Mathematical group matters

[The Statman]

Mention was made that Ken Doherty has gone out having lost his first three matches.

Am I therefore right in thinking, therefore, that it is mathematically impossible to qualify if a player has lost 3?

I'm pretty sure it's unlikely but not certain that, dependent on other results of course, it is 100% impossible.

Any advice gratefully received, since the 2nd round prediction thread is so long!

[zombieman]

I believe it is possible to qualify with only two wins.

Example: Ken Doherty's group.

Consider the following:

Results so far:
Judge bt Doherty
Walden bt Doherty
McCulloch bt Doherty
Allen bt Williams
Williams bt McCulloch
Williams bt Judge
Walden bt McCulloch
Judge bt Allen
Walden bt Allen

Hypothetical results:
Doherty bt Allen
Doherty bt Williams
Walden bt Williams
Walden bt Judge
Allen bt Judge
McCulloch bt Williams

Walden - 5 wins
Doherty - 2 wins
Williams - 2 wins
Allen - 2 wins
Judge - 2 wins
McCulloch - 2 wins

Besides Walden, one player with 2 wins would qualify. Also proven that, Ken Doherty is (mathematically) not out of it yet.

EDIT: A more pedantic reader could remark, that in order for Ken Doherty to qualify, he'd also need to have the best frame difference in the group of players with equal wins. Indeed. However, I don't think that's a problem. Ken Doherty's current frame difference is -6, in optimal case (two 4-0 wins) he could make that into +2, and it's possible (again, mathematically) for the others to have less.

It's also possible (in another scenario) to not qualify with 4 wins, I'll leave the proof for later

[The Statman]

I believe it is possible to qualify with only two wins.

...

Also proven that, Ken Doherty is (mathematically) not out of it yet.

Many thanks. I am right to hesitate before removing impossible pairings, then, as I suspected.