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]

Horse Race
• 5 - Race 5 groups of five horses. Send all fourth and fifth place horses home. Leaves 15 horses
• 6 - Race all five first place horses. Send the fourth and fifth place horse home along with the horses they defeated in the first round. Since the winner of this race is the fastest horse in the whole group, we only need to identify the two fastest in the remaining horses. So, send home the horse who got third place against the second place horses from this race and both of the horses who lost to the third place horse. Leaves 6 horses, one of which is already known to be the fastest.
• 7 - Race the remaining horses, with the exception of the one we already declared as the fastest and keep the top two horses.

I will try to draw a picture to explain the description above. Red number are horses you can eliminate due t being defeated by at least three other horses.

1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5

1.1 1.2 1.3 1.4 1.5
2.1 2.2 2.3 2.4 2.5
3.1 3.2 3.3 3.4 3.5

[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!

Since the speed is constant, you can compare the time taken by each horse and pick the top three.

So each horse needs to run once, i.e. 25/5 = 5 different races.

[Guest]

Seven races.

1. Divide the horses into five groups

2. Do another race with the horses that stood first of each group. Obviously, the winner in this race is the fastest.

3. Do one more race of the horses which stood 2nd and 3rd of the group of the fastest horse, 2nd and 3rd of the race in step 2, and the horse that stood second from the group 2nd of the heat in step 2.

From step 2, we know the fastest horse, and from step 3 the second and third fastest horses.

[Guest]

Seven races.

1. Divide the horses into five groups

2. Do another race with the horses that stood first of each group. Obviously, the winner in this race is the fastest.

3. Do one more race of the horses which stood 2nd and 3rd of the group of the fastest horse, 2nd and 3rd of the race in step 2, and the horse that stood second from the group 2nd of the heat in step 2.

From step 2, we know the fastest horse, and from step 3 the second and third fastest horses.

But what if the 3 fastest horses are in one group in step 1. Then the 2nd and 3rd fastest horses get thrown out.

[Guest]

Seven races.

1. Divide the horses into five groups

2. Do another race with the horses that stood first of each group. Obviously, the winner in this race is the fastest.

3. Do one more race of the horses which stood 2nd and 3rd of the group of the fastest horse, 2nd and 3rd of the race in step 2, and the horse that stood second from the group 2nd of the heat in step 2.

But what if the 3 fastest horses are in one group in step 1. Then the 2nd and 3rd fastest horses get thrown out.

They will run on the third step above. If they are among the three fastest horses, as you suggested, they will stand first and second in the race of step 3.

Edited February 21, 2008 by brhan

[Guest]

They will run on the third step above. If they are among the three fastest horses, as you suggested, they will stand first and second in the race of step 3.

OK, I see what you're saying now. Step 3 is a bit confusing. So the last race will have the horses that finished 2nd and 3rd to the fastest overall horse in step 1, the horses that finished 2nd and 3rd in step 2, and the horse that initially finished 2nd to the 2nd place winner in step 2. That is hard to explain.

This way takes into account if all 3 fastest horses are in one race in step 1, if only the 2nd and 3rd fastest horses are in one race in step 1, and of course if they are all in different opening races.

Good job. I think you got it.

[Guest]

OK, I see what you're saying now. Step 3 is a bit confusing. So the last race will have the horses that finished 2nd and 3rd to the fastest overall horse in step 1, the horses that finished 2nd and 3rd in step 2, and the horse that initially finished 2nd to the 2nd place winner in step 2. That is hard to explain.

This way takes into account if all 3 fastest horses are in one race in step 1, if only the 2nd and 3rd fastest horses are in one race in step 1, and of course if they are all in different opening races.

Exactly, that is what I mean.

[roolstar]

well...

Click here for a surprise

[Guest]

The answer should be two.

The first five to get to the tracks from a starting point will abviously be the fastest ones. You make the top five race and Voila, the top three will the winners. Two races. Or, am I wrong?

