Question 476890
If there are n = 2 tickets only, the probability is 1.  So we exclude this case.  Assume that {{{n>=3}}}.
There are 2(n-1) possible pairs of consecutive integers in two drawings.  There are n(n-1) ways of drawing two pairs of tickets.  Hence the probability is  
{{{(2(n-1))/(n(n-1)) = 2/n}}}.