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What is the expectation of the last number in random permutation of numbers (1,2,3,4)
Intuitively I think it should be (1+2+3+4)/4= 2.5 because all of the numbers should have the same probability of ending up last, and expectation is sum of their probabilities* the number, so its going to be the sum $10* (1/4)$ , however I'm not sure if this is correct and how to prove this.
Any suggestions? Thanks.
This is correct and you've certainly proved it to my satisfaction. If you want to see it more clearly, just note that of the $24$ permuations, $6$ end with $4,$ $6$ end with $3,$ $6$ end with $2,$ and $6$ end with $1$ so the expectation is $$ \frac{6}{24} (4+3+2+1) = 5/2.$$ If you like, the "$6$" comes from the $3!$ permutations of the other three. Actually I'm not positive this is any more clear than your argument, but it gets you down to the individual permutations rather than grouping them, which seemed to be causing you some doubt.
