Probability at Plato's Plateau

In 1798, we're told, Napolean Bonaparte took an army to Egypt. What we are not told, however, is the incident at 'Plato's Plateau', apparently constructed by Alexandrians to honour Plato's solids.

Nineteen of the enterprising invaders happened upon a large tetrahedral pyramid in the desert, with three of its four points on the ground. Sitting at the top was an old man. Three soldiers, wanting to capture the edifice, called upon the old man to surrender. They then positioned themselves so that one soldier was at each of the three lower vertices, and the old man was at the top. If each of the three soldiers, and the old man, picked a random adjacent vertex of the pyramid, and traveled, by climbing or walking, to it, then what is the probability that no on met someone else on the way? What is the probability that the old man was caught?

There were nineteen soldiers so that this same scene could be performed at the hexahedral, octahedral, dodecahedral and icosahedral edifices on Plato's Plateau, with one old man at each. What are the corresponding answers at each of these?

Unfortunately, Napoleon's soldiers were unable to capture any old men (chances?) and in frustration destryoed Plato's Plateau with cannon fire.

Beats me too long😴

I don't know, but I dare you to reply without including " -1 " in the answer.😵

I forgot to say that the old man is caught if and only if he is met by a soldier in transit from one vertex to another.

how long is the old guys beard?

Originally posted by fearlessleader
how long is the old guys beard?

Some of those edifices sound quite hard to walk around. Do the man and soldiers have climbing gear?

Originally posted by iamatiger
Some of those edifices sound quite hard to walk around. Do the man and soldiers have climbing gear?

Anyone 😳?

i spose the old man chooses an edge and descends. the soldier at the bottom of that edge has 3 edges to choose from, so 1 in three that he is caught in transit.
that there is no soldier arriving at the same point is different, but you seem to have excluded that.

no bangs at all:
label the corners 1234, the dude at 1 can choose any corner of 2, 3 or 4 to go to: 3 choices. that dude has 2 choices left, and then of course only 1. so 6 different ways with no bangs.


the total number of ways is 3x3x3x3 = 81 ways

so 6/81 = 2/27 chance of no bangs.


at first i thought : if there are K korners of the shape then there will be (K+1-2)!/(K+1-2)^k ways with no bangbangs and none of those minus one thingys either

but those bigger shapes bring in problems that dudes can't walk to all other korners directly. - someone elses turn now. ...