So here is my solution. I know I said I would post it last night but was too knackered to bother at the end. The solution assumes that the reader understands the basics of probability.

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Let's start with two people. For argument's sake let's label the first person P1, and yourself P2, with P1's allocated seat S1, and yours S2. As the first person to board, P1 gets to choose his seat at random, so obviously the probability of picking S1 is 1/2, hence the probability of not picking S2, thereby leaving P2 with his allocated seat 1/2.

Probability of P2 ending up with S2:

1/2

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Now let's consider the situation with 3 people. P1 gets to choose his seat at random, so the probability of choosing S1 is 1/3. In that case, P2 has to choose S2, so the conditional probability of P3 (you) ending up with S3 in that scenario is 1.

Of course P1 could equally have chosen S3, at a probability of 1/3. In that case, the conditional probability of P3 ending up with S3 is 0.

We now have to consider the scenario of P1 choosing S2, at a probability of 1/3. In this particular case, P2 has two choices, since his allocated seat is gone. He could choose either S1 or S3 equally at a conditional probability of 1/2 for either. Look at it another way, this is equivalent to the situation where we have only two people, the first person having a random choice of seats. We know that the probability of you ending up with your allocated seat in that situation is 1/2, should we consider the case with only two people in isolation of the condition that P1 already chose S2.

Probability of P3 ending up with S3:

1/3 x 1 + 1/3 x 0 + 1/3 x 1/2 = 1/2

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Let's look at the situation with four people next. P1 has a 1/4 probability of choosing S1, in which case P4 ending up with S4 has conditional probability of 1. Likewise P1 has a 1/4 probability of choosing S4, leaving the conditional probability of P4 ending up with S4 0.

Should P1 choose S2, we end up with the equivalent situation of the previous scenario, where with 3 people, the first person is given a free choice. We know that in that case, the probability of the last person ending up with his allocated seat is 1/2.

Should P1 choose S3, P2 will have to sit in S2, and we end up with the equivalent situation of having only two people (P3 and P4), with the first person (P3) having a free choice. We also know from previous analysis that the probability of the latter person (P4) ending up with his allocated seat is 1/2.

Probability of P4 ending up with S4:

1/4 x 1 + 1/4 x 0 + 1/4 x 1/2 + 1/4 x 1/2 = 1/2

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From the above, hopefully a pattern should emerge. I won't write out the complete proof by induction with the "n" and "n + 1".

Hence the answer, whatever the number of people, is actually 1/2.