find the 3 fastest horses

25 not 15, a redo

5 races of 5 fresh horses eliminating 2 slowest in each race. Now you have 15 left.

Add five races to the 15 total (13 races) and the whinniers will have to run one more race.

Rap

I get nine races.

Round 1: 5 races of 5 (eliminate 4th and 5th place finishers)
Round 2: 3 races of 5 as follows:
- all 3rd place finishers from round 1
- all 2nd place finishers from round 1
- all 1st place finishers from round 1

Only the winner of the 3rd place race has the potential for survival.
Only the top two of the 2nd place race have the potential for survival.
The top three from the 3rd place race all survive.
There are six potential survivors.

For the winner of the 3rd place race to survive, the two who beat him in the first round must win their second round races. Call these 1A, 2A, and 3A (top three from race A in first round). Otherwise he's eliminated because he's slower than at least three horses.

That means the runner up in the second place race (call him 2B because he placed second in race B in the first round) is slower than 1A, 2A and 1B, so he's out.

In either case, we're down to at most five horses. Race them (ninth race) and the top three are your fastest three.

If we're lucky, the only survivors from round two are the top three from the first place race. In that case, there is no ninth race.

First race with 5 horses.
Keep the fastest 3 in for the next race and take 2 other horses.
After this race keep the fastest 3 and replace the last 2 with 2 more.
Keep doing this and it will take 11 races.

In that way, every horse will race against the fastest three.

thanks for all the posts.....when I did it, I also got 9 races!
Unfortunately, a trusted source has informed me that it is less than 9.
I can't figure the trick out yet, but will keep trying.....any other help is welcome!

Thanks,
Zipote

I now get seven races

First round same as before.

Sixth race: All of the first place finishers from the first round. Let's say they finish 1A, 1B, 1C, 1D, and 1E.

At this point there are only six contenders:
1A, 1B, 1C, 2A, 2B, and 3A.
But, we already know that 1A is the fastest.

Seventh race: 1B, 1C, 2A, 2B, and 3A. The top two join 1A as the fastest three.

Due to the wording of the riddle, you could have a situation where the last horse in the first race is faster than the first horse in another race.

Otherwise, you could have the first 5 races and the 5 winners would run in the 6th to find the fastest 3.

Only if the fastest three all came from the first race.

Markr!!! You got it! Seven is the fewest # of races in which you find the three fastest horses.

Good job!

Mark wrote, "Only if the fastest three all came from the first race."

Please excuse my pour English. When I wrot;

"Keep the fastest 3 in for the next race and take 2 other horses.
After this race keep the fastest 3 and replace the last 2 with 2 more."

Wat I ment to say was;

"Keep the fastest 3 in for the next race and take 2 other horses.
After this race keep the fastest 3 and replace the last 2 with 2 more."

After the first race ALL the other horses get to race, and the two losers (4th and 5th) drop out in each subsequent race.

The first race does no more than eliminate the first two.

The next nine eliminate eighteen more.

The last (11th) gives the final 1,2,3.


Quod erat demonstrandum. :wink:

How do you know that 1A is one of the fastest though?

Wait a minute....why isn't 2C a contender? And why 3A and not 3B or 3C?
7 is the correct answer, but can you explain it a bit more....perhaps I'm just extra slow as I'm half asleep at the moment, but I follow the rational upto the 6th race, but then get lost with the asumptions you are making!

sorry.... I woke up and finally got it! Please ignore my ignorant bable!

I understood your English in spite of the accent. I had bolded the erroneous text. It is the statement about every horse racing against the fastest three that is incorrect. If the fastest three are all in the last race, then only two horses get to run against them. The fastest three would have to be in the first race for all horses to run against them.

To cast the light of reason into the darker recesses of this problem. May I offer a rebuttal to your contention?

I am sure we would agree that if you raced all 25 horses in one go, the first three would be considered the fastest.

Therefore, the first race of 5 only eliminates the two that are not 1/2/3.

All the other horses get to race, and if they come in the 1/2/3., they stay, if not they too are eliminated.

Assuming the first race 1/2/3, were slower than the other 20 horses waiting to race, they would be replaced two by two by faster horses. I repeat, it makes no difference in what order the ultimate fastest horses race, if they are not beaten they stay.

10/4

The algorithm is sound for determining the fastest three. But it is very unlikely that "every horse will race against the fastest three." I interpret that statement to mean that when the process is complete, each of the 22 slowest horses will have raced against each of the three fastest horses.

2.5

I don't know what you think you know, and I know that you think you now know what I know. However, I know what I now know, and I know that you think you know what I said. Nevertheless, I'm not sure whether you understood that what you thought is not what I meant.

1.25

I'll excuse you...pour me a glass :wink:

Doesn't that first sentence contradict itself.

Correction: What you said is not what you meant.