However, A horse beating another horse in a race is in no way a guarantee for that horse beating other horses or even hat same hose a second time. So if horse a wins a race against horse b. And horse c beats horse b. Horse a could still beat horse c.

@@john_mckinney True. And I mean simply because A won against B once, doesn't mean that A would always win against B. If they race again, B could win this time. This really is just a logic puzzle that doesn't apply in the real world.

KineticManiac But let’ssay A1 won the first race n 1 minute ad A4 and A5 came in 5th in 1:10 A4 and A5 are eliminated In the later race A1 races against B1-E1. A1 finishes first in 1:30 B1 second in 1:31. B1 does not get eliminated even though it is considerably slower than A4 and A5

@@john_mckinney As I said we don't eliminate horses because they're slow, we eliminate them because it's confirmed they're not in top 3. For the solution shown here to work, we need to assume the following: 1) If X is faster than Y, X will always win against Y. 2) If X is faster than Y, and Y is faster than Z, then X is faster than Z. The example you gave is actually a bit problematic, because you said that A1 has completed the race in 1:30 against B1, but A4 and A5 completed the race in 1:10 against A1 (which had completed that race in 1:00). Now, does this mean that A4 and A5 could potentially win against A1 if they race again? 1:10 is faster than 1:30 after all. A better way to think about this is to simply assume that each horse finishes the race in the exact same time. For example, A1 always finishes in 1:00, A5 always finishes in 1:10 and B1 always finishes in 1:30. Now in the first race A5 would be eliminated. In the second race B1 would not be eliminated. At first it seems as if we have a problem, because we eliminated a horse that finishes in 1:10, yet let a horse that finishes in 1:30 continue. This, seemingly a problem, gets solved when you think about how quickly A2 and A3 would finish the race. Surely they are slower than A1 (longer than 1:00), and faster than A5 (shorter than 1:10). So let's assume A2 finishes in 1:05 and A3 finishes in 1:06. So once B1 races against A2 and A3, it's going to lose, and get eliminated. So if B1 is slower than A5, it will eventually get eliminated, just that it will happen after A5. This seems to be what's confusing most people, the objective of the question is to find the fastest 3 horses. It doesn't claim that you have to eliminate the slower ones first, you can eliminate them in any order you like, so long as you can guarantee that you can find the top 3. So here, we eliminate them once we're sure they're not in the top 3.

KineticManiac my example is not problematic, since te horse A1 could have run his fastest race the first time, eliminating A4 and A5, who were (in that race) faster then B1 and C1 in the second race.

@@DougRayPhillips I didn't reconstruct anyone's meaning, thank you very much. I arrived at the optimum solution myself, AND I went a step further to point out additional possibilities. Thinking outside the box and going a little further than the rest of the crowd actually landed me the BEST jobs, I retired at 52 with a full pension and a seven-figure 401K. Don't lecture people when you know nothing about them.

While it is nice to potentially solve in six races, I think where this falls down is if the horse passed down wins the 6th race. There would be a total of 12 horses still in contention for 2nd and 3rd (to adequately test for if 2nd and 3rd came from the same preceeding race). That would require 3-4 more races to tease out (the third race would test the final two of the 12 remaining horses, against the 2 leaders of the first, and the leader of the second. If the leader of the second wins then a fourth race between the leader of the first and the second of the second).

But you need the top 3, not just the top 1.

Thanks Santosh.

You are making things up. They didn't say anything about how much time between races. Could be a week between races. That's like saying... Condition: The horses are well fed and remains constant in all races

Since b2 is for sure faster than b3, even if b3 was faster than c1, b3 would still not take 3rd because b2 is faster.

At first, I had the same doubt but later, I realized. On 3:40, we know that B3 is slower than at least 3 horses; A1, B1 and B2. Since only top 3 matters to us , we can eliminate it.

It doesn't matter. If B3 is faster than C1, than A1, B1, and B2 are all already faster than C1. B3 is never going to be in the top three, so it's never worth considering.

Even any fourth cant be eliminated.

@@FirojAnsari-xo4hh how so?

Thanks Manuel :)

Because if you want top 3 horses only hypothetically it can only be 3 from a. a1 a2 a3 can be fastest 3 theres no way a4 or a5 can be included in the top 3

@@kevinnijjar5238 i think you do not understand my comment. Obviously a4 and a5 are slower than a1,a2,a3. But that may not be same when compare with b1 or c1 or d1 etc...

Now here it was the matter of luck that houses of group B nd C got more chances to preforme but at the last we had to find the fastest 3.... so chill we ar giving nothing to houses who came 4th-5 th..... sooo don't worry their is no discrimination

@@kevinnijjar5238 yup ur right

@@smslca but it doesn't matter because if they are slower than a1,a2,a3 then obviously there are more than 2 faster horces than a4 and a5 and it also goes with b4,b5 c4,c5 d4,d5 e4,e5 . Because there are 3 horses faster than them means they cannot be in any condition among top3.

The premise is that all horses will always complete a race in the exact same time. So if B1 finishes one race ahead of C1, he will finish all races ahead of C1. There is no scenario in which C1 will finish ahead of B1

Speed of horses in each race is same...

