[bonanova]

Given that ...

1. Twenty-five Robots, all named after BrainDenizens, are placed at random on a rail 1 mile long.
   
2. The robot named Prime begins as the thirteenth robot from the North end of the rail at its midpoint.
   
3. Each robot faces North or South with equal probability, and travels at 1 mile/hour in the direction it faces.
   
4. When two robots meet or reach the end of the rail, they reverse direction.
   

On average, what is the net distance Prime has traveled after one hour?

[Guest]

Given that ...

1. Twenty-five Robots, all named after BrainDenizens, are placed at random on a rail 1 mile long.
   
2. The robot named Prime begins as the thirteenth robot from the North end of the rail at its midpoint.
   
3. Each robot faces North or South with equal probability, and travels at 1 mile/hour in the direction it faces.
   
4. When two robots meet or reach the end of the rail, they reverse direction.
   

On average, what is the net distance Prime has traveled after one hour?

A mile , since they do not pause or accerlerate

[Guest]

Given that ...

1. Twenty-five Robots, all named after BrainDenizens, are placed at random on a rail 1 mile long.
   
2. The robot named Prime begins as the thirteenth robot from the North end of the rail at its midpoint.
   
3. Each robot faces North or South with equal probability, and travels at 1 mile/hour in the direction it faces.
   
4. When two robots meet or reach the end of the rail, they reverse direction.
   

On average, what is the net distance Prime has traveled after one hour?

Similar to the last Robots on a rail - we know that he will travel for 1 hour - at 1mph that would be 1 mile - unless I am reading the OP incorrectly...?

[unreality]

Is this different?

[Guest]

The OP says "net distance" which I believe means "displacement" which would then mean change in distance from his original starting point.

be zero?

[Prime]

Similar to the last Robots on a rail - we know that he will travel for 1 hour - at 1mph that would be 1 mile - unless I am reading the OP incorrectly...?

Is this different?

It is different. And, at the first glance, looks much harder.

The previous problem with 25 robots asked when can we be certain, Prime falls off the rail. Meaning what's the longest time Prime can stay on.

This one requires to find the average distance Prime has travelled before falling off the rail. (We know he does fall off the rail for sure after one hour.)

For example, if all robots were crowded up at one edge of the rail and all headed toward that edge, Prime would travel very short distance.

[Prime]

The OP says "net distance" which I believe means "displacement" which would then mean change in distance from his original starting point.

be zero?

Net distance needs clarification. Is it the distance from Prime's initial position to the edge of the rail at which he exited? Or is it the total distance Prime walked back and forth before falling off?

[Guest]

edit nvm prime said it.

Edited November 6, 2008 by lemonymelon

[Guest]

Maybe it's 1 mile, but, think.

25 robots spread evenly on a line within 1 mile.

Each Robot is then 211ft apart ( 5280ft = 1 mile)

Now, each Robot I am assuming is 1 facing north and the one aside to it facing south .

Robot 1 --> < ----Robot 2 robot 3 ---> <--- Robot 4

Now they start walking to each other, that means Robot 1 and Robot 2 will meet halfway between their distances, so 211/2 = 105.5 ft

Now when they hit each other they have to go back to where they came from ( + 105.5) then start traveling to the other robot ( another 115,5+) Creating the whole 211.

Thus repeating over and over until the 1 hour time limit is expired.

Now the only thing left to do is translate the distance traveled in time. Family guy is on right now, but after that I'll give it a whirl.

when he says total net distance, I am assuming your adding up each individual length covered.

Edited November 6, 2008 by Ionno

[Prime]

Given that ...

1. Twenty-five Robots, all named after BrainDenizens, are placed at random on a rail 1 mile long.
   
2. The robot named Prime begins as the thirteenth robot from the North end of the rail at its midpoint.
   
3. Each robot faces North or South with equal probability, and travels at 1 mile/hour in the direction it faces.
   
4. When two robots meet or reach the end of the rail, they reverse direction.
   

On average, what is the net distance Prime has traveled after one hour?

As usually, I was not paying due attention when reading the OP. So unlike the first problem, here robots do not fall off the rail.

The second paragraph seems to contradict the first. If they all were placed at random, how come Prime is at the midpoint? I suspect, Prime was intentionally set in the middle, and 12 robots on either side of him were placed randomly thereafter.

In veiw of that, the question about net distance is whether it means displacement from the origin (midpoint) or total distance traveled.

[Guest]

[spoiler='Hi all!!

This is my first reply so let me know if I'm doing anything out of place.

Thanks and take care.

Fun2Live4 (TY)']The average is (Max + Min)/2.

What's the Max? Since it starts from the middle the Max is 1/2 a mile.

What is the Min? it would be equal distance between each robot so 1/24 of a mile. There are 25 robots but there are 24 gaps between the robots.

So (1/2 + 1/24)/2 = (13/24)/2 = (.5416)2 = 0.271 mile

[Prof. Templeton]

Prime Robot starts in the middle of the mile long rail, the best case scenario would be 1/2 a mile. Let's say he and the twelve robots North of him all are facing North, they would all walk North and off the end of the rail. Worst case scenario would involve Prime Robot traveling 1 mile. So the answer must lie in between.

