K. Swan vs K. Scott — prediction

US Open

SWANWIN PROBABILITYSCOTT

64%

model prob.

@1.65

odds · 61% impl.

🎾Serve 63%

CONDITIONS OF THE MATCH◆ in the model◇ context

Surface◆

Hard

Consistent bounce, medium-fast: neutral conditions, no style favored.

Temperature◇

25°C

Warm: the ball flies a little more and fitness counts.

Humidity◇

52%

Dry air: the ball travels normally.

Wind◇

15 km/h

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.

OUR MODEL'S REASONING

›Ranking: #206 vs #244 (better ranked)

›Recent form: 4/10 in recent matches

WATCH FOR

!Returning from a long layoff (54d) — possible rustiness

Calibrated model probability (~64% out-of-sample accuracy, validated specifically on WTA). Not a guarantee: the model ≈ the market on average, so the odds already capture almost all the edge. 18+ · gamble responsibly.

@1.55

fair odds

+6.2%

expected value

BREAKDOWN · MODEL ANALYSIS

The model makes K. Swan the favorite with a 64% win probability, against K. Scott's 36% — a solid favorite, though Scott keeps real chances. Converted to odds, that probability is worth about @1.55; the offered odds are around @1.65 (a 61% implied), slightly above the market, so the model is a touch more optimistic.

Several factors explain the number: #206 vs #244 (better ranked); 4/10 in recent matches.

Read it with perspective. Our probability is calibrated — when the model says 64%, that outcome happens roughly that percentage of the time, with ~64% out-of-sample accuracy — but being the favorite is not being the winner: roughly 36 out of every 100 times Scott 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. Watch out for: Returning from a long layoff (54d) — possible rustiness. This is informational analysis, not a betting recommendation. 18+ · play responsibly.

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