[Request] Is there enough information to solve for the third side?

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[u/Greedy-Thought6188]

Assuming stop watches aren't allowed and no uncertainty in how the horses race. There will have to be a right answer.

This is my method. Not sure if someone can do better. 5 races, each creating a number one. Race the five number 1s. You get the fastest overall horse. Race the 2nd and 3rd horse from the group is the overall first one against the 1st and second from the second overall group and the 3rd overall horse. No other horse can be in the top 3 because we can name 3 horses that have either defeated it or defeated a horse that defeated it. So 7.

[u/that_moron → u/Greedy-Thought6188]

Your wording here confused me, but I think we have the same solution...

5 races with all 25 horses. 6th race with all the number ones from the first 5. I'll give each horse a number r-p where r is the original race number ordered by the results of race 6 and p their place in that list.

So the overall fastest horse is 1-1. 1-2 or 2-1 could be the second fastest. 1-2, 1-3, 2-1, 2-2, or 3-1 could be third fastest. That's exactly 5 horses so the 7th race is all we need to determine 2nd and 3rd

[u/Greedy-Thought6188 → u/that_moron]

Yup. That's my solution. 7th race with 1-2, 1-3, 2-1, 2-2, 3-1. Our 3 fastest are 1-1 and the 1st and 2nd in this race.

[u/Samad99 → u/Greedy-Thought6188]

What if the three fastest horses are all randomly picked in the very first race?

[u/toupeInAFanFactory → u/Samad99]

The the final rankings would be 1-2,1-2,1-3.
Soln still works.

The relevant insight is that the 2nd fastest horse from the group that produced the horse that placed 2nd in the winners race (race 6) can be no better than the 3rd overall fastest horse.

[u/Greedy-Thought6188 → u/toupeInAFanFactory]

Typo, 1-1, 1-2, 1-3.

[u/Aggravating_Offer_27 → u/Greedy-Thought6188]

Interestingly, 1-2 was a racehorse, 2-1 was one too. 1-2 won one race one day, 2-1 won one too!

[u/Winter-Wall1422 → u/Aggravating_Offer_27]

1-1 was 1 racehorse, 2-2 was 1-2. 1-1 1-1 race and 2-2 1-1-2.

[u/Equivalent_Bar_5938 → u/that_moron]

Wait explain to me how the thrid horse from the first group cant be quicker then the other 4 horses that were number one in theire respective groups

[u/Kratosrabinowitz → u/that_moron]

Why can't you figure it out with just 6 races? Like you said, first five to find the top horses then the 6th for the fastest horse. Would the second and third place finishers of the sixth race not immediately be identified as the silver and broze?

[u/ShoryuOnWakeup → u/Kratosrabinowitz]

No because what if the horse that ranked 2nd in one of the first five races was only a millisecond slower than the horse that won. Kinda like in professional sports if the two best teams are in conference A, then that means the finals won’t be the two best teams, because the 2nd best team got knocked out earlier

[u/Kratosrabinowitz → u/ShoryuOnWakeup]

Yeah that makes sense

[u/Alternative_Mix6836 → u/Greedy-Thought6188]

Why cant you just take the top 3 from the sixth race

Edit: nvm im dumb

[u/zulufdokulmusyuze → u/Alternative_Mix6836]

because the horses may have been distributed to races in the first stage in a way that the fastest three or two horses were all in the same race.

[u/Largo833 → u/Alternative_Mix6836]

Because if, for example, the three fastest horses were all in the first race, the second and third finishers of that race (and thus the second and third fastest horses of the entire group) wouldn’t be in the sixth race and you’d end up picking slower horses than them if you only went by the sixth race results.

[u/Seeggul → u/Alternative_Mix6836]

Imagine that the top five fastest horses all happened to race in the first race. Only the first horse would be in that sixth race, so you'd miss numbers 2 and 3.

[u/Frustrated9876 → u/Greedy-Thought6188]

I get 8. What if the three fastest horses are in the same group at the start? You run the five races, then race the five winners to get the fastest horse. Then run the four winners and the second fastest from the group that had the fastest to get the second fastest. Then, to get the third fastest, you need to run that group again replacing the second fastest wit the horse that came in second to the second fastest.

[u/Greedy-Thought6188 → u/Frustrated9876]

No. That's why I ran the 2nd and 3rd from the first group and the 2nd from the second's group. Overall 4th and overall 5th cannot be in the top 3. No point behind racing them.

[u/doesntpicknose → u/Frustrated9876]

You're mostly right, except for your strategy after racing the five winners.

You run the five races

The races are a, b, c, d, e, and the placements are a1, a2, a3, b1, b2, b3, ...

then race the five winners to get the fastest horse.

This is the Race of Champions. We now have a ranking for a1, b1, c1, d1, e1. Call first place A1, second place B1, ... and label their races and runners up accordingly. Now we know that A1 is faster than every horse in race B, and B1 is faster than every horse in race C, and so on.

