So, about this math problem...

[ichabod]

Let me try:

Set up the matrix of transition probabilities. This is a matrix such that the i, j entry is the probability that the dragon will move
from room i to room j, given that they are in room i. Note that the rows must all sum to 1 (he has to be somewhere). Now look at the columns as a system of equations. If the first column has a 0.5 in the second row and a 0.75 in the fifth row, then m1=0.5*m2+0.75*m5. Once you have all the equations (you should have one for each room), set m1 to 1, and calculate all the other m's. Now, since these are supposed to be probabilities, they should add up to 1. So sum all of the m's, and divide each m by the total. Now mi is the probability that the dragon is in room i, at some random point in time long after he started moving around.

 

[LazarusLong42]

Ay ay ay. For any complex network of more than a few rooms, you're no longer talking about a simple statistics problem; you're talking about chaos theory. However, for some simple networks we should be able to figure out actual formulas.

The easiest is a ring of N rooms, connected 0-1-2-3-4-.... with the last room connected back to room 0. As t approaches infinity: if N is odd, the probability is 1/N for all rooms; if N is even, the probability is 2/N for odd rooms at odd time increments, 0 for even rooms at odd time increments, and vice versa for even time increments.

(This is assuming the N rooms are labeled 0, 1,... , N-1 and that the dragon starts at t=0 in room 0.)

The simple case here--that the probability is simply 1/N for each room--would also be the case for any system in which every room was connected to every other--for instance, the room placed on the corners of a tetrahedron with tunnels along the edges, or, assuming tunnels can cross and not create "rooms", a pentagram.

The Sigil has already answered for your network. It's actually pretty easy to set up in Excel and run out for a hundred iterations or so.

And now, you have me setting up random ones because, yes, I'm a dork...

 

[FyreHowl]

ANd becuase I feel like it

AND Because I feel like being a spoiled brat.

In the normal table top game of 6 players assuming they comprise the stereotypical intelligence of any given D&D group. Also assuming that they do not know the layout of the dungeon, nor do they understand what the Dm is rolling for. Now, if this is their weekend relaxation- what are the odds that any given player will a. notice if you use the wrong percentage. and if a is true, what are the odds they b. CARE. and if they B care. what are the odds c they can argue with GM Fe-aht.


ONE IN A MILLION BILLION JILLION.

*ahem* thank you. This is from the reality-check system of guardinals on guard against pointless math.


Ie: unless yo'ure asking for curiosity's sake, at which, have at it, does it matter? make it up!

 

[LazarusLong42]

Jeez, FyreHowl... us dorks are having fun with math here!

After trying a few randomly-created networks, it appears Sigil's intuition is correct for the general case, and that the general case is not nearly as chaotic as I thought.

To wit: The probability P(i) of the dragon being in room i is equal to the number of connections attached to room i, C(i), divided by twice the total number of connections.

UNLESS there are no rings in the network with an odd number of rooms, in which case the probabilities are bistable, and you need to (1) determine which rooms are separated by an odd number of steps from the starting room, and which are separated by an even number of steps, then (2) apply the same formula, separately to each set of rooms.

 

[FyreHowl]

Heh heh, Then forgive me. Somehow after 5 hours of complex circuit analysis today, meaning inductors with stupid imaginary numbers...math...doesnt seem fun!

 

[LazarusLong42]

Heh heh, Then forgive me. Somehow after 5 hours of complex circuit analysis today, meaning inductors with stupid imaginary numbers...math...doesnt seem fun!

LOL!!

Ah yes, negative imaginaries for the inductors... positive imaginaries for the capacitors. That was a fun class. (It was also almost eight years ago... man I feel old.)

 

[Shemeska]

LOL!!

Ah yes, negative imaginaries for the inductors... positive imaginaries for the capacitors. That was a fun class. (It was also almost eight years ago... man I feel old.)

As a person who graduated with a chemistry degree, and Fyrhowl's DM, let me say the following... that A) math blows (which is why I'm going to grad school for Molecular Biology and not Chemistry), and B) GM Fiat. At this rate I'm approaching chosen status of the almighty Lord Rule 0.

Thank you. That is all.

Oh, and one more thing... *points* NERDS!!!! (and from me thats saying something...)

 

[LazarusLong42]

Shaneska: As a molecular biologist (MS, 2000) I can tell you that molecular biology blows much harder than math ever can. Math never has cells that say "These LB plates have 3 micrograms too much ampicillin in them, so we're not going to grow tonight."

Or proteins that will only interact on Tuesdays when Venus is rising. Or antibodies that work in the first seven bleeds you get from the company, get paid for, and then cease to work in any way, shape or form. Or restriction enzymes that, from one company, work really well, and from another company, work not at all. Or sequencing reactions that come back from the core with no data, or...

Crud. Now I'm just venting. Sorry

 

[tarchon]

Let me try:

Set up the matrix of transition probabilities. This is a matrix such that the i, j entry is the probability that the dragon will move
from room i to room j, given that they are in room i. Note that the rows must all sum to 1 (he has to be somewhere). Now look at the columns as a system of equations. If the first column has a 0.5 in the second row and a 0.75 in the fifth row, then m1=0.5*m2+0.75*m5. Once you have all the equations (you should have one for each room), set m1 to 1, and calculate all the other m's. Now, since these are supposed to be probabilities, they should add up to 1. So sum all of the m's, and divide each m by the total. Now mi is the probability that the dragon is in room i, at some random point in time long after he started moving around.

It's a standard-issue Markov chain. Generally, the easy way to find the steady state is to just multiply the initial state vector by the transition matrix, and keep multiplying the result by the transition matrix until it converges.

 

[tarchon]

AND Because I feel like being a spoiled brat.

In the normal table top game of 6 players assuming they comprise the stereotypical intelligence of any given D&D group. Also assuming that they do not know the layout of the dungeon, nor do they understand what the Dm is rolling for. Now, if this is their weekend relaxation- what are the odds that any given player will a. notice if you use the wrong percentage. and if a is true, what are the odds they b. CARE. and if they B care. what are the odds c they can argue with GM Fe-aht.

ONE IN A MILLION BILLION JILLION.

Well, our half orc barbarian has a masters in operations research, so I'm pretty sure he knows the hidden Markov chain problem.