# Probability of pair of gloves selection

In his wardrobe, Fred has a total of ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that any set of eight gloves has an equal chance of being chosen.

I am told to calculate the likelihood that these 8 gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same pair.

This is what I came up with, that is, the probability of success for each choice:

$$\frac{20}{20}×\frac{18}{19}×\frac{16}{18}×...×\frac{6}{13}=\frac{384}{4199}≈0.09145$$

At first, I was a little confused by the wording but I believe this seems about right.

Is there an alternative way to get the desired probability, e.g. with $$1-...$$?

Thanks in advance for any feedback.

There is a more general formula for this.
Here you are asked that no pair is selected, but this formula will take care of any number of pairs selected

With $$10$$ be the number of pairs, and $$k$$ the number of pairs selected from $$8$$ gloves, the formula is

$$\dfrac{\dbinom{10}{k}\dbinom{10-k}{8-2k}\cdot2^{8-2k}}{\dbinom{20}{8}}$$

For the particular case for $$k=0$$, it simplifies to

$$\dfrac{\dbinom{10}{0}\dbinom{10-0}{8-2\cdot0}\cdot2^{8-2\cdot0}}{\dbinom{20}{8}}$$

$$= \dfrac{\dbinom{10}{0}\dbinom{10}8 \cdot2^8}{\dbinom{20}{8}}$$

• Explaining briefly for OP: this can be justified by an extension of my argument where we first choose $k$ pairs from the original $10,$ then from the remaining $10 - k$ pairs we pick the remaining $8 - 2k$ gloves and choose whether each is the left or right glove. We can actually generalize this for all numbers of original pairs of gloves and gloves selected as well, but we need to be a bit careful regarding bounds. (for instance, notice that this formula is clearly only valid when $0 \leq k \leq 4$: if $k = 5$ then the probability is $0$ because we'd need to pick at least $10$ gloves) May 16 at 7:28
• I was about to point out that $\dbinom{20}{16}$ as a denominator would seem impossible as it yields $2.3777...$, but seems like it's already corrected. Anyways, this formula is exactly what I was looking and I feel like this should be the answer. I would have definitely marked it as an answer if you would have responded 10 seconds earlier @trueblueanil May 16 at 7:29
• @nimen55290: No matter, the important thing is that I have been of help ! May 16 at 8:05
• I really appreciate it @trueblueanil May 16 at 8:16

We can use a combinatoric argument if you like: there are $$20 \choose 8$$ ways we could possibly choose $$8$$ gloves from the $$20,$$ neglecting order.

To see how many of these will involve us choosing no pairs, we can think about first choosing which pairs we will take one glove from, and then from that choosing what glove to pick from each pair. There are $$10$$ pairs so we have $$10 \choose 8$$ ways to choose our pairs, and then for each set of pairs there are $$2^8$$ ways that we can choose to take the left or right glove from each.

So, if all possible sets of gloves are equally likely to be taken, the probability of taking no pairs of gloves should be $$\frac{{10 \choose 8} \cdot 2^8}{20 \choose 8} = \frac{\frac{10!}{2! 8!} \cdot 2^8}{\frac{20!}{12!8!}} = \frac{(10 \cdot 9 \cdot \ldots \cdot 3) \cdot 2^8}{20 \cdot 19 \cdot \ldots \cdot 13} = \frac{20 \cdot 18 \cdot \ldots \cdot 6}{20 \cdot 19 \cdot \ldots \cdot 13}$$

There are $$\binom{20}{8}$$ ways to select the $$8$$ gloves, all of which we assume are equally likely. Let's say a selection has "Property $$i$$" if it includes both gloves of pair $$i$$, for $$1 \le i \le 10$$, and let $$S_j$$ be the total probability (with over-counting) of the selections with $$j$$ of the properties, for $$1 \le j \le 4$$. So $$S_j = \frac{\binom{10}{j} \binom{20-2j}{8-2j}}{\binom{20}{8}}$$
By inclusion-exclusion, the probability of a selection with none of the properties, i.e. with no pair of matching gloves, is $$1-S_1+S_2-S_3+S_4 = 0.0914503$$