.5 steps, he must make __at least__ 2 steps. The probability of leaving the area after one step is zero, so the first step does not add anything to the average steps; whereas it does add its full size to the average distance.

One way to run a computer simulation is to make random steps until the drunkard wanders past the designated radius, tally up the steps it took for each experiment and calculate the average. That should show how many steps it takes on average to leave the area.

Another way is to have n steps in each experiment, tally up the distances traveled, calculate the average. That would show the mean distance from origin after n steps.

Past prime, actually.

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