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<blockquote data-quote="The Sigil" data-source="post: 1207345" data-attributes="member: 2013">Intuitively, I came up with the same answer a few minutes after my post above - by which time our internet connection at work had been severed by the local SBC crew. :bI expressed it as M/N where M is the number of &quot;entrances to the room&quot; and N is the total number of entrances to all rooms (in case you have a funny &quot;one-way&quot; movement method such as a teleporter), since a &quot;one-way&quot; passage does not need to be double-counted as a connection.&nbsp; (I figured since this is fantasy, a &quot;one-way&quot; move is a possibility).I didn't prove it, but it looked intuitively correct and the math works out (i.e., Probability across all rooms sums to one) so I was fine to assert it was so.&nbsp; As you mentioned, there are two cases to consider - one with a &quot;bistable&quot; solution (of steps only from point A to point B regardless of path is always odd or always even) and one &quot;stable&quot; solution (where the number of steps from point A to point B can be odd or even depending on which path you take).&nbsp; To the layman...If you have no case where you can move one step at a time from Point A to Point B to Point C to Point A, you have a &quot;bistable&quot; solution and will have one set of probabilities for all the &quot;even&quot; rooms and one set for all the &quot;odd&quot; rooms (i.e., rooms that are reached from the start in an even/odd number of steps).&nbsp; No room will be in both the &quot;Even&quot; and &quot;Odd&quot; sets... your solution is then for t=odd it is M(odd)/N(odd) and for t=even it is M(even)/N(even).&nbsp; If you DO have any point where you can move from A-B-C-A then it's &quot;general&quot; solution of M/N at any time t.--The Sigil</blockquote>

[QUOTE="The Sigil, post: 1207345, member: 2013"] Intuitively, I came up with the same answer a few minutes after my post above - by which time our internet connection at work had been severed by the local SBC crew. :b I expressed it as M/N where M is the number of "entrances to the room" and N is the total number of entrances to all rooms (in case you have a funny "one-way" movement method such as a teleporter), since a "one-way" passage does not need to be double-counted as a connection. (I figured since this is fantasy, a "one-way" move is a possibility). I didn't prove it, but it looked intuitively correct and the math works out (i.e., Probability across all rooms sums to one) so I was fine to assert it was so. As you mentioned, there are two cases to consider - one with a "bistable" solution (of steps only from point A to point B regardless of path is always odd or always even) and one "stable" solution (where the number of steps from point A to point B can be odd or even depending on which path you take). To the layman... If you have no case where you can move one step at a time from Point A to Point B to Point C to Point A, you have a "bistable" solution and will have one set of probabilities for all the "even" rooms and one set for all the "odd" rooms (i.e., rooms that are reached from the start in an even/odd number of steps). No room will be in both the "Even" and "Odd" sets... your solution is then for t=odd it is M(odd)/N(odd) and for t=even it is M(even)/N(even). If you DO have any point where you can move from A-B-C-A then it's "general" solution of M/N at any time t. --The Sigil [/QUOTE]

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