Probability of having at least one pair by drawing 4 shoes from 12 pairs. There are $12$ pairs of shoes in a cupboard. $4$ are drawn at random. What is the probability that there is at least one pair?
My first attempt: If we chose a pair at first and then draw any two at random from the rest, then there will be at least one pair. 
We can choose one pair in ${12}\choose 4$ ways and chose any $2$ from the rest in ${22}\choose 2$ ways.
Therefore, the required probability= $\frac{{12\choose 4} \times {22\choose 2}}{{24\choose 4}}$ = $\frac{6}{23} =\frac{42}{161}$ 
But the given answer is $\frac{41}{161}$.
Another attempt: Each of the 4 shoes we choose, will come from one of the pairs. We can choose the four pairs in ${12\choose 4}$ ways and can select a shoe from each of the pairs in $2$ ways so that no pair is obtained. Therefore, required probability =$1-$ $\frac{{12\choose 4} \times 2^4}{{24\choose 4}}$ = $\frac{41}{161}$
What is wrong with the first attempt?
 A: Your first method may double-count the possibility of getting two pairs:  when you "chose any $2$ from the rest in ${22 \choose 2}$ ways", you may be choosing another pair and these two pairs are also counted when chosen in the other order.
In your second method of looking at $1-$ the probability of choosing from different pairs, a similar method is to say that each time you choose a shoe its pair becomes undesirable, making the result $$1-\frac{24}{24}\times \frac{22}{23}\times \frac{20}{22}\times \frac{18}{21} =\frac{41}{161}.$$
A: Just to build on Henry's answer, suppose we put the pairs of shoes in order, 1 through 12. By your method, you are over counting picking pair 1 and another pair 11 times. Similarly, you are over counting picking pair 2 with another pair (other than pair 1) 10 times. Continuing on, you are over counting by precisely 11th triangular number, $T_{11} = 66$. Correcting for this,
$$
\frac{12\cdot {22 \choose 2}-66}{24\choose 4} = \frac{41}{161}
$$
as desired.
A: This might be a bit late.
I was reading Feller's vol 1 chp 4 and got into this question from a very similar exercise.
http://www.amazon.com/Introduction-Probability-Theory-Applications-Edition/dp/0471257087

The ways you can do it is of course various, and easiest way to think is to use 
1 - Prob(0 pairs selected) 
and this is just 
$$1 - \frac{{12\choose 4} \times 2^4} {{24 \choose 4}}  $$

The other way you are considering, of course is correct. This is from inclusion exclusion principle.
https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
The way you calculation is done by:
$$P_1 = S_1 - S_2 + S_3 + - \ldots $$
where $S_1 = \sum_{i=0}^n p_i$, $S_2 = \sum_{i,j=0}^n p_{ij}, \ldots$ and $p_i = P(A_i)$, $p_{ij} = P(A_iA_j),\ldots$
So, in this problem you are looking at:$S_1 = \sum_{}p_i$ where $$p_i = \frac{22\choose 2}{24\choose4}$$ because you are considering 2 slots are already picked, and the rest can go anywhere in the 22 shoes.
And this has ${12\choose 1}$ number of ways happening for your first 2 slots.
Now consider $p_2$. This is calculated as
$$p_2 = \frac{20\choose 0}{24\choose4} $$
This is because, once you set 2 pairs of shoes, you are left with 0 to pick from the rest 20. 
And there are ${12\choose2}$ in summation for $p_2$

In summary
$$P_1 = S_1 - S_2 = \sum{}p_i - \sum{}p_{ij}=\frac{{12\choose1}\times{22\choose2}-{12\choose2}\times{20\choose0}}{24\choose4}$$
A: Calculate $1$ minus the probability of the complementary event:
The number of ways to choose $4$ out of $24$ shoes is:


*

*Choose the $1$st shoe out of $24$ shoes

*Choose the $2$nd shoe out of $23$ shoes

*Choose the $3$rd shoe out of $22$ shoes

*Choose the $4$th shoe out of $21$ shoes


The number of ways to choose $4$ out of $24$ shoes with no pairs is:


*

*Choose the $1$st shoe out of $24$ shoes

*Choose the $2$nd shoe out of $22$ shoes

*Choose the $3$rd shoe out of $20$ shoes

*Choose the $4$th shoe out of $18$ shoes


So the probability of choosing $4$ out of $24$ shoes with at least one pair is:
$$1-\frac{24\cdot22\cdot20\cdot18}{24\cdot23\cdot22\cdot21}$$

Please note that I've essentially taken into account the order of the shoes.
If I chose not to take it into account, then I would need to divide each result by $4!$.
But since this factor appears in both the numerator and the denominator, I can ignore it.
A: In your first counting method you have two counting stages. In the first one you have $12$ different possible results: $\{P_1, P_2,\ldots,P_{12}\}$, where $P_i$ is a particular pair. In the second stage you would have $\binom{22}{2}$ results for each result or pair of your first stage. That gives you a total of $12\binom{22}{2}$ possible results. Now, let's take a loot a two different results of the first stage, say $P_1$ and $P_9$. For $P_1$ in the first stage, is possible to have $P_9$ as a result in the second stage, and for $P_9$ in the first stage, it is possible to $P_1$ in the second stage, and since the final result $P_1P_9$ is the same as $P_9P_1$, you are effectively counting two times the same result.
A: To select a pair, the order is irrelevant.
1. Select one shoe
2. Select one shoe - chance this is a pair of first = 1/23 = .04
3. Select one shoe - chance this is a pair of first = 1/22 or chance this is pair of second = 1/21 = 1/22+1/21 = .09
4. Select one shoe - chance this is a pair of first = 1/21 or of second = 1/20, or of third = 1/19 = 1/21+1/20+1/19 = .15
After four selections, probability is .04+.09+.15=.28