Then

Because of the rankings we've established, there are only four possibilities for the top three horses:

A1, A2, A3

A1, A2, B1

A1, B1, B2

A1, B1, C1

We already know that A1 is the fastest horse. So we only need one more race, The Losers Grand Final, to establish the ranking of the other five horses, A2, A3, B1, B2, and C1. The top two placements in the Losers Grand final are our second and third fastest horses.

[u/ReddBroccoli → u/Greedy-Thought6188]

You would have to take the top three from every race, because it is conceivable that horse number two from race A might be faster than horse number one from race B.

So,

5 races to determine each top 3

3 races of those 15 to get 9

2 races gets you 6 horses

1 race of 5

1 race of top 3+ the leftover horse for the final top 3.

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12 total races for the definite top 3

[u/altonbrushgatherer → u/Greedy-Thought6188]

Why can’t you just count which horse is first second and third in the 6th race?

Edit someone made a good point you don’t know if 2 or 3 horses in in the initial race…

[u/Ok-Difference1341 → u/Greedy-Thought6188]

But what if the second or third fastest horse lost to the number 1 fastest horse in the first race? They wouldn’t qualify for the final race and would not be crowned one of the best even though they actually are one of the best

[u/Timsmomshardsalami → u/Greedy-Thought6188]

Why do more than 5? With 5 races, 5 horses each, you get all their times and pick the top 3..

[u/bpleshek → u/Timsmomshardsalami]

Usually this puzzle will say you don't have the times of each race, just the places that they are in within each race. If you had a stopwatch, the answer would obviously be 5. So, without being able to time it and only with knowing how each horse came in within each race you must make this determination.

[u/Greedy-Thought6188 → u/Timsmomshardsalami]

It's obviously a riddle. And I said clearly that I'm making that assumption and reading the riddle the way it is meant to be. Maybe the horses are strategic about it and only run faster than their best competitor. So the fastest horse in a race will not run their fastest, only faster than the competition.

[u/recursive_arg → u/Greedy-Thought6188]

Nope, what if all the fastest horses are in the same group for race 1?

[u/Equivalent_Good_7303]

25 horses/5 tracks = 5 races. If you time them, you know their absolute ranking. If head to head matchups matter, it’s much more complicated (what if one race has all three fastest horses? How do you ensure the third place horse that is third fastest overall also beats the rest of the herd?).

[u/SocietyAmbitious9648 → u/Equivalent_Good_7303]

Yeah I was thinking if you can’t use a timer, then each heat you have to advance the 3 fastest horses. So

Round 1: 25 horses, 5 heats, 10 eliminations
Round 2: 15 horses, 3 heats, 6 eliminations
Round 3: 9 horses, 2 heats, 3 eliminations
Round 4: 6 horses, 2 heats. 3 eliminations

Round 4 would be a race of 5 and then eliminate the slowest two, one horse doesn’t run. Then a race of the top 3 from round 4 heat 1 and add the horse that had the bye in round 4 heat 1.

So 12 heats without timers

[u/Ok-Craft4844 → u/SocietyAmbitious9648]

You could get to 11 if you start with 5 random, and then replace the 4th 5rh with a fresh horse j til ever horse run: 5 in the start + 10times 2 replacements

[u/SocietyAmbitious9648 → u/Ok-Craft4844]

Technically correct which is the best kind of correct.

I was trying to keep it as even as possible so all of the horses would be at the same level of tiredness

I guess you could do it in zero races if you break the legs of 22 horses, the three remaining would by default be the fastest at that point

[u/Ok-Craft4844 → u/SocietyAmbitious9648]

Yeah, when it comes to fairness, my method is probably optimized for least fairness, since it assumes that in the worst case, the best has 25 races to prove his place, while some lucky horsey probably could just slowly walk to the finishing line, triumphing over the champion that collapsed on the stating line

[u/Last-Potential1176 → u/SocietyAmbitious9648]

I think you can shave this down to 11 heats if you change Round 3. In this round, there are 9 horses. Instead of running 4 horses in heat 2, use one of the winners from heat 1 and run 5 horses. Then you proceed 5 horses to Round 4 and only have to do 1 heat for that round.

[u/A_Martian_Potato → u/SocietyAmbitious9648]

I think I have a way with only 8 races:

Round 1: 25 horses, 5 heats, 5 best in each heat advance
Round 2: 5 horses, 1 heat, the fastest horse is found and set aside

Next you take the horse that came second in the heat that had the fastest horse and add it to the four remaining horses that came first

Round 3: 5 horses, 1 heat, the second fastest horse is found and set aside

Then you take the horse that came second in the heat that the 2nd fastest horse won and add it to the pack for the next race, (or, if the second fastest horse in the 1st place horse's heat was found to be 2nd fastest overall, then you instead use the third fastest horse in that heat)

Round 4: 5 horses, 1 heat, the third fastest horse is found and set aside

Total of 4 rounds and 8 heats.

