The simulation actually follows a binomial distribution, meaning the standard deviation can be calculated as sqrt(npq), where n is the number of trials, p is the probability of success (pi/4), and q is 1-p. For a simulation of 10.000 iterations, the standard deviation is approximately 41. You say you observed 3.135 after 10.000 iterations, meaning you observed about 7837 hits. The expectation was to observe 7854 hits, meaning there was only a difference of 17 hits - well within 1 standard deviation.

edit: naturally you can also use this information to calculate the number of trials needed get a desired probability that the estimation falls within certain bounds from pi.