You can solve it with a linear recurrence relation [0]: the halting probability from position n is ((sqrt(5)-1)/2)^(n+1), where n is twice the number of odds minus the number of evens. (In fact, this +2/-1 random walk is precisely how the machine implements its termination condition.) The expected value of n is 1/3 the number of iterations. At the end of the longest simulation that has been computed, n is greater than 2^37, so the halting probability is less than 10^(-10^10).