Horse Race

[Guest]

Ok, so there are 25 horses and the race track only allows 5 horses to race at a given time. Given that there is no stop watch available your task is to determine the fastest 3 horses. Assume that each horses speed is constant in different races, what is the minimum number of races to determine the fastest 3?

Hint: Its single digit number!

[Guest]

Dear anyone,

My guess would be 8 races.

the fastest three horses all come from one single race.

can anyone who share my answer tell me if im correct?

[Guest]

Im sorie,

i changed my answer to 7 races. jus realised need not go on the 8th race if the three fastest horse comes from the same initial race.

[Guest]

It is 6, 5 races to determine the fastest among the 5 different groups, and then the last one to see who are the fastest horses.

[Guest]

In all of my calculations, I cannot reduce the number of races below 8 (except one scenario, which is not guaranteed). What I would like to know is how the solutions posted account for the third fastest horse being the runner-up to the second fastest horse in the original heat. That is, if the winner of the first heat in step one is the fastest horse, the winner of the second heat is the second fastest horse, and the runner up to the winner of the second heat is the third fastest, then an eighth race is needed to determine the top three horses.

1) Run five separate races of five horses each. those in fourth and fifth place cannot be one of the three fastest.

2) Race the five winners against each other. The first place horse is the fastest horse. The fourth and fifth place horses cannot be one of the three fastest, neither can the third place horses from the original heats except the horse which ran third against the fastest horse. Likewise, the horses that ran second (in the original heats) to the fourth and fifth place horses in the 'winners' heat cannot be one of the fastest three horses.

3) This leaves us with 7 horses. If we race the top three finishers of the heat in step two against the second and third place finishers to the fastest horse in the original heat and the results indicate that the top three horses are the fastest horse and the two runners-up from the same original heat are the three fastest in this race, then we can declare them the three fastest horses. However, if the results of this race are different, one last race is needed to determine the three fastest horses.

So, the least number of races that must be run is 7 if and only if the three fastest horses are in the same heat in step 1. Otherwise, an eighth race is needed to determine the top three.

[Guest]

1. Take five horses and race them.

2 Take the winner of the 5 and put him up against four new horses.

3. So on and so on.

4. In the end the fastest is left over. IT should take a total of 6 heats. (1×5 + 5×4)

Edit: Srry, misread. Top 3.... darn...

Edited August 7, 2008 by Mr_ME

[Prime]

It can be done in 7.

1. Make 5 races, 5 different horses each. -- That's 5 races.

2. Take the winner from each of the 5 groups and race those 5 horses. The winner of that race is the fastest horse of all. That's race #6

3. Name groups according to places their winners took in the winners race: 1, 2, 3, 4, 5. And name horses in accordance with their group name and their place inside the group. I.e., the horses from the group of the fastest horse are 1.1, 1.2, 1.3, 1.4, and 1.5. The horses from the group of the horse, which took 2nd place in the winners race are: 2.1, 2.2, 2.3, 2.4, and 2.5. And so on.

Run the race to determine the overall 2nd and 3rd place between: 2.1, 3.1, 1.2, 1.3, and 2.2. The first horse in that race is in second place, and the second -- in the 3rd place overall. And that's the race #7.

[Guest]

There are 25 horses, so that would be 5 races to figure out the 5 fastest. Then you take the 5 that won the 5 races, and race them in a 6th race. After that you eliminate the 2 slowest and get the 3 fastest.

Edited August 15, 2008 by Kait

[Guest]

i think it's possible in 8 races:

number all horsed from 1 to 25.

race 1: race horses 1 - 5

race 2: race horses 6 - 8 with the 2nd place and the 3th place of race 1 => make new top 3

race 3: race horses 9 - 11 with the 2nd place and the 3th place of race 2 => make new top 3

... keep doing this until ...

race 8: race horses 24 and 25 with the 2nd place and the 3th place of race 7 => now you got the top 3 all all horses

Edited August 16, 2008 by dfhwze

[Guest]

i think it's possible in 8 races:

number all horsed from 1 to 25.

race 1: race horses 1 - 5

race 2: race horses 6 - 8 with the 2nd place and the 3th place of race 1 => make new top 3

race 3: race horses 9 - 11 with the 2nd place and the 3th place of race 2 => make new top 3

... keep doing this until ...

race 8: race horses 24 and 25 with the 2nd place and the 3th place of race 7 => now you got the top 3 all all horses

i couldn't seem to edit, so new reply:

forget this, this isn't 100% accurate.

[Guest]

ok so here is how I did it and it is not 6 races because what if the 2nd and 3rd place horses are in the first group?