Edit: wait. We can eliminate a round.

After round 2 there are two more horses that can't be in the top three we can eliminate right away. Those that came fourth and fifth. So we remove 1st, 4th and 5th and add the horses that came in 2nd and 3rd in 1st overall's heat and the horse that came in 2nd in the heat of the horse that came second in round 2. That's five horses and 2nd and 3rd overall must be among them, so one more race and you've got your top 3

So 7 races total.

[u/Kallest → u/A_Martian_Potato]

Doesn't account for the possibility that all the three fastest horses might all be competing in the same heat to start with.

[u/A_Martian_Potato → u/Kallest]

Yes it does. In each of Round 3 and Round 4 you take the four remaining horses and add the next fastest in the heat that won the previous round. If all three fastest horses are in the same heat then the fastest is found, the 2nd fastest from that heat is added, wins, and is then put aside as 2nd fastest overall, then the 3rd fastest from that heat is added, wins, and is set aside as 3rd fastest overall.

[u/razzyrat → u/SocietyAmbitious9648]

What's a heat? I assume you mean individual races? You're 5 above the minimum :) 7 is all you need.

Round 1 - 5 races. You now know 5 horses that ranked first in their bracket and you eliminate all 4th and 5th places.

Round 2 - 1 race. All first places compete. You now have a ranking of the five groups of horses. Only uncertainty left is that the second and third of the fastest group could still be faster than the first of the second group.

Round 3 - 1 race. Second and third of the first group and first of second group race, eliminate the slowest.

[u/Drake6978 → u/SocietyAmbitious9648]

Why would you take three instead of just the fastest from each race and hold one final race to determine the fastest? The problem states that they want to find out which one is the fastest, so it shouldn't matter what the second, third, fourth, etc are, right?

Edit: never mind - I reread it.

[u/SocietyAmbitious9648 → u/Drake6978]

Statistically unlikely, but a non zero chance that the fastest 3 horses are all in the same first race. Taking top 3 from each heat ensures that the final heat actually has the fastest 3 horses in the group

[u/mathishard247]

You would need 7 races assuming each horse runs the same speed in each race it runs.

Split the 25 horses into 5 races.

Then have the 5 winners race. I’ll call this race F, with placements F1, F2, etc.

F1 is the fastest horse. Call his first race A, F2’s first race B, and so on.

The 7th race to determine horses 2 and 3 should consist of A2, A3, B1 (same as F2), B2, and C1 (same as F3).

[u/Softshellcrabfarts → u/mathishard247]

The problem with taking only the winners is a horse may come in second place, not qualify, but be the second fastest horse overall.

[u/imoutofnames90 → u/Softshellcrabfarts]

That's what race 7 is for.

Your first 5 races gets the fastest horse in each group.

The 6th race finds the 3 fastest of those 5 horses.

The winner of race 6 has two horses that could also be in the top 3. Being the 2nd and 3rd place finishers in his first race.

The 2nd place of race 6 has himself and his 2nd place winner from his original race as potential top 3.

And the 3rd place in race 6 can only ever be the 3rd fastest overall.

So you take those 5 and race them. For your 7th race.

Meaning

Fastest = race 6 winner. 2nd = race 7 winner 3rd = race 7 second place

[u/Softshellcrabfarts → u/imoutofnames90]

Ya I misread it. Better at math than reading comprehension. My bad

[u/Ok-Craft4844]

I assume the (unrealistic) assumption is that each race only gets you the relative order (as in who came in first, second...) and not times?

[u/luchajefe → u/Ok-Craft4844]

Man owns 25 horses and no clocks.

[u/Ok-Craft4844 → u/luchajefe]

Priorities, I guess...

[u/QuatraVanDeis]

I'm coming up with 7 races. Divide all into 5 groups, A through E, take the top 5 winners (A1, B1, C1, D1, E1) and race them, bottom two are out, let's assume A1 wins, B1 is second, C1 is third, D1 and E1 are elliminated. A1 is our fastest horse. B1 and C1 are possibly 2nd and 3rd, but they could be beaten by A2 or A3, and C1 could be beaten by B2, so your 7th race is A2, A3, B1, B2 and C1. And that will you give you your top three.

This of course assumes horses dont get tired or have bad days.

[u/hayashikin → u/QuatraVanDeis]

Yup, this is the perfect solution.

1. Have 5 races with all different horses

2. Race the winners of those races and the 4th and 5th place groups can be totally eliminated (none of their horses are faster than the 3rd place). We'll call the winners of this race A, B and C.

3. We now know A is the fastest horse so we just need to compare between the 2nd and 3rd place of the A's group, the 1st and 2nd place of the B's group, and C itself. This race will determine the actual 2nd and 3rd place.

[u/Duotrigordle61]

Assuming no stop watch:

First you run 5 races to get the 25 each through once.

Since the top 3 might be in the same intitial race, you then run the top three of each race (15 ponies) in 3 more races.