[Guest]

Your seeing it as how many rounds there are, though its actualy says heats. There have got to be 5 heats to race all 25 horses with 5 per a heat. Hope you understand now.

[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!

A SIMPLE QUASTION TO EVERYONE-WITHOUT STOP WATCH HOW DO YOU KNOW,THAT THE HORSE LET SAY WICH FINISHED 4th IN IT'S GROUP IS NOT FASTER,THAN THE HORSE WICH FINISHED FIRST IN OTHER GROUP? IMAGINE THAT YOU HAVE 5 PROFESSIONAL SPRINTERS IN ONE GROUP AND 5 PEOPLE RANDOMLY TAKEN FROM THE STREET.HOW BIG IS THE CHANCE THE FIRST FROM THE RANDOM GROUP TO BEAT THE 5th FROM THE SPRINTERS?

[Guest]

If you read people's answers, Ash, you'll notice that everyone has taken that into consideration.

Riddari is right...you only need five.

[Guest]

The statement is that the horses speed will be constant in all races therefore if you run the 5 heats and take the winners from those to run the last race the 1st, 2nd and 3rd place finishers are the fastest 3

[Guest]

I say 6 races

take the fastest horse from each of the 5 heats, then pick the top three as the best of the best.

[Guest]

When i did this, i found that they only needed six races, while many other people are saying seven...please someone correct me if im wrong, which i probably am...

You have to run 5 races first, take the top five from there, race them, and then you get your top three...did i just oversimplify that? because it should work...

[Guest]

When i did this, i found that they only needed six races, while many other people are saying seven...please someone correct me if im wrong, which i probably am...

You have to run 5 races first, take the top five from there, race them, and then you get your top three...did i just oversimplify that? because it should work...

I think your case doesn't handle Enlightened's case on post #5 of the thread.

[Guest]

I don't understand why you eliminated the 2.3-3.5 with the exception of 3.1 (Well, I get 3.1).

[Guest]

I don't understand why you eliminated the 2.3-3.5 with the exception of 3.1 (Well, I get 3.1).

Not sure I understand your notation, but I hope I understand your question. I also had difficulty at first with bhran's solution the last time he posted it. After charting it out I realized that seven heats is sufficient because the third place finisher of each of the first heats could not possibly be among the top five, except for the horse that placed third after the first place finisher in the sixth heat. Does that help, or am I just confusing things?

[Guest]

I say six races. 5x5=25.

Take the five winning horses, put them against each other in the sixth race thaen take the top three out of those.

[Guest]

I say six races. 5x5=25.

Take the five winning horses, put them against each other in the sixth race thaen take the top three out of those.

ok, not so different...the same came up in between the time I posted.

[Guest]

I say six races. 5x5=25.

Take the five winning horses, put them against each other in the sixth race thaen take the top three out of those.

If the horses are randomly placed in the original heats (they have to be since we don't know their speeds) then we can't assume that the 2nd fastest and 3rd fastest each won a heat, they may have been placed in the same original heat as the fastest horse.

Take this as an example, let us assume we know before hand that the fastest horses were A1, A2, and B1, which were randomly spread into the 5 different starting heats. A1 and A2 would be in the same heat, B1 would be in the second heat, etc... the original 5 heats stack up like this:

Heat__A B C D E

---------------

Place_1 1 1 1 1

______2 2 2 2 2

______3 3 3 3 3

______---------

______4 4 4 4 4

______5 5 5 5 5

If we only did 6 heats, with the last being the race of the winners above, the results might be something like this:

Heat__F

--------

Orig__A1

Place_B1

______C1

______--

______D1

______E1

And we would assume that the fastest 3 horses were A1, B1, and C1 but we know from our fore-knowledge that A2 is faster than B1 so we came up with false results. We have to correct for this problem, and it takes a seventh race to do so.

We know that any horse that placed 4-5 in heats A-E can not be in the top 3, so they go home.

