I'm not sure how to calculate it, but the odds of 3 matching (supposedly random) numbers in consecutive positions must be fairly low.
The exact chances depend on the numbers in question, for example your actual tickets:-
05, 17, 21, 36, 39, 45
xx, 17, xx, xx, 39, 45
The first number could take any of 16 numbers (1-16) except 05, and the 3rd and 4th numbers any between 18-38 (except for 21 or 36).
Take these numbers for the first ticket:-
xx, 03, xx, xx, 08, 09
And you've only got a very limited number of possible matches for the 1st, 3rd and 4th numbers as they appear on the ticket in order.
Going back to: xx, 17, xx, xx, 39, 45
If we take n2=18 then n3 can be any of the 17 numbers between 19 and 38 inclusive except 21 and 36
If we take n2=19 then n3 can be any of the 16 numbers between 20 and 38 inclusive except 21 and 36.
If we take n2=20 then n3 can be any of the 15 numbers between 22 and 38 inclusive except 36.
(Spotted that the numbers probably very slightly out here but I can't be arsed to fix it. It's close enough...)
17+16+15...+3+2+1 = (17*18)/2 = 153
Multiplied by the 15 possible numbers for the first number means we have 15*153 = 2295 tickets that are of the form: xx, 17, xx, xx, 39, 45 (and don't include any of 05, 21 or 36)
So that's 2295 of the 49C6 (13,983,816) possible tickets.
So about a 1 in 6093 chance that you'll get a ticket with numbers matching those constraints for another ticket (if your original ticket was xx, 17, xx, xx, 39, 45 ).
For:-
01, 03, 04, 05, 08, 09
xx, 03, xx, xx, 08, 09
Then there's only one other ticket:
02, 03, 06, 07, 08, 09
As any other of that form would have to have one of the numbers 01, 04 or 05 and match a 4th (or 5th or 6th) number from the previous line.
Only 1 ticket means we're back to a 1 in 13983816 chance of that happening.