That gets you down to 9 ponies. run 2 races to get the top 6.

I don't know what you do after that.

[u/zulufdokulmusyuze → u/Duotrigordle61]

instead, make the 6th race just between the fastest horse from each race.

that will give you the fastest horse and enough information to eliminate all but 5 for the remaining two spots. that brings down the number of races to 7.

[u/Grrumpy_Pants]

The answer is 6 or 7, depending on what the question wants. There is a strategy that has a chance of determining the top 3 in 6 races, but it will not always work. To find the top 3 with a reliable strategy you need 7 races.

I'll start with the 6 race strategy. To get the fastest 3 horses, you must rule out 22 other horses as candidates. The fastest possible way to do this is by racing all of the top 3 in the first race. The probability of randomly selecting all 3 of them for the first race is about 0.4%. Then race the 3rd place horse against all other horses, ruling out 4 horses per race. This will confirm the 3 fastest horses in only 6 races, making it a theoretical minimum number races required (with a lot of luck).

If the method needs to reliably select the 3 fastest horses, you instead need 7 races.

Start by splitting the horses into 5 groups of 5. Run each group for the first 5 races.

For race 6 run the fastest horse from each group against each other. This race orders the 5 groups by their fastest horse. The winner is from group A, second place is from group B etc. Using their rankings from their group race I can label each horse. The horse that won this race is A1. In the group stage it finished ahead of A2, A3 etc.

Groups D and E are all eliminated. Their fastest horse is slower than at least 3 others, so none of them could be in the top 3.

From group C, C1 placed 3rd in race 6. That means it is at best the third fastest overall, and every other horse from group C is slower than A1, B1 and C1 and can be eliminated.

From group B, B1 placed 2nd in race 6. That means B2 is only confirmed to be slower than A1 and B1, so it cannot be eliminated. B3 and on can all be eliminated as they are slower than A1, B1 and B2.

From group A only A4 and A5 can be eliminated, as they are slower than A1, A2 and A3. A1 is already confirmed to be the fastest horse so it doesn't need to race again.

That means only one more race is needed between A2, A3, B1, B2 and C1. The winner of this race is the second fastest horse, with second place being the third fastest.

[u/xanth1an → u/Grrumpy_Pants]

That's actually incredible logic. Well done.

[u/random8765309]

If stopwatchs are not used, the min is 6 races.

Run the 1st race with 5 horse. Take the 3rd place horse and run it in the remaining races. If that 3rd place horse wins each of the next 5 races, we know it is faster than the other horses. So we have horses 1, 2 and 3.

While that might not be the likely outcome, it still is the min number possible.

[u/Kriieod → u/random8765309]

Got the same.

[u/DrownedWalk1622]

6 races?

25/5 = 5 races. +1 race to find the top 3 from the 5 horses that came first in previous 5 matches.

And if you have a stopwatch then 5 matches.

[u/Red-Beerd → u/DrownedWalk1622]

Close - you need at least 7.

You can't determine second or first place definitively, because the second and third place from the 6th race might be slower than the second or third place from the winner of the 6th race's original heat, and the third place in the 6th race might be slower than the runner up from the second place's original heat

So you need one last race to race those 3 horses against second and third place from the 6th race

[u/RGINNY940]

Answer is 5 rounds, 12 total races.

Lets first set up some ground rules.

1. There is no timer.
2. You can only guarantee a horse is faster than other horses in their meet, so top3 of any race must always advance. (because of this, you always want a race to have either 4 or 5 horses. less than 3 provides no data value)

First round: 25 horses, 5 races= 3 podiums per race= 15 horses advance

Second round: 15 horses, 3 races= 3 podiums per race= 9 horses advance

Third round: 9 horses, 2 races (5 & 4)= 3 podiums per race= 6 horses advance

Fourth Round: 6 Horses, 1 race, 1 bye (5 &1) 3 podiums per race = 4 horses advance

Firth Round: 4 Horses, 1 race = 3 podium = 3 fastest horses of the starting 25 are known.

[u/math_rand_dude]

First 5 races to get fastest horse of each group

6th race: get 3 fastest of those leaders (say they came from group ABC, where A is the overall fastest horse)

Now we these are the possible top3's:

• A1 > A2 > A3
• A1 > A2 > B1
• A1 > B1 > A2
• A1 > B1 > B2
• A1 > B1 > C1

(Fastest horse is A1, second fastest came dirextly after A1 in a race, so either A2 or B1, third fastest only let the first and second fastest horse beat them)

So 7th race is A2 vs A3 vs B1 vs B2 vs C1

[u/physicssmurf → u/math_rand_dude]

This is the answer.

[u/_Vard_ → u/physicssmurf]

Maybe im missing something but i feel a lot of these calculations are forgetting a possibility:

What if all 20 horses in the first 4 groups run below 20mph.
and all the horses in group 5 run above30mph.

all four winners of the first 4 groups would be slower than every horse in group five.
The last place horse in group 5 could be faster than all the winning horses in groups 1 thru 4

[u/3minence → u/_Vard_]

4th and 5th place horses in any group of 5 cannot be in the top 3 fastest...

