Bruteforce thinking works in this case, given that there's only ~12*2^12 total states and transition matrix is very sparse, 1/11 is quick to calculate.

But not all of these states are valid, visited set is just defined by 2 markers on the circle (and the start position), so now state count is much smaller.

Ladybug needs to be on 7 or 5 while having a nice (7,5) visited state to reach 6, movements inside (7, 5) don't really matter, so state count gets to 12*11/2=66. Quite small and enough to do by hand.

edit: been thinking a bit on finding a short proof, as 1/11 (or 1/(N-1) in general case) sounds like there could be a nice short proof, but it only made me realize how these constructive proofs are so clean and any attempts to formalize this gets me into graph theory vibes where I just feel like proof is making nonsymbolic leaps in reasoning that I just can't feel are true.

> After 5000 runs, they were all 8.4-9.7%

This sample size is really small. I ran 100 million simulations in Nim[0] (takes around a minute). And distribution converges toward 9.09% on all positions equally:

    Average turns: 65.99609065001634
    Final position distribution:
     4: 9.095%
    11: 9.093%
     7: 9.091%
     3: 9.091%
    10: 9.090%
     9: 9.090%
     1: 9.090%
     8: 9.090%
     2: 9.090%
     6: 9.090%
     5: 9.089%
     0: 0.000%

Shouldn't the code say:

    position = (position + direction + 1) % 12;