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

ok first step each column races which eliminates the bottom 2 from each column

5 races

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

k so now the 1st row races eliminating the last 2 places because they could not get 3rd place or second and the first place of that race is the winner since it beat it's column and the first place winners of the other columns

so now 6 races

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

Now we can eliminate the columns of the two horses that came in last place of the last race without a race because they lost to the horses who lost and we can eliminate 2c because 1a, 2a, and 2b are all faster than it, and the same with 3c beacuse 1a, 3a, and 3b are faster

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

ok now we can race 3a, 1b, 2b, 1c, and 2c and the fastest 2 from that race can advance to race 2a and lets say 1b beats 3a and 3 a comes in second

so 7 races so far

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

now we can determine the last race and this is the 8th race so lets just say 1b comes in first and 2a comes in second out of that so we have our top 3

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

so 8 races to determine top 3, 1a, 1b, 2a in my race

(edit, tried to make c row even, would not work)

Edited September 8, 2008 by DaGriller

[Guest]

ok so here is how I did it and it is not 6 races because what if the 2nd and 3rd place horses are in the first group?

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

ok first step each column races which eliminates the bottom 2 from each column

5 races

1a2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

k so now the 1st row races eliminating the last 2 places because they could not get 3rd place or second and the first place of that race is the winner since it beat it's column and the first place winners of the other columns

so now 6 races

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

Now we can eliminate the columns of the two horses that came in last place of the last race without a race because they lost to the horses who lost and we can eliminate 2c because 1a, 2a, and 2b are all faster than it, and the same with 3b and 3c beacuse 1a, 2a, and 3a are faster

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

ok now we can race 3a, 1b, 2b, 1c, and 2a and the fastest 2 from that race come in 2nd and 3rd, lets say 1b beasts 2a and 2 a comes in second

7 races

1a 2a 3a 4a 5a

1b 2b 3b 4b 5b

1c 2c 3c 4c 5c

1d 2d 3d 4d 5d

1e 2e 3e 4e 5e

so 7 races to determine top 3, 1a, 1b, 2a in my race

(edit wouldnt work) (typed in my rough draft, this is final)

Edited September 8, 2008 by DaGriller

[Guest]

Start with 5 heats named Apple, Bear, Cat, Dog, Elf

A horse's name is then referred to as the heat plus their place ... so Apple1 is the first heat first place.

The 6th race is as everyone has said the 5 winners. The winner of this 6th heat is the fastest horse.

Now take the 2nd and 3rd horse from the 6th race lets say bear1 and cat1. The 2nd place horse of the 6th race can be beat by any horse from the Apple heat. Since the 4th and 5th finishers of the apple heat are automatically out of the runnings we will take Apple 2 and apple 3. Now the 3rd place finisher can be beat by any horse from heat apple or bear. Now remember we're only looking to fill 2nd and 3rd place apple1 is the fastest. We will include bear2 in the final race too. The only place bear2 can place is in 3rd place after bear 1

the 7th heat is Apple2,Apple3,Bear1,Bear2,Cat1

these are all the possible outcomes

Apple1, Apple2, Apple3

Apple1, Apple2, Bear1

Apple1, Bear1, Bear2

Apple1, Bear1, Cat1

[Guest]

First I thought six, but maybe for a complete check, 9?

1. split into groups of 5 and race = 5 races

2. Collect the 1st 2nd and 3rd from all races!!

3. Race all of the 1st places and get a winner from that. = 1 race

4. Race all of the 2nd places and get a winner from that. = 1 race

5. Race all of the 3rd places and get a winner from that. = 1 race

6. Finally race the winner from 4. 5. and 6. against each other to see who is 1st 2nd and 3rd!!

Maybe too extreme??? Has this already been answered??

(I was also, for a full check, thinking 11, but that is not a single digit number, so 9!!)

Edited November 1, 2008 by NintendoMad

[Guest]

Damn it it's seven!! There is another topic with the same question!! Roolstar has the link!!! His solution and PDR's makes perfect sense!!

[Guest]

I had a way that took 11 tries, haha. Then I saw I was supposed to get single digits.

My way was do the original 25, then take the top three from each race and have a new race for 1st place winners 2nd place winners and 3rd place winners. (this is 8 now) then two more races for the 1st-3rd from each of those races, then another to get it all done... but everyone else's way is much better.

[Guest]

Shouldn't the answer be 6? First run five groups of five. (25 total horses) Now you have the fastest 5 horses. Run one more race and the horses that come in first, second, and third are the fastest three as all the horses run the same speed in every race. Can someone explain why this wouldn't work?

[fabpig]

Shouldn't the answer be 6? First run five groups of five. (25 total horses) Now you have the fastest 5 horses. Run one more race and the horses that come in first, second, and third are the fastest three as all the horses run the same speed in every race. Can someone explain why this wouldn't work?

Because the 2nd and 3rd fastest horses may have been 2nd (or even 3rd) in the fastest run of one of the original 5 races.