No, yours doesn't work, my friend The fastest of each heat might not necessarily be the 5 fastest because the heats are random. Two of the five fastest could have run in the same initial heat. My solution is the best one so far, from what I can tell. Bonanova never did give his answer

... and your solution was ...

* Split them into five heats. Take the 3 fastest in each heat, meaning you now have the 15 fastest.

* So do 3 heats now. Take the 3 fastest in each heat. You have the 9 fastest horses.

* Do a single heat with 5 random horses of the 9, cutting out the 2 last-place horses, leaving you with 7.

* Take another random 5 horses and do it again, cuttnig out 2, leaving you with 5 horses.

* Do one final heat and take the top 3 from that race.

Those top 3 horses will only have to race 5 times each.

It works ... but a lot races ....

The top 3 horses need to run 3 times, and in total seven races. If you don't understand my approach, there are other posts with the same solution but different approach. You may need to optimize your procedure ...

**Edited by brhan, 17 March 2008 - 11:50 AM.**