1913 Kentucky Derby

The 1913 Kentucky Derby was the 39th running of the Kentucky Derby. The race took place on May 10, 1913.[1] Horses Prince Hermis, Sam Hirsch, Flying Tom, and Floral Park scratched before the race. The winning time of 2.04.80 set a new Derby record.[2] With odds of 91–1, winning horse Donerail is the longest odds winner in Kentucky Derby history.[3]

39th Kentucky Derby
Kentucky Derby
Grade I stakes race
Donerail after winning the 1913 Kentucky Derby
LocationChurchill Downs
DateMay 10, 1913
Winning horseDonerail
JockeyRoscoe Goose
TrainerThomas P. Hayes
OwnerThomas P. Hayes
SurfaceDirt

Full results

Finished Post Horse Jockey Trainer Owner Time / behind
1st 5 Donerail Roscoe Goose Thomas P. Hayes Thomas P. Hayes 2:04.80
2nd 4 Ten Point Merritt C. Buxton Calvin Banks Anthony L. Aste 12
3rd 3 Gowell John McCabe John T. Weaver John T. Weaver 1 12
4th 8 Foundation Johnny Loftus William L. McDaniel C. W. McKenna Head
5th 6 Yankee Notions Buddy Glass William H. Karrick Harry K. Knapp Neck
6th 1 Lord Marshall T. Steele J. Oliver Keene J. Oliver & G. Hamilton Keene 5
7th 2 Jimmie Gill Charles Borel Shelby West L. P. Doerhoefer & Shelby West 8
8th 7 Leochares Charles Peak John F. Schorr John W. Schorr 15
  • Winning Breeder: Thomas P. Hayes; (KY)

Payout

Promotional flyer for the Selig Polyscope Company's documentary, The Kentucky Derby at Churchill Downs (1913)
Post Horse Win Place Show
5 Donerail $ 184.90 41.20 13.20
4 Ten Point 3.50 3.30
3 Gowell 14.10
  • The winner received a purse of $5,475.
  • Second place received $700.
  • Third place received $300.[2]
gollark: Against the random one it rapidly decides to not trust it and probably does well for it, against tit for tat it cooperates, against tat for tit it soon apifies it, against devil it also soon apifies it, against angel it's nice to it (suboptimal, can't really fix it easily), against time machine it cooperates, against grudger it cooperates, and that's basically it.
gollark: It probably isn't optimal but you know.
gollark: ```scheme(define forgiving-grudge (lambda (x y) (let* ( (defection-count (length (filter (lambda (m) (= m 1)) x))) (result (if (> defection-count 3) 1 0)) ) result)))```As far as I can tell this consistently wins.
gollark: I fixed it except now my thing plays itself at some point and recurses infinitely.
gollark: It has a child process. This is ridiculous. It lies.]

References


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