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fresh_42 said:Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases.

Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes:

Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.

When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all