A. Fery vs M. Rottgering — prediction
Consistent bounce, medium-fast: neutral conditions, no style favored.
Strong heat: warm air speeds the ball up and physical wear tells in long matches.
Dry air: the ball travels normally.
Light wind: no noticeable effect.
Surface feeds the model (surface specialization is one of its factors). Weather and altitude are context we publish for you — they do NOT move the probability.
›Ranking: #37 vs #389 (better ranked)
›Recent form: 7/10 in recent matches
›Model 75% vs market 68% → the model sees it as MORE likely than the odds
The model makes A. Fery the favorite with a 75% win probability, against M. Rottgering's 25% — a conviction read: the model sees the match clearly leaning one way. Converted to odds, that probability is worth about @1.33; the offered odds are around @1.46 (a 68% implied), slightly above the market, so the model is a touch more optimistic.
Several factors explain the number: #37 vs #389 (better ranked); 7/10 in recent matches; model 75% vs market 68% → the model sees it as more likely than the odds.
Read it with perspective. Our probability is calibrated — when the model says 75%, that outcome happens roughly that percentage of the time, with ~65% out-of-sample accuracy — but being the favorite is not being the winner: roughly 25 out of every 100 times Rottgering wins. The model also tends to agree with the market, so the odds already capture almost all the edge: don't take it as a sure value. This is informational analysis, not a betting recommendation. 18+ · play responsibly.