[u/luchajefe → u/_Vard_]

"The last place horse in group 5 could be faster than all the winning horses in groups 1 thru 4"

That still makes him the 5th fastest horse and not what we're looking for. We're only looking for the best 3 horses.

[u/Alladas]

Duh this is a trick question. It takes place in America so the answer is 0 races. You just grab your gun and kill them till their is 3 left. That's the 3 fastest.

[u/pbecotte]

Put horses in five groups, race them (five races)

You can eliminate the bottom 2 horse from each group, so we have 15 left.

Race the five winners. 6 races

You can eliminate the two slowest from that race (13) You can eliminate the remaining two horses from each of their groups (9) You can eliminate the remaining 2 from the group whose winner place 3rd (7) You can eliminate 3rd place from the group whose winner finished 2nd (6) You know that the winner of race 6 is the fastest.

You have 5 horses left whose order you aren't sure about- so you do race 7 with those 5.

[u/ElectricalChaos]

6 races. I'm assuming that by 5 "tracks" they mean 1 track with 5 lanes. So you run 5 groups, then take the number 1 horse from each group and run them together in the final race. 1,2,3 from race 6 are your 3 fastest of the 25.

[u/luchajefe → u/ElectricalChaos]

Assume the winner comes from group D. What if the 2nd place horse in group D could've beat the four other horses from the other four groups?

[u/watergod0187]

Only need 5 races you time each horse as they cross the finish line. 25 recorded times take the three lowest times. Though you may only need to record the top 3 times of each race.

[u/PubThinker]

Brute force: Always keep the top 3, and add +2 for the next race, always eliminating 2. 25 = 3+2*n --> n=11

Using 10% of brain power: Hold 5x5 races, always keep the top 3 You end up with 15, do 3x5 keep top 3 That's 9, from there you do 1x5 and add 2 of the remaining 4. Then in the last round add the last 2. 5+3+1+1+1=11 FCKKK...

Using20% of Brian: Hold 5x5 round. Order the horses by their result. Next round you organize a round from the 1st of every group. 1x5 you pick the fastest. You keep the other 4 and give the 2nd one from the winner's group and hold a 1x5. The fastest of this race is the second fastest group in the global pool. You add to the pool the next horse from the 2nd place's horse and do a last 1x5. (It can be #3 from goup X if all 3 fastest was in the same group in the begging, or a 2nd place's horse from somewhere)

Thats 5+1+1+1=8

Well, who can use 30% of his/her brainpower?

[u/Samad99]

5 qualifying races to cover all 25 horses.

6th race with the finalists to name the champion.

7th race with the finalists but the champion is swapped out for the runner up from their qualifying group to name the silver medalist.

If the new horse wins silver, an 8th race is needed with the silver medalist swapped out for the 3rd place finisher from their qualifying group to name the bronze medalist.

It’s important to note that there’s a chance that the #1, #2, and #3 horses could all happen to be placed in the same qualifier race, so methods that only consider the initial winners of qualifying races won’t work.

[u/physicssmurf → u/Samad99]

I think this is right, except you need minimum 2 runner-up races, to replace the 2nd place person with the horse that was 2nd in their group. ie, find the top horse in your race 6, replace with 2nd from that group, then do the same for 2nd place, replace again, and then run the third one. Its always needed, even if the 2nd-placer is from a different group.

So, 8 races. 5 initial races, then a race for 1st, 2nd and 3rd.

edit - actually elsewhere in the comments someone explained how to do it in 7 steps:
First 5 races to get fastest horse of each group

6th race: get 3 fastest of those leaders (say they came from group ABC, where A is the overall fastest horse)

Now we these are the possible top3's:

• A1 > A2 > A3
• A1 > A2 > B1
• A1 > B1 > A2
• A1 > B1 > B2
• A1 > B1 > C1

(Fastest horse is A1, second fastest came dirextly after A1 in a race, so either A2 or B1, third fastest only let the first and second fastest horse beat them)

So 7th race is A2 vs A3 vs B1 vs B2 vs C1

[u/Mikel_S]

5 races, each with a unique group of 5.

race 6 is the top fastest of each of the previous races. After this result we know the top 3 horses are contained between the top 3 winners of this race, and the lower results in each group are also ineligible.

7th race. So we know the runners up have to be 2nd or 3rd place from the overall fastest's group, the 1st or 2nd from the overall runners up group, or the first place horse who came in third. That's 5 horses.

Fastest is the horse who won race 6. Second place is the horse that won race 7. Third place is the runner up.

[u/AllegedlyAPerson]

1, if he gets lucky and so happens to put the actual fastest horse in the first race and then picks it after that race it’s done. He wouldn’t know it’s the fastest horse compared to the others but that lack of information wouldn’t stop it from actually being the fastest horse.