[Guest]

Another thing to consider is, if we are talking about displacement, is .25 of a mile South considered "-.25 miles," or is everything positive?

Because if it could be positive or negative

zero might still be a plausible average.

However, if the final change in distance must be considered positive,

zero is obviously incorrect.

But I was always bad with probability anyway...

[Guest]

Prime Robot starts in the middle of the mile long rail, the best case scenario would be 1/2 a mile. Let's say he and the twelve robots North of him all are facing North, they would all walk North and off the end of the rail. Worst case scenario would involve Prime Robot traveling 1 mile. So the answer must lie in between.

When the robot reaches the end of the rail it turns around, it does not fall off.

[Prof. Templeton]

When the robot reaches the end of the rail it turns around, it does not fall off.

Oh, I was confusing this with the last Prime on a rail question.

[Prof. Templeton]

If Prime Robot travels at 1 mile/hour and doesn't fall off after an hour he's traveled 1 mile.

[Prof. Templeton]

If Prime Robot travels at 1 mile/hour and doesn't fall off after an hour he's traveled 1 mile.

Well that wouldn't be his net distance. The farthest from his starting point he could get would

Less then half a mile.

[Guest]

Prime will always be exactly where he started.

[Guest]

Prime will always be exactly where he started.

I say Prime always returns to his original position only because it is stated in the OP that he begins at the "midpoint." However, it is also stated that all 25 robots are placed at random along the rail... O.o

[bonanova]

Let's get rid of any caveats.

Prime is correct that the OP is weirdly stated. My bad.

Let these statements replace the corresponding ones in the OP

• Prime Robot is placed at the center of the rail.
  
• Twelve robots are then placed at random on one side, and twelve more are placed at random on the other side of Prime.
  

etc...

d3k3,

Why will Prime "always be exactly where he started."?

[Prime]

If the question was, what's the total distance traveled by Prime. Then Prof. T. definitely answered that -- 1 mile in one hour. (For Prime robot never rested.)

If the net distance was a vector (South -- negative, North -- positive), then as Rossbeemer stated several times, the average would be zero. (There is nothing in the OP to give preference to North or South direction.)

Bonanova, purposefully avoided clarifying NET distance -- he expects us to know. Since Bonanova acknowledged neither Prof. T.'s nor Rossbeemer's answers, the NET must be always positive by definition.

Given that, I expect the average net distance (displacement off center) to be different from zero. Because at the very least, the initial direction in which Prime moved must account for something.

[bonanova]

I grabbed an online dictionary and looked up net as an adjective...

Quick definitions (net)

▸ adjective: remaining after all deductions ("Net profit")

▸ adjective: conclusive in a process or progression ("The net result")

Not meant as a clue, just a clarification.

And ... I just love it when people show their work.

It's more fun to comment on that.

[Guest]

The robots should be an average of 0.04 miles away from each other.

And their arrangement should be (averagely) like this: <--.....<--.....-->.....-->.....<--.....-->.....<--.....--> (if one neighbour is faced to it, the other is not faced)

When I examined their behavior, I realized that all of them are at their start point after 0.08 miles of travel. And after 0.04 miles of travel, half of them are 0.04 miles far from start point and others are at start point. Since 1/0,12 gives a remainder of 0.04, half of them will be at start point after 1 miles, while half of them are 0.04 miles away. And averagely, (0+0.04)/2=0.02 miles away from start point. But in fact non of them will be 0.02 miles away, eather 0 or 0.04 miles?????

[Guest]

Late to the party again, but let me try to offer some insight...

My thoughts on 'net distance' would be the total distance of where Prime finishes from the midpoint, regardless of positive or negative. If the robot inevitably winds up back at the midpoint, it would be zero.

Since it is also impossible for the robot to be at either end of the rail (12 robots on each side blocking), it must be less than a half mile net distance.

I think multiple answers exist based on the direction of the robots. I haven't done any mathematical work on this, but assume that all but one robot on the South side (the end one) is headed South. Prime also is headed south. All except one (that directly north of Prime) on the North side are headed North. This would allow Prime to move some distance away from the centerpoint and then be "pinballed" back and forth between the others.

I'm sure there are cases where my logic fails, or exceptions can be made, but I just wanted to offer some thoughts on the matter

[Guest]

Based on some rough maths and an excel model of the whole thing I think the answer is in the region of 60 feet.

[Guest]

Let these statements replace the corresponding ones in the OP

• Prime Robot is placed at the center of the rail.
  
• Twelve robots are then placed at random on one side, and twelve more are placed at random on the other side of Prime.
  

etc...

In that case, it will always be exactly zero.

A distant observer will not be able to tell whether the robots turn about upon meeting each other, or simply continue on their original course. It will therefore appear as though the only thing that causes them to change direction is reaching an end of the rail. Thus, after one hour, it will appear as though the robot that started at a position x from the north end of the rail will end up at a position x from the south end, regardless of which direction it was initially traveling. In other words, the starting position will be reflected about x=0.5 (but with different robots), and because a robot began at x=0.5, there will be a robot at that same position after 1 hour. Because no positions were actually exchanged, that robot must be Prime.

Note that after two hours, all robots will have returned to their starting positions, regardless of the initial distributions of position and direction.