Probability of getting the same number three times.

Question

If I have a set of numbers $\{1 \ldots n\}$ where $n \ge 1$ and I pick $3$ numbers from the set independently and uniformly. Whats the probability I'll get all $2$'s, the probability I get all the same numbers and the probability that my first two numbers are the same.

I've I compared this to rolling a fair die three times. Wouldn't probability of getting three $2$'s and getting all the same numbers be $\left(\dfrac 16 \right)^3$ since its uniform. And the probability of getting two of the same numbers be $\left(\dfrac 16 \right)^2 \left(1-\dfrac{1}{6} \right)$??

Any help would be appreciated.

Answer 1

1. You're right, the probability of all 2s is $$P(222) = P(2)P(2)P(2) = \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} = \left(\frac{1}{6}\right)^3= \frac{1}{216},$$ where use the product by independence.

2. When calculating probability that they are all the same is, you have to account for all the choices you have for a match, which is $\binom{6}{1}$. So, $$P(\text{All match}) = \binom{6}{1}\left(\frac{1}{6}\right)^3 =\frac{1}{36}.$$

3. For the first two to match you have again $\binom{6}{1} = 6$ choices for this pair. After, you will only have $\binom{5}{1} = 5$ choices for the remaining single. Lastly, you need to account for the order of the first two since for example $221$ is the same as if I switched the first two $221$, which is $\binom{3}{2,1} = 3$. So the probability of interest is $$\binom{3}{2,1}\binom{6}{1}\left(\frac{1}{6}\right)^2\binom{5}{1}\left(\frac{1}{6}\right) = 90\left(\frac{1}{6}\right)^3 = \frac{5}{12}.$$

Answer 2

You are partially correct.

The probability of three 2's is $\left(\frac16\right)^3=\frac{1}{216}$.

The probability of all three the same is $1\times\left(\frac16\right)^2=\frac{1}{36}$ as the first number can be any value.

The probability of two the same and the third different is: $1\times\frac{1}{6}\times\frac{5}{6}=\frac{5}{12}$. Same argument applies to the first dice as before. $$$$

Generalizing this to $n$ sided dice you get respective answers of:

$$\frac{1}{n^3}$$

$$\frac{1}{n^2}$$

$$\frac{n-1}{n^2}$$

Note: For your third problem your wording says first two the same. If you meant just two the same and one different then you need to multiple this answer by 3 to account for different orderings, i.e. AAB, ABA, BAA.