Assuming all megaballs are equally likely to be chosen, 1/25 = p = probability that a draw prediction is correct due purely to chance. Let Y = number of correct predictions in n=322 independent draws. The probability distribution of Y = P(Y=y) is binomial with parameters (p, n = 322). Since n is large, P(y <= Y) = cumulative distribution function of Y = cdf of Y is very well approximated by the Central Limit Theorem as
P(y <= Y) = N[(y - n*p) / sqrt(n*p*(1 - p))],
where N(x) is the cumulative standard normal distribution = (1/2)*erfc(- x / sqrt(2))
The probability that a prediction method gets 25 or less megaball predictions correct out of 322 trials
= P(y <= 25) = N[(25 - 12.8) / 3.52)] = N(3.47) = 0.999736
==> P(y >= 26) = 0.0264% = probability that 26 or more predictions were correct due to pure chance.
That is, it is sigma = 3.47 unlikely to be due to chance, or 1 in 3790 due to chance.
Your simulation to calculate this probability was off by a factor of about 60.