KalshiChiusoVerifica disponibilitàKYC richiestoDati di risoluzione verificatiBen calibratoQuesto mercato si è risolto: A.J. Ewart beats De La Fuente and Ford
Risolto: 29 ott 2026, 19:56
Risoluzione verificata dal provider
CoinRithm dispone di prove del provider sul vincitore e sul momento di regolamento di questo mercato.
Avviso sulla qualità dei datiDati obsoleti
Dati al 29 ott 2026, 18:42 UTC · criterio pm-quality-4
Trend
David Ford beats De La Fuente and EwartPersa
Santiago De La Fuente beats Ewart and FordPersa
A.J. Ewart beats De La Fuente and FordVinta
Regole
If A.J. Ewart is determined to have won the 3-ball matchup among A.J. Ewart, Santiago De La Fuente, and David Ford in the 1st round of the 2026 VidantaWorld Mexico Open, then the market resolves to Yes.
- Tie Resolution If multiple participants are tied for the 3-ball lead at the end of the round: Yes payout: $1/N, rounded down to the nearest cent (N = number of tied participants) No payout: $1 minus the Yes payout Withdrawal Rule If a golfer withdraws prior to teeing off, all markets will resolve to fair market value. Kalshi is not affiliated, associated, authorized, endorsed by, or in any way officially connected with the PGA. All trademarks, logos, and brand names are the property of their respective owners.
Regole per ogni esito
- David Ford beats De La Fuente and Ewart
- If David Ford is determined to have won the 3-ball matchup among David Ford, Santiago De La Fuente, and A.J. Ewart in the 1st round of the 2026 VidantaWorld Mexico Open, then the market resolves to Yes.
- Santiago De La Fuente beats Ewart and Ford
- If Santiago De La Fuente is determined to have won the 3-ball matchup among Santiago De La Fuente, A.J. Ewart, and David Ford in the 1st round of the 2026 VidantaWorld Mexico Open, then the market resolves to Yes.
- A.J. Ewart beats De La Fuente and Ford
- If A.J. Ewart is determined to have won the 3-ball matchup among A.J. Ewart, Santiago De La Fuente, and David Ford in the 1st round of the 2026 VidantaWorld Mexico Open, then the market resolves to Yes.