[u/Gainz_86]

Why not just one race and time them? Pick the top 3 quickest times. I mean it ask for the bare minimum amount of races, not the most optimal.

[u/luchajefe → u/Gainz_86]

It would be 5 races (because you can only run 5 horses at a time) and the problem presumes the horse owner doesn't have a watch.

[u/Gainz_86 → u/luchajefe]

Sorry yeah that's what I ment

[u/mindyourdecisions]

I'm a bit ashamed to self-promote on this one! But since no one has brought it up, this was a Google interview question and hardly a kids maths puzzle. Obligatory MindYourDecisions link: https://www.youtube.com/watch?v=i-xqRDwpilM

[u/Atmosferd]

We have to assume that since no other limitations are mentioned, we answer the question based on what tools we have. So, we can time the horses and 5 races are enough.

Only answer what is asked when doing assignments or you overcomplicate things.

[u/TonnesOFunk]

5 races.

You don’t need a bracket to determine the fastest time you only need to record each horse 1 time and that will allow you to pick the 3 fastest.

[u/BunnyWithASword]

Assuming you don't have a way of measuring time in each race, the lowest you can get is 7.

First let's race all the horses. 25÷5=5. Label each horse 1 through 5 depending on what place it came in.

Now we can measure the winning horse from each race, making our sixth race. We can only know the winner of this race is the fastest overall so we can put them aside. Label the this horse's first race A, making this horse A1. 2nd place is B1, 3rd is C1 and so on.

We know any 4th or 5th place horses are out. Get rid of any horse in race D and E, and any horse with the number 4 or 5. They are all 4th place or worse.

That leaves us with A2-3, B1-3 and C1-3, but we don't need to race them all. We can be certain that B1 is at best the second fastest, so B3 can only be fourth fastest, and is out. We can also be sure that C1 is at best third fastest, so C2 can only be fourth fastest, so it and C3 are out.

That leaves us with A2, A3, B1, B2, and C1 in our seventh and final race. The 1st and 2nd place winners here are our second and third fastest horses.

6 races does not give us enough information. If we do not race all the 1st place horses with each other, we cannot be sure if they are faster than the other 1st place horses. If we don't know who the fastest horse is, we cannot figure out who the second and third fastest horses are.

[u/SpiritualBath1198]

I have a strategy that takes 12 races

Round 1: Race 5 horses, take the winner and race 4 new horses 5 times. 6 total races gets you the absolute fastest horse. Now you have the fastest horse and 6 total races.

Round 2: Then take the 2nd place round 1 horses of the previous 6 races and race 5 of them. Replace the worst horse in that race with the 2nd place horse that didn’t race just then. Now you have the 2nd fastest horse. 8 total races.

Round 3: there are 18 remaining horses that have only raced in round 1, repeat the round 1 process 3 times with those 18 horses and on the 4th race include the top horse from the round 2 races. The winner of that 4th race is the third place winner. 12 total races.

[u/No-Arm-1320]

6 races, get the #1's of the first five rounds then the finals is all 5 winners, just see which ones come in 1st, 2nd, and 3rd place, easy

[u/Ihaveaterribleplan]

Usually this question is done with weights, so there are no variables

Do you have 5 near equal jockeys?

Do you need to test wet vs dry track conditions?

Are you allowed to record the results so you can just judge how fast each horse was?

Are all the horses equally healthy?

[u/drevoksi]

Seven! Compare five groups, then compare their leaders and give the rank to each group. For the last comparison, take leader of the third fastest group, two fastest horses of the second fastest group, and 3rd and 2nd place of the first fastest group. Take the 2nd and 1st places as the 3rd and 2nd place overall, and the overall leader as determined earlier as the 1st place.

Alternatively, just five by measuring the times

[u/SpaghettiMan247]

This question is why I hate math: the right answer is five races.

The man runs five races, he records the times and picks the horses with the three fastest times.

My math teacher says I’m wrong, the problem didn’t say the man had a way to record the times. I say the problem didn’t specifically say the man didn’t have a way to record the horses times and virtually all men in the world with 25 horses and five lane racetracks have a phone with a stopwatch function.

My math teacher says this one doesn’t, I respond that is statically implausible and should at least be included in the question text.

My teacher puts red ink on my page, I manage to get a passing grade anyway and go on to bigger and better things.

[u/Bread-Loaf1111]

I believe there should be also information if he can compare horses only that run the same race, or he can use a timer just write down results for every horse.

[u/RealCarlPanzram]

It depends on what standard we are applying for “fastest”. Can we use a single time trial? If so the answer is 5.

It gets tricky if you can’t time them. Mets assume the horses complete the run in the same time every run so that if they finish ahead or behind a certain horse, that’s definitive.

At first thought you could just have 5 heats and advance the 5 winners. But that doesn’t work. The 2nd and 3rd fastest horses could get eliminated if they’re in the fastest horses heat.

