I do not know what the price will be but the log brownian model does NOT say that (I am just repeating myself here) ...
Look at a call closed formula price, N(d2) represents the probability of the call being in the money, if you put S=K (our example here) you are left with N(something negative) which means that the probability that the price ends up above current value is ... less than 50% !
And below current value is ... more than 50%.
I don't understand your notation,
here is mineBasically, in the log-Brownian model the difference between successive values of Z(i) = log(P(i)) are independent random variables with probability distributions that are symmetric about zero. Therefore after any number n of steps the probability distribution of Z(i+n) will be symmstric about the starting value Z(i). That means Z(i+n) wil be less than Z(i) with 50% probability. Since log is monotonic, it preserves cumulative probabilities, therefore P(i+n) will be less than P(i) with 50% probability. What is wrong with this argument?
(Strictly speaking, "Brownian" requires a normal distribution of increments with zero mean and fixed variance. In practice the variance varies slowly and the distributions have fatter tails than the norma; but by the law of large numbers they become near-normal for large n.)