Sir Alex Ferguson should be a happy manager, with his team through to the knockout stage of the Champions League and having lost only once in 17 matches in all competitions.
No such luck: the Manchester United manager is decidedly unhappy about the 'handicap' his team have faced in the Premier League this season. And, if he loses tomorrow's derby at Manchester City, Ferguson will feel once again that the fates are conspiring against him.
Ferguson's biggest gripe is that, in his view, United have been handicapped by a tough fixture list in the first half of the season, which has allowed Chelsea and Liverpool to pull clear at the top of the table. He points out, correctly, that United have had to play away in the Premier League after all six matchdays at the group stage of the Champions League, and he adds that United have the top 10 teams from last season away from home in the first half of the season.
The second claim is not wholly accurate, but the gist of Ferguson's argument is valid, except that it does not take into account that some of last season's top-10 teams have under-performed this season. One of those away games - directly after a Champions League match too - was at Blackburn, who finished seventh last season but are currently second-bottom. United didn't find that match too difficult, winning 2-0.
Another under-performing team from last season's top 10 are Manchester City, who had 29 points at this stage last season but have only 17 so far this time. Chelsea and Liverpool have already won at Eastlands this season, both scoring three goals as they took advantage of City's slow transition under Mark Hughes.
Ferguson, though, might be a little perturbed that United's visit coincides with City showing signs of finding form at last. Having beaten Arsenal 3-0 at home last weekend, City followed up with a fine 2-0 victory at Schalke on Thursday night in the Uefa Cup. Just as United come to visit, City are looking like a top-10 team for the first time in weeks. Just my luck, Ferguson will be inclined to think.