7 plays
First divided into 5 groups of a,b,c,d,e, 5 games
Get a1,b1,c1,d1,e1, assuming a1>b1>c1>d1>e1 (here you can change the serial number but not change the order)
Launched A1 as the first horse, D1,e1 could not be the first three horses, so d1,e1 do not take part in the last one.
At the same time, the reason may be the second third horse is: A2,A3,B1,B2,C1, (reasoning, there are more than two faster than it automatically out)
Then a2,a3,b1,b2,c1 another race, of which the first two horses are the second and third horses.
It is also necessary to prove that
7 times is the least answer, which requires proving 6 times not possible . Although I can be sure the answer is 7 times, but the idea of proof is not well organized.
If you want 25 horses to pick the fastest 3, only 5 horses run at the same time, at least a few times to race?
