We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice? We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice?
My attempt:
For the first draw of 2 dice together, the sample space has 36 possible outcomes and the ones with the same number in both dice are $(1,1), (2,2)...(6,6)$ so a total of 6. The probability for one throw is $P_1 = \frac {6}{36} = \frac {1}{6}$
Therefore for 2 such throws, the probability to get doubles is $(\frac {1}{6})^2$.
The probability to get doubles in 3 out of 6 throws is
$(\frac {1}{6})^3*(\frac {5}{6})^3$ ?? Is this correct?
or maybe $\frac {6^3}{36^6}$?
Thank you!
 A: As you observed, the probability of obtaining doubles on any one throw of a pair of dice is $1/6$ since $6$ of the $36$ possible outcomes result in doubles.  Since the probability of obtaining doubles is the same for each of the six trials, this is a binomial distribution problem.
The probability of obtaining exactly $k$ successes in $n$ trials is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ counts the number of ways in which exactly $k$ of the $n$ trials could result in a success.
Here, we will define a "success" as a double, so $p = 1/6$.  Since there are six trials, $n = 6$.  Hence, the probability to get exactly three doubles in six trials is
$$\Pr(X = 3) = \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$
In your attempt, you did not take into account the number of ways exactly three of the rolls could result in a success.
If you were instead interested in the probability of at least three doubles in six trials, you would add the probabilities of three or more successes.
$$\Pr(X \geq 3) = \sum_{k = 3}^{6} \binom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6 - k}$$
