Question: > 
> What is the probability of rolling three twenty-sided dice and having at least two of the same number appear?
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I thought that the probability would be:

$$1 - \bigg(1 - \frac{1}{20}\bigg) \bigg(1 - \frac{1}{20}\bigg) \bigg(1 - \frac{1}{20}\bigg) = 1 - \frac{19^3}{20^3}$$

That is one in twenty is the probability of rolling two of the same number in two dice, and then simply sub it into the formula for the probability of at least one of three unrelated events occurring. This leaves an answer of $14.2625\%$, but the experimental probability is consistently around $14.1449\%$. What is wrong with my calculation?

Edit:
The probability of one of any unrelated events occurring is $$Pt=1 -\bigg( 1-P1 \bigg)\bigg( 1-P2 \bigg)\bigg(1- P3 \bigg) ...\bigg(1- Pn \bigg)$$

So it would seem that the probability would be as visualized above, where one in twenty is the probability of any two die rolls being the same. Does this mean that the three probabilities are not unrelated as I previously believed?

Answer: 
Consider the complement namely all three dice are distinct. There are $20\times19\times18$ possible such outcomes out of $20^3$ possible outcomes. Hence the probability is 
$$
1-\frac{20\times19\times18}{20^3}=1-\frac{20}{20}\times\frac{19}{20}\times\frac{18}{20}.
$$
Assuming that the rolls are independent, the event X of rolling 3, 20-sided dice may be considered as a binomial distributed variable with n=3 and p=1/20. You are looking for P(X>1) which is the probability that you will get at least 2 outcomes the same. This is 0.00725. However, since there are 20 possible outcomes, you multiply this by 20 which gets you 0.145 (14.5%)