A line of 100 airline passengers are waiting to board a plane. Everyone has an assigned seat. Assume we know that the first, drunk passenger did not choose his own seat. All other passengers will go to their proper seats unless it is already occupied. In that case, they will randomly choose a free seat. You are person 100. What is now the probability that you end up in your own seat?
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"with a twist" What twist? This sounds like the normal airline problem to me. – JMoravitz Sep 23 '20 at 23:52
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This problem is telling us that the drunk passenger definitely chose the wrong seat. The other problems say that there is a possibility he chose the correct seat. – math2020e Sep 23 '20 at 23:57
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So, we know in the original problem the probability you end in the correct seat is $0.5$. We know that the drunk passenger in the original problem chooses his intended seat with probability $\frac{1}{100}$ and that if he did, you definitely end in the correct seat, so $0.01$ of the overall probability and $0.01$ of the probability (additive-wise, not fractionwise) of you ending in the correct seat is accounted for by this. This leaves $0.99$ of the time he is not picking his intended seat and $0.49$ of that (additive-wise) you end in the correct seat. – JMoravitz Sep 24 '20 at 00:00
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So... the probability you are after? $\dfrac{49}{99}$. This was only one very small step further than the original problem. – JMoravitz Sep 24 '20 at 00:00