# Probability and gloves

We have five different pairs of gloves. Three people choose at random one left and one right glove. What is probability, that each person doesn't have a pair. My attempt: $$|\Omega| = {{10}\choose{6}}$$ First person can choose the left glove in five ways, right glove-four ways. Second person : left glove-4 ways and right glove-4 ways. Third person : left glove:3 ways and right glove - 3 ways. So, $$|A|=5\cdot 4+ 4\cdot 4+ 3\cdot 3$$ What do You think about it?

• Go for finding probability of complement by means of inclusion/exclusion and symmetry. Commented Oct 9, 2018 at 19:55

Since the gloves are distinct, there are $$5 \cdot 4 \cdot 3 = 60$$ ways for the three people to choose three of the five left gloves and the same number of ways for them to choose three of the five right gloves. Hence, there are $$60^2 = 3600$$ selections in the sample space.

Suppose they each choose a left glove first. As stated above, this can be done in $$60$$ ways. From the $$60$$ ways, they could each choose a right glove, we must subtract those selections in which one or more people select a matching glove.

There are $$\binom{3}{1}$$ ways to select a person who chooses a matching glove and $$4 \cdot 3$$ ways for the other two people to select two of the remaining four gloves.

However, subtracting $$\binom{3}{1} \cdot 4 \cdot 3$$ from the $$60$$ ways in which two people could select right gloves subtracts too much since we subtract each case in which two of the people select matching gloves twice, once for each way we could designate one of them as the person who selected a matching pair of gloves. Hence, we must add those cases back.

There are $$\binom{3}{2}$$ ways to select two people who select a matching glove and $$3$$ ways for the other person to select one of the three remaining right gloves.

However, if we add $$\binom{3}{2} \cdot 3$$ to $$60 - \binom{3}{1} \cdot 4 \cdot 3$$, we will have added too much since if all three people select matching gloves, we have not counted them at all. This is because we counted them three times, once for each way we could have designated one of the three as the person who selected a matching glove, and subtracted them three times, once for each of the $$\binom{3}{2}$$ ways we could have designated two of the three people as the pair of people who receive matching gloves. Thus, we have not excluded them at all, so we must subtract the number of ways all three people could select matching gloves from the total.

There is only one way for all three people to select matching gloves.

Hence, by the Inclusion-Exclusion Principle, the number of ways none of the three people can select a right glove that matches his or her left glove is $$60 - \binom{3}{1} \cdot 4 \cdot 3 + \binom{3}{2} \cdot 3 - \binom{3}{3} = 60 - 36 + 9 - 1 = 32$$

Hence, the probability that no person selects a matching glove is $$\frac{60 \cdot 32}{60 \cdot 60} = \frac{32}{60} = \frac{8}{15}$$ Note that the factor of $$60$$ cancels out since what matters is whether each person's left glove matches his or her right glove.

• I really admire the patient teaching in this answer. Commented Oct 10, 2018 at 8:50

Let $$L$$ be a set of left and $$R$$ a set of right shoes. Then the sample space is $$\Omega = \{(a,a',b,b',c,c');a\ne b\ne c\ne a\in L,\;a'\ne b'\ne c'\ne a'\in R\}$$ so $$|\Omega|=5^24^23^2$$. Now our event is $$\overline{A\cup B\cup C}$$, where $$A$$ first man has a pair,... Since $$|A| = |B| = |C|= 5\cdot 4^23^2,$$ $$|A\cap B| = |A\cap C| = |B\cap C| = 5\cdot 4\cdot 3^2$$ and finally $$|A\cap B\cap C| = 5\cdot 4\cdot 3$$

So $$|A\cup B\cup C| = 3(5\cdot 4^23^2)-3( 5\cdot 4\cdot 3^2)+5\cdot 4\cdot 3$$ so $$P' ={3( 4\cdot3)-3(3)+1 \over 5\cdot 4\cdot 3} = {7\over 15}$$ so $$P = {8\over 15}$$

Let $$G_{ir}$$ and $$G_{il}$$ be the right and left gloves of pair $$i$$. Then we can seperate right and left pairs as two sets such as $$L = \{G_{1l}, G_{2l}, G_{3l}, G_{4l}, G_{5l}\}$$ and $$R = \{G_{1r}, G_{2r}, G_{3r}, G_{4r}, G_{5r}\}$$. Also, let $$X$$ be the event that no person chooses a pair.

