Hi,
you should have a look at this other nice problem and the solution I gave, since this is highly similar to your own problem. I didn't think it through, so I can't tell if this solves your problem completely but it should help greatly.
Hi, i need help with one interesting exercise.
So, i have 100 meters long rod and there are 100 walking robots. Every robot moves with speed 1 m/s. When two robots meet, they turn round and continue from each other. When a robot gets at the end of the rod, it will turn round too. The robots are located and directed on the rod randomly and independently.
In the middle of the rod, there is also robot R and it is directed to the right side. His behavior is same as the others. What is the probability, that the robot R will be after 100 s exactly in the middle of the rod.
Thank you for your advices.
Hi,
you should have a look at this other nice problem and the solution I gave, since this is highly similar to your own problem. I didn't think it through, so I can't tell if this solves your problem completely but it should help greatly.
Hi, I already know that idea, but thanks a lot.
I have a feeling, that the probability could be zero, because I can construct infinite uncountable many cases, when the R is not in the middle. (all other robots are located at meter 75 and directed against the robot R). Maybe the positive cases have also infinite uncountable count. I am not sure yet.
I think the answer is if there are robots, with even, and 0 else.
Why? Because I claim the robot in the middle will be back to its first place iff there are equally many robots at his right and left.
The argument I pointed at in my previous post shows that at the end one robot will be in the middle (that's already something!), but we don't know which one.
I claim more precisely the following: if at the beginning there are robots on the left of the middle and on the right (plus 1 at the middle), then at the end there will be robots on the left and on the right (plus one at the middle). And I even claim that the positions at the end are simply symmetric to the ones at the beginning with respect to the middle. This implies what I said above about equally many on left and right. And the proof of this fact is very similar to that in the post I quoted. You may want to try to draw sketches to find this proof.