What is the expected number of tyres that are installed in their original positions?

Question

After summer, the winter tyres of a car (with four wheels) are to be put back. However, the owner has forgotten which tyre goes to which wheel, and the tyres are installed `randomly', each of the $4! = 24$ permutations being equally likely.

What is the expected number of tyres that are installed in their original positions?

Expected no. of tyres installed in original position = $P(1\ \text{tyre}) + 2P(2\ \text{tyre}) + 3P(3\ \text{tyre}) + 4P(4\ \text{tyre})$

I'm stuck after that. When counting $P(1\ \text{tyre})$ am I allowed to take the combination that all 4 tyres are in their right position ? Because that also includes 1 tyre in its original position...

If yes then $P(1\ \text{tyre}) = 13/24$

Or must I find the probability that exactly 1 tyre is in the original position ? If so, then what is the probability that 3 tyres are in their original position ? Because having 3 tyres in the original position would mean the 4th tyre must be in the original position ?

Answer 1

If $X$ is the number of tires installed in their original positions, then $$ X=1_{A_1}+1_{A_2}+1_{A_3}+1_{A_4}, $$ where $A_i$ is the event that the $i$th tire is installed in its original position, and $1_A$ is the indicator that the event $A$ occurs (it is $1$ if the event occurs, and $0$ otherwise). So, by linearity of expectation, $$ \mathbb{E}[X]=\mathbb{E}[1_{A_1}]+\mathbb{E}[1_{A_2}]+\mathbb{E}[1_{A_3}]+\mathbb{E}[1_{A_4}]=P(A_1)+P(A_2)+P(A_3)+P(A_4). $$ So, you need only find the probability of a fixed tire being installed in its original position.

But, these are straight forward: of the $24$ possible arrangements, there are $6$ that install tire 1 in its original position, so that $P(A_1)=\frac{1}{4}$. Similarly, $P(A_2)=P(A_3)=P(A_4)=\frac{1}{4}$.

So, you have that $$ \mathbb{E}[X]=4\cdot\frac{1}{4}=1. $$

Answer 2

Think about the possible cycle shapes for the symmetric group on $4$ elements.

There is $\color{red}{1}$ identity element which will cause all $\color{blue}{4}$ wheels to be in their correct positions.

There are $\color{red}{6}$ element with cycle shape $(ab)$ which will cause all $\color{blue}{2}$ wheels to be in their correct positions.

There are $\color{red}{8}$ element with cycle shape $(abc)$ which will cause all $\color{blue}{1}$ wheels to be in their correct positions.

There are $\color{red}{3}$ element with cycle shape $(ab)(cd)$ which will cause all $\color{blue}{0}$ wheels to be in their correct positions.

There are $\color{red}{6}$ element with cycle shape $(abcd)$ which will cause all $\color{blue}{0}$ wheels to be in their correct positions.

Now tot these up and we have \begin{eqnarray*} E(W) =\frac{\color{red}{1} \times \color{blue}{4} +\color{red}{6} \times \color{blue}{2} +\color{red}{8} \times \color{blue}{1} +\color{red}{3} \times \color{blue}{0} +\color{red}{6} \times \color{blue}{0} }{24} = 1. \end{eqnarray*}

Answer 3

First note that $P(4\ \text{tyres})=1/24$ and $P(3\ \text{tyres})=0$.

For $P(2\ \text{tyres})$, we must count how many permutations leave exactly two tyres in their correct positions. There are $\binom{4}{2}$ ways to choose the lucky two tyres to be positioned correctly, and given any such choice, there is exactly one way in which the other two would be positioned incorrectly. Consequently $P(2\ \text{tyres})=\binom{4}{2}/24=1/4$.

Finally we need $P(1\ \text{tyre})$. There are $4$ ways to choose which lucky tyre is going to be positioned correctly. For each of these choices, how many ways can the other three tyres be arranged so that none of which are in the correct positions? Given a second tyre, there are two (of the remaining three) positions that it can be placed incorrectly. Afterwards there is only one way to place the remaining two tyres. Thus there are $(4)(2)=8$ arrangements having exactly one tyre in its correct position, so $P(1\ \text{tyre})=8/24=1/3$.

Answer 4

$0$ tires in their correct positions (derangements): The number of ways $k$ of the $4$ tires are in their correct positions is $\binom{4}{k}$. The remaining $4 - k$ tires can be arranged in $(4 - k)!$ orders. Hence, by the Inclusion-Exclusion Principle, the number of arrangements in which no tires are in their correct positions is

$$4! - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0! = 9$$

exactly $1$ tire is in its correct position: By the generalized Inclusion-Exclusion Principle, the number of arrangements in which exactly one tire is in its correct position is

$$\binom{1}{1}\binom{4}{1}3! - \binom{2}{1}\binom{4}{2}2! + \binom{3}{1}\binom{4}{3}1! - \binom{4}{1}\binom{4}{4}0! = 8$$

where the factor $\binom{m}{1}$ counts the number of ways we can designate $1$ of the $m$ tires that are in their correct positions as being the one in its correct position.

exactly $2$ tires are in their correct positions: By the generalized Inclusion-Exclusion Principle, the number of arrangements in which exactly two tires are in their correct positions is

$$\binom{2}{2}\binom{4}{2}2! - \binom{3}{2}\binom{4}{3}1! + \binom{4}{2}\binom{4}{4}0! = 6$$

where the factor $\binom{m}{2}$ denotes the number of ways we can designate $2$ of the $m$ tires that are in their correct positions as being the ones that are in their correct positions.

exactly $3$ tires are in their correct positions: If three tires are in their correct positions, so is the fourth. Hence, there are no arrangements with exactly three tires in their correct positions.

exactly $4$ tires in their correct positions: There is only one arrangement in which all four tires are in their correct positions.

Hence, the expected number of tires in their correct positions is $$\frac{9 \cdot 0 + 8 \cdot 1 + 6 \cdot 2 + 0 \cdot 3 + 1 \cdot 4}{4!} = 1$$

Answer 5

Label wheels as $(A)(B)(C)(D)$ in correct order.

The numbers of outcomes for different correct tires: $4 \ tires: (A)(B)(C)(D) \Rightarrow n(4)=1.$

$3 \ tires: (A)(B)(C)(D) \iff 4 tires \Rightarrow n(3)=0.$

$2 \ tires: (A)(B)DC; (A)D(C)B; (A)CB(D);D(B)(C)A;C(B)A(D);BA(C)(D) \Rightarrow n(2)=6.$

$1 \ tire: (A)CDB;(A)DBC;C(B)DA;D(B)AC; DA(C)B; BD(C)A; CAB(D); BCA(D) \Rightarrow n(1)=8.$ I think you can calculate the expectation.