tag:blogger.com,1999:blog-8847433.post3867051256332316409..comments2016-01-31T05:21:31.465+00:00Comments on Tony's Blog: Paine ReliefTony Deblinghttp://www.blogger.com/profile/17324997608113897906noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-8847433.post-15953890914162013592012-01-22T13:19:22.654+00:002012-01-22T13:19:22.654+00:00Hi Tony,
If you modify the problem so that the fir...Hi Tony,<br />If you modify the problem so that the first passenger takes any seat arbitrarily, then the probability that each passenger k>1 finds his seat already taken is 1/(n-k+2), where n is the number of passengers.<br />So the probability that the last passenger finds his seat taken is, rather elegantly, 1/2.<br /><br />The extra condition that passenger 1 cannot take his own seat makes the result a little more complicated.<br /><br />Being a bit of a slogger, I didn't get the answer intuitively but by calculating the probability that passenger k finds his seat already taken as a sum of the probability that passenger 1 took it, passenger 2 took it, ... etc. These sums look quite fearsome when first written down, but simplify very nicely.Stephen Clarkehttp://www.blogger.com/profile/04246458521808235249noreply@blogger.com