What is the probability of rolling three twenty-sided dice and having at least two of the same number appear?

Question

What is the probability of rolling three twenty-sided dice and having at least two of the same number appear?

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 1

You're very close. Once the first one is rolled, there is a $(1-1/20)$ chance it is not replicated by the second die. Once the second has also been rolled, assuming it did not replicate the first, there is a $(1-2/20)$ that the third die replicates neither of the original two. Thus the probability is $$1 - (1-1/20)(1-2/20).$$ This gives an exact result of $29/200 = 0.145$.

Answer 2

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 $$ 1-\frac{20\times19\times18}{20^3}=1-\frac{20}{20}\times\frac{19}{20}\times\frac{18}{20}. $$

Answer 3

We will consider the probability that two of the same number do not appear. After the first role, this is of course $1$. Then, after the second roll, there is a $\frac {19}{20}$ chance that the die role is not repeated. On the third role, there is a $\frac {18}{20}$ chance that the role is repeated. Multiplying together and subtracting from one we get: $$1-\left(\frac {19}{20}\right)\left(\frac {18}{20}\right)=0.145$$ You have to remember to consider all previous die roles, and not just the one immediately before.

Answer 4

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%)