It's that time of the season again! Maths and scenarios and calculations galore.
Group B of the 2014 World Tour Finals ended its second day of matches, and we can already calculate most of the possibilities for the semifinal qualifications. (Also check out the similar
scenarios for Group A)
Thursday's matches are Federer-Murray and Nishikori-Raonic. Here are the qualification scenarios for every possible outcome of those matches (winner + number of sets). Where it says "GR", it means we're going to have to calculate the game ratios (games won out of total played games) of all players to know who qualifies (Raonic can only qualify if he wins in 2).
ETA: Of course, Raonic's withdrawal changed everything, as Ferrer can't qualify no matter what he does. I'm writing this as the Nishikori-Ferrer match is at 6-4 4-6, so whoever wins, does it in 3 sets. This leaves us with the following scenarios:
| Ferrer in 3 | Nishikori in 3 |
| Federer wins | | |
| Murray in 2 | 1. Murray
2. Federer | 1. Murray
2. Federer |
| Murray in 3 | 1. Murray
2. Federer
| |
The old scenarios, for those curious:
| Nishikori in 2 | Nishikori in 3 | Raonic in 3 | Raonic in 2 |
| Federer in 2 | | | | |
| Federer in 3 | | | | 1. Federer
2. Murray |
| Murray in 3 | | | | |
| Murray in 2 | GR (see below) | | | |
Game ratio scenarios:
If Andy and Kei both win in 3 sets, should the 2nd qualifier be decided on games ratio??
ReplyDeleteJill.
No. If they both win in 3, this is what happens:
DeleteRF, KN, AM all have 2 wins, MR has 0. So we have a 3-way tie, and we go to set ratio.
RF is 5-2, KN and AM are both 4-3.
So RF qualifies first, and we're down to KN and AM. Now that we're down to 2 players, what decides is their H2H - and since Nishikori beat Murray, he's the one who qualifies as 2nd.
Thanks, I'd forgotten the 2 players = H2H rule. Jill
ReplyDelete