How does variance change as sample size increases Situation:
n possibilities each have their own probability of happening, and their own payout when they do. 
So expected payout $E_n$ is $\sum\limits_{i=1}^n \text{probability}_i*\text{payout}_i$
And the expected variance is $\sum\limits_{i=1}^n ( \text{payout}_i- \text{expected  payout})^2* \text{probability}_i$
Question:
If this activity is repeated 5 times, the expected payout is $5* \text{expected payout}$
But the variance is not 5 times the expected variance; it should get proportionally smaller as this activity goes on. But how do I calculate the variance through iterations?
 A: The variance of a sum is the sum of the variances plus twice the sum of the covariances.
e.g.   $\,\, \operatorname{Var}(X+Y+Z)$
$$= \operatorname{Var}(X)+\operatorname{Var}(Y)+\operatorname{Var}(Z) \\+ 2 [\operatorname{Cov}(X,Y)+ \operatorname{Cov}(X,Z) + \operatorname{Cov}(Y,Z)]$$
If your events are independent, those covariances are zero, and that reduces to "the variance of a sum is the sum of the variances":
$$\operatorname{Var}(X+Y+Z)= \operatorname{Var}(X)+\operatorname{Var}(Y)+\operatorname{Var}(Z)$$
So if you repeat it 5 times, yes, the variance of the total is indeed 5 times larger.
The standard deviation therefore increases as the square root of the number of repetitions, which may be what you're anticipating.
A: If you are interested in the sum of payouts then, as @Glen_b suggests, variance gets larger with more repetitions of the experiment: $kVar_n$ after $k$ repetitions.
If you are interested in the average payout, then variance is $\frac{Var_n}{k}$ after $k$ repetitions. Since you say:

[variance] should get proportionally smaller as this activity goes on.

I guess you are interested in the average. This also assumes independence.
