Now, what about time 2? 3 exits leaves probability of 1/3 in rooms 6, 4, and 3, and a probability of 0 everywhere else. I think.
Assuming I have that right, the problem for me is turning this from a recursive problem (ie, if I know the time, and the room, I can compute the probability at time + 1 of him being in a given room) can some limit be taken, as time goes to infinity, which will give a probability per room at any time?
Well, this should be fairly straightforward... Let P
be the probability of him being in room n at a given time; the sum of the probability figures at any given time will be one.
At t=0, since you stated such, P(6) = 1.
The way to tackle this is as follows:
1.) What is P
for t=n each room given that he is in room n at t - n-1?
For example, if he was in room 1 before he last moved,
P(2) = 1
If he was in room 2 before he last moved,
P(1) = 1/3
P(3) = 1/3
P(4) = 1/3
If he was in room 3 before he last moved,
P(2) = 1/2
P(5) = 1/2
If he was in room 4 before he last moved,
P(2) = 1/2
P(5) = 1/2
If he was in room 5 before he last moved,
P(3) = 1/3
P(4) = 1/3
P(6) = 1/3
And of course if he was in room 6 before he last moved,
P(5) = 1
By successively multiplying probabilities, we can see what room he's in at a given moment in time... multiply his probability of being at a certain location based on his location last turn (by the chart above) by the probability of him being in the location the turn before... for instance, the probability of his moving from 6 to 5 on turn 1 is one. The probability of him moving from 5 to 4 on turn 2 is 1/3, so his probability of BEING in room 4 on turn two is 1 (probability of him making the move from 6 to 5) *1/3 (probability of him making the move from 5 to 4). Repeat ad nauseum.
t=0 -> P(6) = 1 by definition
t=1 -> 1*possibilities from room 6 above -> 1 * P(5) = 1
t=2 -> 1*possibilities from room 5 above ->
P(6) = 1/3*1
P(4) = 1/3*1
P(3) = 1/3*1
t=3 ->
1/3 chance of being in P(6) time 1/1 chance of moving from 6 to 5 which leads to:
P(5) = 1/3 * 1
1/3 chance of being in P(4) which leads to:
P(5) = 1/3 * 1/2
P(2) = 1/3 * 1/2
1/3 chance of being in P(3) which leads to:
P(2) = 1/3 * 1/2
P(5) = 1/3 * 1/2
So the sum for t=3 is:
P(5) = 1/3*1 + 1/3*1/2 + 1/3*1/2 = 1/3 + 1/6 + 1/6 = 2/3
P(2) = 1/3*1/2 + 1/3*1/2 = 1/3
Now we can consider t=4, again making a tree from P(5) and P(2):
P(3) = 1/3 * 2/3
P(4) = 1/3 * 2/3
P(6) = 1/3 * 2/3
P(1) = 1/3 * 1/3
P(3) = 1/3 * 1/3
P(4) = 1/3 * 1/3
P(3) = 1/3*2/3 + 1/3*1/3 = 2/9 + 1/9 = 1/3
P(4) = 1/3*2/3 + 1/3*1/3 = 2/9 + 1/9 = 1/3
P(1) = 1/3*1/3 = 1/9
P(6) = 1/3*2/3 = 2/9
Now for t=5
P(2) = 1/2 * 1/3
P(5) = 1/2 * 1/3
P(2) = 1/2 * 1/3
P(5) = 1/2 * 1/3
P(2) = 1 * 1/9
P(5) = 1 * 2/9
Or, quickly, P(2) = 1/6+1/6+1/9 = 1/3+1/9 = 4/9 and P(5) = 5/9
It is interesting to note that on odd-numbered turns (t=1, 3, 5), etc., the dragon will always be found in either room 2 or room 5, while on even-numbered turns, the dragon will always be found in room 1, 3, 5, or 6.
Intuition tells me that the probability of the dragon being in a particular room as t-> infinity for this particular layout is directly proportional to the number of "entrances" into a room; i.e., that as t -> infinity we have:
Odd-numbered turns: 50% chance of being in room 2, 50% chance of being in room 5 (3 entrances to each).
Even-numbered turns: Room 1 has one entrance, Room 6 has one entrance, room 3 has 2 entrances, as does room 4. Hence we have 6 total "entrances" which tells me that on even numbered turns:
Room 1: 1/6 chance of being in room 1. 1/6 chance of being in room 6. 1/3 chance of being in room 3. 1/3 chance of being in room 4.
Can't empirically prove it off the cuff, but if you work out the next few turns, you'll see the numbers starting to get closer and closer to those distributions.
More in a bit.
--The Sigil
EDIT: On even-numbered turns... if the odds for odd-numbered turns are in fact 50-50 for large quanitites of t, then it seems patently obvious that the odds are 1-2-2-1 on even numbered turns...
P(1) = 1/3*1/2
P(3) = 1/3*1/2
P(4) = 1/3*1/2
P(3) = 1/3*1/2
P(4) = 1/3*1/2
P(6) = 1/3*1/2
P(1) = 1/6; P(3) = 1/6+1/6; P(4) = 1/6+1/6; P(6)=1/6
How do we know that we'll get to the 50/50 split on large quantities of t? Without actually doing the math, suffice to say that we have converging series. I have done enough exercises like this to trust my own judgement, though hopefully it is self-evident that over time, if he can be in only one of two locations on an even-numbered turn, and the physical setup is a diagonal reflection, the odds will go to 50-50. Just for sake of argument, though, let's run another turn or two...
t=5
P(2) = 4/9
P(5) = 5/9
So at t=6
P(1) = 1/3*4/9
P(3) = 1/3*4/9
P(4) = 1/3*4/9
P(3) = 1/3*5/9
P(4) = 1/3*5/9
P(6) = 1/3*5/9
or
P(1) = 4/27
P(6) = 5/27
P(3) = P(4) = 1/3*4/9 + 1/3*5/9 = 1/3
so at t=7
P(2) = 1 * 4/27
P(2) = 1/2 * 1/3
P(5) = 1/2 * 1/3
P(2) = 1/2 * 1/3
P(5) = 1/2 * 1/3
P(5) = 1 * 5/27
And we have P(2) = 4/27 + 1/6 + 1/6 = 4/27 + 1/3 = 13/27
P(5) = 5/27 + 1/6 + 1/6 = 5/27 + 1/3 = 14/27
These are converging towards 1/2.
If anyone feels like tackleing it, I'd be interested in the answer. (Yay! Answers!)
since there is no facing in this edition.
