If we had a 4 sided dice (numbers 1,2,3,4 on the faces) and rolled it 5 times and recorded the results. What would be the probability that we rolled the same number exactly 3 times? What about 3 or more times?
For finding out the probability of getting 3 matching numbers in our sequence of rolls, I came up with the following.
$${5 \choose 3}×(4/4×1/4×1/4×3/4×3/4) = .3516$$
It does not intuitively seem correct at first, but I am trying to model that the first roll can be anything (4/4) and the next roll must match whatever it got, thus (1/4), and same thing for the next roll (1/4). The last two must not match the first, so I am finding (3/4) for both of those rolls. I am multiplying the whole thing by the different ways that those matching numbers could have come up in the sequence (5 choose 3). The result seems kind of high to me, but I understand that probability is often not intuitive for most people.
Given that my logic above is correct, could this be expanded in some way to account for the second part of my question about 3 or more matches?
Best Answer
Your logic is correct. The probability of rolling exactly three $4$'s in five rolls of the die is $$\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2$$ where $(1/4)^3$ is the probability of rolling a $4$ three times, $(3/4)^2$ is the probability of rolling a number other than $4$ twice, and $\binom{5}{3}$ represents the number of ways that three of the five rolls could produce a $4$. The same calculation applies to rolling a $1$, $2$, or $3$ exactly three times in five rolls, so the probability that the same number is rolled exactly three times is $$4\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2$$ which agrees with your result.
The probability of rolling the same number at least three times in five rolls is found by adding the probabilities of rolling the same number exactly three times, exactly four times, and exactly five times, which is $$4\left[\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2 + \binom{5}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^5\right]$$