We also know that horse D1 and E1 can go home because they are slower than A1 B1 and C1. Also, any horse slower than D1 and E1 can go home, so there goes D2 D3 E2 and E3.

C1 placed at best 3rd place, so C2 and C3 can go home because they are at best 4th and 5th respectively.

B1 placed 2nd at best, which means B2 may be 3rd but B3 is 4th at best so B3 can go home.

We can't eliminate A2 or A3 because there is still the possibility that 1 or both are faster than B1 so we have to keep them both.

We are left with 6 horses, A1 A2 A3 B1 B2 and C1. We know A1 is the fastest so we don't need to race him. We are left with a final seventh heat between A2 A3 B1 B2 and C1. Which from our fore-knowledge we know ends up like this:

Heat__G

--------

Orig__A2

Place_B1

______--

______A3

______C1

______B2

A1 is the fastest because he won Heat F. Heat G was about claiming 2nd and 3rd which horses A2 and B1 did.

[Guest]

I got 6, race all 25 horses in groups of 5.

5 races have gone by.

take the first place riders of each group and race them. 5 people in that race. this is the sixth race.

the top 3 of that race are the fastest.

am i right??

[Guest]

I got 6, race all 25 horses in groups of 5.

5 races have gone by.

take the first place riders of each group and race them. 5 people in that race. this is the sixth race.

the top 3 of that race are the fastest.

am i right??

I thought this originally too but it's not correct because the question is to get the 3 fastest horses.

now lets take for example the 3 fastest horses all compete in the first race. Then at the end of 6 races you would not have the correct answer as the 2nd and 3rd fastest horses were eliminated in the first heat.

If you ran your races such that you did the first heat with 5 horses and always advanced the top 3 you could do it in 9 heats. However it is impractical as the horses would get tired since there is a potential for up to 3 horses to run 9 heats. There may be a better solution, ill think about it.

Edited June 19, 2008 by Lizard450

[Guest]

Brhan is right. At first I didn't quite understand what does he mean for the seventh heats. After a long thought the explanation is really there to justify why 7 heats will do.

First, we can't say 6 heats will do because by doing this we will ignore the chances that 2nd and 3d fastest horses from the group where the winner horse in heat 6 can run faster than the 2nd and 3rd horses in the heat 6.

For example, if we DO have stop watch, the result for the first 5 heats (let's say heat A, B, C, D and E) may be as such:

A ( 10sec, 10.5 sec, 11 sec, 12 sec, 13 sec)

B (10.2 sec, 10.6 sec, 11.1 sec, 11.8 sec, 12.1 sec)

C (9 sec, 9.8 sec, 10.1 sec, 10.4 sec, 10.7 sec)

D (10.9 sec, 11.3 sec, 11.7 sec, 12.2 sec, 12.4 sec)

E (12.8 sec, 12.9 sec, 13.1 sec, 13.3 sec, 13.5sec)

But because you don't have the stop watch so in the heat 6 you choose the best from each heat A to E and that which are the 10sec, 10.2 sec, 9 sec, 10.9 sec and 12.8 sec.

Notice that 2nd and 3rd horse in Heat C can run even faster than the 1st place from heat B, D and E.

So, 6 heats is not good enough to find the top 3 fastest horse from the 25.

But from the heat 6, you will find the result will be

C (9), A (10), B(10.2), D(10.9) and E(12.8)

and you will suspect whether 2nd and 3rd place for heat C can run faster than A (10) and B(10.2) ?

So that's where Brhan's solution applied.

Put 2nd and 3rd place of heat C: C(9.8), C(10.1) and 2nd place of Heat A (A 10.5) over and compete with A(10), B(10.2) ... C(9) is not necessary as it has been sure be the first among 25!

And here the heat 7 go to find the 2nd and 3rd among 25!

[Guest]

my guess is 6

split into groups of 5 and do 5 races then the fastest from each race eachover and the one that wins out of the 5 is the fastest