Carlos Alcaraz outplayed unbeaten Jannik Sinner, and now it's about finding his best against Daniil Medvedev again.
These two players met in last year's Indian Wells final, so fans will be treated to the same final for the second year in a row. It's not surprising because both of them have been really strong on this court, even though Medvedev hates it.
It's a slow and high-bouncing hard court, one that helps the Russian drag out points, while for Alcaraz, it allows him to unleash his devastating power. He was able to do that against Sinner often in the semi-final, scoring his biggest win in a very long time.
Alcaraz needed it because his confidence had been trending downward for a while, and handing the Italian his first loss of the year would certainly help him with that.
Head-to-head:
Another thing that should give him confidence is the way this matchup unfolded at last year's Indian Wells Open. Many expected a close match, but Alcaraz absolutely demolished Medvedev in two simple sets.
He was aggressive and took the ball early, leaving Medvedev with no chance to really drag out the rallies. I expect a similar approach from Alcaraz this time around, especially with the way he's been swinging the racquet in the last few sets.
The question is whether Medvedev will have a counter to that. He didn't have it back then, but he's beaten Alcaraz in the past. It's 3-2 overall for the Spaniard, with the most recent meeting at the ATP Finals going the Spaniard's way as well.
| Carlos Alcaraz | Daniil Medvedev | |
|---|---|---|
| 3 | H2H | 2 |
| 2 | Rank | 3 |
| 20 | Age | 28 |
| 11-3 | 2024 W/L | 14-2 |
| 166-46 | Career W/L | 347-138 |
Prediction:
I think the courts will help Alcaraz quite a bit. They leave Medvedev as a sort of a sitting duck at the mercy of the Spaniard, and I think he'd do way better on a quicker court like the one at the US Open.
It would allow him to put Alcaraz under pressure, which he shouldn't feel here. That's why I'll back the Spaniard to defend his trophy.
Prediction: Carlos Alcaraz to win in two sets.
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