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Really basic probability question that I want answered,

There is a class room with 3 students and 3 seats. Each student selects a seat and sits on it. The next day, the instructor randomly picks a seat for each student. What is the probability that each student is assigned a different seat from the one they sat on yesterday? (i.e: Everyone is in a different seat).

The way I approached the problem was to find the probability of the students sitting in the same seat. If there are 3 students, A, B, C, and the instructor is assigning them to the seats one by one, then

P(A is assigned a different seat than yesterday) = 2/3

P(B is assigned a different seat than yesterday) = 1/2

P(C is assigned a different seat than yesterday) = 1

P(Everyone is assigned a different seat) = 2/3 * 1/2 * 1 = 1/3

Is this the correct way of doing this?

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You have to be careful, because the parts aren't independent - certainly the probability that A gets a different seat is $\frac{1}{3}$. But there are two possibilities - either he gets B's seat (in which case B has a 100% chance of getting a different seat, and C has a 50% chance) or he get's C's seat (in which case B has a 50% chance of getting a different seat, while C has a 100% chance).

Now in this situation it still works, because you can choose to look at the person whose seat A didn't get, and you get the answer you've found. But for larger groups, this won't work - for example, with 4 people you can have it so that there's a loop of people sitting in each other's seats (e.g. A sits in B's seat, B sits in C's, C sits in D's and D sits in A's), or you can have it so that two pairs of people each swap seats (so A and B swap, and C and D swap), and a simple approach to the probability won't work.

In fact, there's a bit of a trick to calculating it, and the details are explained in the Wikipedia article on Derangement, which is what this is usually referred to as.

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Your method fortuitously gives the correct value, but is not actually correct.   Student $A$ can be in the wrong seat by being in either seat B or C.   If A is in seat B then B may equally likely be in seat A or C, otherwise if A is in seat C then B may equally likely be in seat A or B.

$\begin{align}\mathsf P(A\in\{B_0, C_0\}) &= 2/3\\ \mathsf P(B\in\{A_0,C_0\}\mid A\in\{B_0, C_0\}) &= {\mathsf P(B\in\{A_0,C_0\}\mid A\in\{B_0\})\mathsf P(A\in\{B_0\}\mid A\in\{B_0,C_0\}) \\ + \mathsf P(B\in\{A_0,C_0\}\mid A\in\{C_0\})\mathsf P(A\in\{C_0\}\mid A\in\{B_0,C_0\}) } \\ & = 1\cdot \tfrac 12+\tfrac 12\cdot \tfrac 12\\ & = \tfrac 34 \end{align}$

And so on.

A simpler method:

There are several equally likely ways to arrange three students, of which, only some have no student in their original position.   Let us have a look.

$$\left\{\rm \underset\checkmark A\underset\checkmark B\underset\checkmark C, \underset\checkmark A\underset\times C\underset\times B, \underset\times B\underset\times A\underset\checkmark C, \underset\times B\underset\times C\underset\times A, \underset\times C\underset\times A\underset\times B, \underset\times C\underset\checkmark B\underset\times A\right\}$$

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