@@musicmantra3606 yes that's the assumption we must keep. Thanks

but may be a2 is faster than b1

Real life and logic puzzles don't always intersect on a Venn diagram

If you answer like this..... Interviewer : get out from here. You : pls sir give me one more chance. Interviewer : in real life you can't get 2nd chance immediately... 😂😂😂😂

@@dwezz100 w

This is what people say when they don't get correct answer

Agree with you ❤️

Hey Tommy... as we know B4 cannot survive the competition so we simply eliminate it. But C1 has a chance to claim the 3rd position of overall competition. Also, we don't need to compare each horse with all other horses, rather we should keep on eliminating the horses which can't survive the competition. I hope that helps !

Thanks for the reply Logically, I actually opened up excel and put horse times on just to see where I was going wrong. I cant believe I didn't see this...Thank you for your response.

you're welcome 😊

I still don't get it. Maybe A4 and A5 were faster than all the horses from groups B, C, D and E. The solution given in the video only allow us to have an ESTIMATION. Reality might be different !! You will tell me: "As we know A4 and A5 cannot survive the competition so we simply eliminate them" However the ASKED question is: Find the 3 FASTEST horses without a chronometer. I might be finicky or trouble maker, but words have a meaning. For me, fastest means fastest. The only ACCURATE solution is to put in competition every horses 2 by 2. and find out which horse has never been defeated, which horse has been defeated only once and which horse as been defeated only twice.

@Greg, "Maybe A4 and A5 were faster than all the horses from groups B, C, D and E." Indeed, A4 and A5 may be faster than B1, C1, D1 and E1. However, if that's the case, then B1 and C1 will simply not finish in 1st or 2nd place in the seventh race; instead, A2 and A3 will win that race and prove to be the #2 and #3 fastest horse overall. There's no estimation, the method in the video is accurate. But there's simply no necessity to include A4 and A5 in race 7, because they wouldn't finish in 1st or 2nd place anyway.

When race 6 ends and its horse wins, the horses of race 5 represent A1, A2, A3

@@rubalsher well it doesnt after 6th race ends we'll simply make race between A1, B1, C1,D1,E1 and E1 will win. Boom E1 is fastest and you can do rest steps as shown in vid

What if A2 is faster than B1 ??

@kiran kumar, No, because that only works if the three fastest horses were all in different groups during the first five races. There's a 45.6% chance that that is not the case.

@@LOGICALLYYOURS what if all the horses of any group are running slower than the rank 5 horse of any other group? Then how and why did you eliminate several horses assuming incorrectly??

@@rochishmopidevi7617 The whole set running slower will be eliminated by the race of 5 winning horses. In fact, both sets that don't place at least 3rd there are holding only horses not faster than the top 3 sets.

​@@LOGICALLYYOURS Same want to ask you what if A4 and A5 where faster than B1 , B2 and C1?

Can u plz explain clearly?

which company?

Thanks Kanimozhi :)

@@LOGICALLYYOURS conduct 5 races to 5 groups and identify the first position of all 5 races..from the 5 horses( which got first position in five groups) conduct additional 6th race to identify first three positions

Thats whats race 6 is for. D1 and E1 are the fourth and fifth horses of the sixth race among winners of races 1-5, and thus eliminated

Real life and logic puzzles don't always intersect on a Venn diagram

Even if you change the order, you'll need to re-arrange the columns, which is just equivalent to the solution shown. It was only for the sake of simplicity by ensuring all combinations are taken care.

Hi Tan, I truly agree with what you just said.... but a Puzzle always aims to identify the tricky things hidden in it. Since the puzzle doesn't mention the tiredness, you can simply ignore it :)

I was told this years ago but it was 'mechanical horses that always run at the same speed'...which gets rid of the issues of real life horses. It was fun to work it out again :)

Would the following formulation make you feel more comfortable? Let S be totally ordered set of 25 elements. If by choosing any 5 element subset of S you could determine the order of its elements (inherited from S), what is the minimal amount of 5 element subsets you have to choose before you can determine the largest 3 elements of S?

I agree with Mr. Tan. The assumptions (horses run the same speed each race, track conditions do not affect horses performance, etc.) must be stated at the intro (when the no stopwatches condition was listed) otherwise, the answer doesn't create a logical solution.

That strategy doesn't work 100% of the time though.

Kindly respond for my question

@@karthiks5722 B3 cannot be among the three fastest horses because A1 B1and B2 are faster.

@@rwb966 Thanks

There is a limit to no of races. It should be minimum. But no cap over the time. So take as much time as needed to make them all fresh.

I just watched a video on this yesterday - same exact problem, the uploader stipulated mechanical horses, so fatigue not a factor.

Hi Archana... or is it the other way around :) a comparison of date published would be useful :)

Because there is a chance for A2 or A3 faster than B1 C1 and so on

A1 has currently won 2 races, the A group race and race 6. Since A1 won race 6, the 2nd and 3rd of the A group race has a chance to compete with the 2nd and 3rd of the 6th race (for cross-checking I think...) to determine which belong to the top 3. For B2 being able to compete in the 7th race, the horse is just there because it is within the range of a top 3 (3rd). idk if this is a plausable answer but i took the shot anyway.

You can't, but it does not matter. Even if A4 is faster than B1, it still cannot be in the top 3 - it will still be slower than A1, A2, and A3.

You don't have to

So happy to see my comment of before 1 year. This time video came in my recommendations, just one day after the interview at DE shaw. I was asked this same puzzle in the interview and I was able to ans correctly. 😇.

Tell us how.

Because there are (at least) three horses faster than B3, they are A1 B1 and B2.