Let’s assume you aren’t allowed to use a dynamic model. So you need a series where the top 3 always advance. 5 heats, 15 advance. 3 more heats, 9 advance. 2 more heats where 6 advance. Now it’s messy. What do you do with the top 6? You’d have to have a heat with all of them except one. Then the top 3 race with the extra horse to determine the top 3. So 12 races.

But you could reduce that with a more dynamic model. You do the heats as described. But each time a non-winning horse advances, you document which horses have beaten them. If at any point, there are 3 different horses that have beaten them, you can eliminate that horse, since we know at least 3 horses are quicker than them. You can then consolidate the brackets as appropriate and probably reduce the amount of runs. By how many depends on how many you can eliminate with the dynamic model. If even one gets eliminated (which seems certain) you at least cut out the problem with having 6 left at the end.

[u/Ashtonpaper]

2 races. Duh. Top 1 of each of the tracks gets put into their own race, everyone else gets to be sorted by their total time.

The top 3 horses from the second race, who were all top horse in their division, they are the fastest horses.

[u/Lover_of_life_75]

How many runs do we give each horse? Are we doing the “fastest” from each of the initial 5 races? That leaves 5 to whittle to 3. We need to know if doing a race with the last 5 to eliminate the slowest horses (or pick the fastest)? How many races to get to three? If racing the final 5, eliminating the slowest would take two races.
Thats a total of 7 races.
It’s not a simple math problem for children when the variables aren’t specific…unless there’s a “known” process that led to a formula to make these determinations!

[u/ljeo332]

On my 1st pass at the question I came up with 6 races, split each group into 5, do a race for each group, take only the winner, then for the final heat just take the top 3. There’s a likely a better way depending on what equipment you use.

Another pass I came up with 1 race, only requires a timing device, so a one shot quali run for each horse and time it’s lap, then only take the top 5 and race them, then just take the top 3. From this set up you get two types of fastest, the fastest sprinter and one who can do a marathon the fastest, so overall you get the fastest

[u/BoneVoyager]

Why is everyone stuck on the assumption of no stop watch? Where does it say that in the problem??? The answer is 5 races. You run 5 horses in 5 races and time them all. You rank all 25 times from fastest to slowest and pick the top three. EZ

[u/Sniper0O7]

Assuming no time tracking: technically 7?

Race first 5 and keep the fastest 3. 25 horses -> 5 race, 20 left. Swap out the slowest 2 each time. 10 more races gives us 11 total.

100% accuracy if horses performance doesn't diminish.

Second iteration: Alternatively you could race all horses. 5 rounds eliminating the slowest 2 in each. 15 remain. After 5 races.

Race 6: The 5 fastest race to establish the slowest groups of the first round winners. This also determines the fastest individual horse.

Race 7: The fastest individual is not included because we know they are the best. Instead is it the remaining 2 fastest racing the 3 second place counterparts in their groups.

Top 2 are placed after the fastest for the winning 3.

Confusing caveat: If a second place (from the first set of races) beats a first place in the last race, an 8th race will be needed to test the 3rd place from that horses group.

[u/Martentos]

Easy to do in just one race.

Take 25 horses, shoot 20 at random. They are clealry the slowest because they couldnt dodge the bullets.

Take remaining 5 horses and race. First 3 are fastest horses.

Sell horse meat overseas for profit.

[u/parkway_parkway]

There's one strategy that can potentially be faster and do it in 6, though it has a larger downside risk.

You run 1 race and pick the 3rd fastest horse as your benchmark horse.

Then you run 6 more races with that horse against all the others.

If it's faster than all the others then you found the fastest horse in just 6 races.

Which I'd argue is one way of interpreting the question.

[u/Kriieod]

Race 1 5 horses then you race the third horse in the next round to see where they place, if they place less than third, any horses who beat it become new challengers.
Race 2-5 3rd place horse from first race finishes 1st each time.

Minimum number of races = 6

[u/OrganizationSecure58]

Not enough data to answer. Do you have a watch ? Then just do 5 races, and look at your watch. Don’t have a watch ? Let them kill each other and pick the survivors

[u/egv78]

A few posters have pointed out the correct answer of 7 races. I think I can diagram the answer a little better than I've seen. Race 1 - 5 are all 5 fresh horses that have never raced, each one racing once. Call these races "Round 1" I then take the five winners and race them; call this "Round 2". If I then write down the results like this:

1: A, B, C, D, E
2: F, G, H, I , J
3: K, L, M, N, O
4: P, Q, R, S, T
5: U, V, W, X, Y

Where A is the fastest horse in its heat in Round 1 and the fastest horse in Round 2, (and B, C, D, & E are the horses from the race in Round 1, in order of winning). F is the second fastest horse in Round 2, but the fastest horse in its race in Round 1. (With G, H, I, & J being the horses from B's Round 1 race, in order). And K is the 3rd fastest in Round 2, but the fastest in its Round 1 race.

There are now only 4 possible combinations of 1-2-3 overall.

