> The worst case of (1,2,3) is respectively 4, 3 and 2, resulting in log_2 24 = 4.32 bits. The first choice isn't picking from 6, it's picking from 4.

Oh, you are correct that the first number has only 4 possible choices. But what about the second number? It can't be 1 (already picked) and 6 (ordering requirement), but other numbers are open to pick. The third number is no different, it can't be 1 and 2 (already picked) but otherwise all other numbers are allowed. The resulting N would be 4, 4 and 4, i.e. 6 bits.

Maybe it helps to see the fuller picture:

  1st  2nd  3rd         2^(expected # of bits)
  
  1    2    3, 4, 5, 6  64
  1    3    4, 5, 6     48
  1    4    5, 6        32
  1    5    6           16
  
  2    3    4, 5, 6     36
  2    4    5, 6        24
  2    5    6           12
  
  3    4    5, 6        16
  3    5    6           8
  
  4    5    6           4

Ideally (and necessarily) the last column should be 24 for all possible cases, but your algorithm can go anywhere from 4 (2 bits) to 64 (6 bits).