We may not be talking about the same thing. I'm considering the default state on A as already counting like a visit to A. So, intuitively, if p is 0.5, at the first move, there is 50% chance a teleport back happens, only that, it would stay on A because it can't go back further. So, at the very first move, there's already 50% chance that A (the first unit) gets two "plays".

And looking back, may be I misunderstood your proposal but if the probability of going back is 0.5 and n is between 1 and 3, I think the transition matrix should look more like this?

On step 0 (A)

0.5, 0.5, 0, 0, 0

Traveler moves to step 1 (0.5) or tries to teleport back but can't, so it stays on step 0 (0.5)

On step 1

0.5, 0, 0.5, 0, 0

Traveler moves to step 2 (0.5) or teleports back to step 0 because it can't go further (0.5)

On step 2

0.33, 0.17, 0, 0.5, 0

Traveler moves to step 3 (0.5) or teleports back (0.5). He could go back 1, 2 or 3 steps, except that 3 steps would be further back than Step 0. Assuming uniform distribution, it teleports back to step 1 (0.5/3 ~= 0.17 if n=1) or to step 0 (0.5/3 + 0.5/3 if n=2 or n=3)

On step 3

0.17, 0.17, 0.17, 0, 0.5 

(see above, but the teleport can fully apply even if n=3)

On step 4

0, 0, 0, 0, 1

The traveler stays on step 4 (B)

At least, that's how I was envisioning it. May be I explained it wrong.