Now, those three people should choose one element from set $$L$$ and one element from set $$R$$. Now, let them choose three elements from $$L$$ first with $$\binom{5}{3}$$, say these are $$G_{il}, G_{jl}, G_{kl}$$, and we can permute them with $$3! = 6$$ ways. Then there are three cases for choosing the right gloves:

Case 1: We might have all three $$G_{ir}, G_{jr}, G_{kr}$$ but since we don't want any of them to have a pair, number of derangements of three elements gives us the size of this event, that is, $$2$$ (namely they are $$\begin{pmatrix} G_{il} & G_{jl} & G_{kl} \\ G_{jr} & G_{kr} & G_{ir} \\ \end{pmatrix}$$ and $$\begin{pmatrix} G_{il} & G_{jl} & G_{kl} \\ G_{kr} & G_{ir} & G_{jr} \\ \end{pmatrix}$$ where $$a^{th}$$ column reperesents what person $$a$$ has chosen, $$a = 1,2,3$$).

Case 2: We might have any two of $$G_{ir}, G_{jr}, G_{kr}$$ chosen with $$\binom{3}{2}$$ and by symmetry, size of events of no person having a pair is equal for all choices. So we can calculate it for a specific case $$G_{ir}, G_{jr}$$ chosen and the other right glove is say $$G_{xr}$$ with $$\binom{2}{1}$$. Then we can have $$\begin{pmatrix} G_{il} & G_{jl} & G_{kl} \\ G_{jr} & G_{ir} & G_{xr} \\ \end{pmatrix}$$, $$\begin{pmatrix} G_{il} & G_{jl} & G_{kl} \\ G_{jr} & G_{xr} & G_{ir} \\ \end{pmatrix}$$, $$\begin{pmatrix} G_{il} & G_{jl} & G_{kl} \\ G_{xr} & G_{ir} & G_{jr} \\ \end{pmatrix}$$ so we have $$3\cdot 2 \cdot \binom{3}{2} = 18$$ possibilities.

Case 3: We might have only one of $$G_{ir}, G_{jr}, G_{kr}$$ is chosen with $$\binom{3}{1}$$ and the rest of the two elements with $$\binom{2}{2}$$. Then again, by symmetry, sizes of these three events are the same. So if we can calculate it for $$G_{ir}$$, $$G_{ir}$$ can have $$2$$ places and for other two elements, they can replace their places with $$2! = 2$$ ways. So there are $$2\cdot 2\cdot \binom{3}{1} = 12$$ possibilities for this case.

And we are done with the cases since we cannot have three right elements from $$R$$ such that all three indexes are different from $$i,j$$ and $$k$$. Then in total, we have

$$P(X) = \frac{3!\cdot\binom{5}{3}\cdot(2+18+12)}{5^24^23^2} = \frac{32}{60} = \frac{8}{15}$$

Yet another way of looking at it:

It doesn't really matter which $$3$$ left hand gloves are chosen, the math can be limited to the selection of the $$3$$ right hand gloves.

When the $$3$$ right hand gloves chosen are the same as the left hand gloves there are only $$2$$ out of $$6$$ permutations that won't match.

When $$2$$ of the $$3$$ match, there are $$6$$ ways to do that and for every one of those there are $$3$$ out of $$6$$ permutations which don't match making a total of $$18$$

When only $$1$$ of the $$3$$ matches, there are $$3$$ ways to do that and for each of those, there are $$4$$ out of $$6$$ permutations that don't match making a total of $$12$$.

The grand total of $$3$$ pairs of non-matching gloves is $$2+18+12 = 32$$

There are $$\binom{5}{3}\cdot 3! = 60$$ permutations of $$3$$ left and $$3$$ right gloves so the probability is:

$$P(0\ \text{match}) = \frac{32}{60} = \frac{8}{15}$$

• Nice approach (+1) Commented Oct 10, 2018 at 8:48

Give the three people numbers $$1,2,3$$ and let them choose gloves one by one.

For $$i=1,2,3$$ let $$E_i$$ denote the event that person $$i$$ chooses a pair.

To be found is $$1-P(E_1\cup E_2\cup E_3)$$ and with inclusion/exclusion and symmetry we find that this equals:$$1-3P(E_1)+3P(E_1\cap E_2)-P(E_1\cap E_2\cap E_3)=$$$$1-3P(E_1)+3P(E_2\mid E_1)P(E_1)-P(E_3\mid E_1\cap E_2)P(E_2\mid E_1)P(E_1)=$$$$1-3\frac15+3\frac14\frac15-\frac13\frac14\frac15=\frac{60-36+9-1}{60}=\frac{32}{60}=\frac8{15}$$

• An elegant approach. Commented Oct 10, 2018 at 8:59