[math-fun] Question-sequence probability

There are 9! sequences for the 9 questions. So there are (9!-1) 'other' sequences. There are 9*8/2 = 36 pairs to flip if one flip is required. So there are 1+36 sequences 0 or 1 'flip' away from the 1-9 sequence. Therefore the probability = 37/(9!-1) = 37/362,879, just over 1 in 10,000 g