You're on the right track zexen but there's a faster way the determine the 3 fastest. I'll give everyone a hint in saying that after the 5 initial races you can eliminate all the horses the placed 4th and 5th because they got beat by 3 or more horses making it impossible for them to be in the top 3.

That's exactly what Zexen was saying, and I put it though the test also and came up with 11 races.

Race 1=1.2.3.4.5.
Race 2=1.2.3.6.7.
Race 3=1.2.3.8.9.
Race 4=1.2.3.10.11
Race 5=1.2.3.12.13.
Race 6=1.2.3.14.15.
Race 7=1.2.3.16.17.
Race 8=1.2.3.18.19.
Race 9=1.2.3.20.21.
Race 10=1.2.3.22.23.
Race 11=1.2.3.24.25.

I don't see how that can break down further.

There is a faster way.

I'll further my hint this way :

The 5 first races go like this :

1-2-3-4-5
6-7-8-9-10
11-12-13-14-15
16-17-18-19-20
21-22-23-24-25

To simplify, let's say they place according to their assigned number. We can eliminate 10 horses that way : 4-5-9-10-14-15-19-20-24-25.

I tried it that way too, thinking it would give me a shortcut, but I haven't found anything yet. I may have one lead I'll test out which may cut it to 10 races, but I can't imagine it being any less than 10.

I'll let you try out stuff and I'll come back later to see if you need more hints!

Gahh, your hints are making it harder :P.

Okay. Idea.

1=1.2.3.4.5.

2=3.6.7.8.9
If any is faster than 3,

3: 1.2.6.7.8
So now you have either 1.2.3 left, and have sent out 6 in 2 races, or have 1.2.6 left (which gets translated to 1.2.3 later) in 3 races.

Wouldn't this get it done faster, at least potentially?

What you gotta realize is that the information gathered from the first 5 races is useful for further races. So racing all 25 horses in the first 5 races has the advantage of giving information about every horse.

What if you race all horses in 5 races first.
Then you race the winners in every race.
This way we can eliminate all except the number 2 and 3 from the race of the horse that one won the winners race, and nr 2 from the race of the horse that came second. Oh, and except the 3 winners of the winners race, ofc.

We now have 6 horses left. No?

That means we would have 5 horses left to test, since we already know which horse came first. How convenient. They all fit in one race! The nr 1 and 2 in that race are nr 2 and 3 globally.

You do it in 7 races!

Is this correct?

And we have a winner! That's exactly it.

Ok, another easy one to keep it going:

2, 3, 10, 12, 13, 20, 21, 22...

What is the next number?