Probability of rolling doubles from 5 dice If I roll five dice, what is the probability that there is a matching pair among them?
This is the way I thought about the problem. Let $X$ be the random variable describing the number of occurrences of a number $a$, where $1 \leq a \leq 6$. Then the probability that $a$ occurs twice is the binomial
$P(X=2) = {5 \choose 2}(\frac{1}{6})^2(\frac{5}{6})^3 = 0.16$
Now there are 6 possibilities for $a$, so the probability of rolling a pair of any number between one and six is $0.16*6 = 0.96$.
I don't know where I went wrong  but the final answer I get seems wrong intuitively to me. I would be grateful for some correction or verification.
 A: You can't simply add the probabilities $P(X=2)$ as $a$ ranges over $1,\dots, 6$, since the events of rolling two of a number aren't mutually exclusive (it is possible to roll two ones and two fives). Furthermore, you would want to look at $P(X\ge 2)$, not $P(X=2)$, since having 3 or more of a number also counts as having a matching pair. 
I think the best way to do this problem is first find the probability there is no mathcing pair, then subtract from 1. For the dice to be all different, there are 6 choices for the first die, 5 for the second, 4 for the third$\dots$, and $6^5$ possible rolls total, so the probability is
$$
1-\frac{6\cdot5\cdot4\cdot3\cdot2}{6^5}
$$
A: I'm going to assume you're asking for the probability that there is at least one pair, not exactly one pair. These are of course different questions. The probability that there are no matching pairs given five dice rolled is
$$P=1\cdot\frac56\cdot\frac46\cdot\frac36\cdot\frac26=\frac5{54}.$$
The probability that there is at least one pair is $1-P=\frac{49}{54}$.
It isn't too much more difficult to work out the probability that there is exactly one pair. By symmetry and mutual exclusion, this is six times the probability that the number $a$ appears exactly twice and there are no other pairs. This is given by
$$Q=\frac{{5\choose{2}}\cdot{5\choose3}\cdot3!}{6^5}=\frac{25}{6\cdot54}$$
and so the probability is $\frac{25}{54}$.
A: The fastest and safest method for solving this would be Mike Earnest's solution or something like it.
If you want to add the probabilities of getting your pair rather than subtracting the probability of not getting a pair, you have to avoid double-counting.
As has already been mentioned, you can roll two $1$s and two $2$s in the same five rolls.
To compute the probability of exactly two $1$s and no other pairs,
you have $5$ choices for the first die that is not in the pair, but for each of those
there remain only $4$ choices for the next die, and at the end there are
only $3$ choices for the last die, so
$$P(\mbox{pair of twos}) = 
\binom52 \left(\frac16\right)^2 \cdot \frac56 \cdot \frac 46 \cdot \frac 36.$$
This event is mutually exclusive from having a pair of $2$s, so now it is OK to
multiply by $6$ to find the probability of having exactly one pair of exactly one number.
If you want at least one pair, then you must also add the probability of
having exactly one pair each of two different numbers
and the probability of having three or more of a single number
(which you could further subdivide into exactly three, exactly four, or five
of the same number).
This is much more tedious and error-prone than the method of subtracting the
probability of no pairs.
