infinite number of tosses of a fair coin. The probability of heads equals tails. So carried to infinity does the fact of equal probabilities for head and tails make the particular random walk from these tosses go to zero.  In other words will the walk tend to zero distance at infinity because the flips obey the probabilities of the fair coin.

I think I kind of understood what you are trying to ask but can't quite rephrase it. If you are talking about the markov chain and want to know that eventually there are equal number of heads and tails on an average, then the answer is yes. Let $X_i$ be a random variable such that $X_i = 1$ if $i^{th}$ toss is heads else $X_i = 0$. Then $E(X) = \sum E(X_i) = 0$. Let me know if this makes sense.