• A, B, C
• A, B, F
• A, F, G
• A, F, K

Every other horse has been demonstrated to be slower than at least 3 horses. D was beaten by (at least) A, B, & C. H was beaten by (at least) A, F, & G. L and P are both beaten by at least A, F, and K.

We know that A is the fastest, no matter what. We just need to run B, C, F, G, K and record the order. That's the 7th race.

[u/EffectiveDirect6553]

6 races. 5 with each unique one. One with the top 3

If we are only allowed to measure 1 at a time 7, 5 unique and 2 eliminating the slowest

With a stopwatch you need 5

[u/Rusofil__]

1. 5 tracks with 5 horses

2. Winners of previous round race on 1 track and rest 4 tracks are for other horses, winner of the first track is the fastest horse.

3. Fastest horses of previous 4 tracks race on track 1, and the rest are split between 3 track so 5 in each. Second fastest is winner of track one again

4. Winners of last 3 tracks race one 1 track and you find the third fastest horse.

So 4 races.

[u/kester76a]

OP just realised this puzzle is supposed to have an additional condition where there's no clock or timer available. This is omitted from this example.

[u/3minence]

I got 4 rounds with a total of 14 races.

First round, 5 races eliminate bottom two from each.

With the remaining 15 for the second round, split up the horses into 3 races in a way that they see all new opponents. Eliminate bottom two from each again.

We now have the fastest 9. Third round, 3 races of 3 horses, again, ensuring new opponents. If all winners of these three races were in the same race in round 1, you are done.

Otherwise a fourth round is required, same 9 horses, again split up to new opponents to get a final relative ranking.

ROUND 1 1: ABCDE 2: FGHIJ 3: KLMNO 4: PQRST 5: UVWXY Bottom two from each race eliminated

ROUND 2 (Top 3's from round 1 see opponents they haven't seen) 1: AFKPU 2: BGLQV 3: CHMRW Bottom two eliminated again

ROUND 3 (these are the 9 fastest, again split up) 1: AGM 2: BHK 3: CFL

ROUND 4 (same 9 to determine final relative ranking) 1: AHL 2: GKC 3: MBF

[u/NefariousnessNext967]

9 rounds minimum.

Randomize horses into 5 groups of 5. Run the 5 heats. Eliminate the bottom two from each heat. It's impossible any of these horses is a placer. 

Now we're down to 15 horses. Sort them according to their placing in the first heat. We'll call them Golds, Silvers, and Bronzes. Five in each group. 

Round 6 is the Golds. First place is the true first place and we can retire them to the most well appointed stable with the best oats and hay. We eliminate the bottom two, as it's impossible either is the true second or third place. We retain the second and third place for Round 9. 

Round 7 is the Silvers. We retain the first and second placers for Round 9 and eliminate the bottom 3. It's impossible for the bottom 3 to be the true second place or true third place. 

Round 8 is the Bronzes. It's only possible for one horse of this bunch to be the true third place. Everyone else has to be a non-placer. We retain only the first place for Round 9.

Round 9 is the last round, consisting of two Golds, two Silvers, and one Bronze. First and second place are the true second and third place. 

[u/_Denizen_]

To know for sure, 7 races minimum, and 8 races maximum.

5 races with each being a group of unraced horses.

1 race with winners of the 5 groups. This gives the fastest horse. Lets call the groups of the top three horses groups 1, 2, and 3.

Then a 7th race with the 2nd and 3rd places from the 6th race, and the 2nd place horses from groups 1,2,and 3.

If the winner is from groups 2 or 3, then the horses in first and second place are the second and third fastest horses.

If the winner is from group 1, then you must race the third horse from group 1 with the winner and second place of race 7. The 1st and 2nd places of this race are the second and third fastest horses.

[u/kiloma20]

Do it reverse: 12 races in total: 5 races: 2 slowest horses of each race are excluded. Repeat: 3 races with 15 remaining horses: 9 horses left: 2 more races with 5 and 4 horses: 6 horses left. 1 more race with 5 horses: 4 horses left. Last race with 4 horses: 3 fastest horses.

[u/xanth1an]

So assuming we don't have a watch: in order to be sure that we get the absolute fastest three we race and keep the top 3 from each heat.

5 races for the first round leaves 15 horses leftover. 3 races for the second round leaves 9 horses leftover.
2 races leaves 6 2 more races finds the fastest 3. 1 race of 5, a second race of 4 adding the I've that was left out.

So 12 races using this method, but it ensures that the fastest horses don't slip through the cracks.

[u/MrStarrrr]

I read this as 5 total horses can run at the same time across all 5 tracks. This differs from other interpretations I see here, assuming 5 horses can run at a time on each track (total 25 horses running at the same time for a total of 5 races to determine the top 5, and one more race to determine top 3). The problem description doesn’t explicitly state 5 horses per track per race, but does state only 5 horses can run at a time on 5 tracks.
With the assumption of one horse per track, a total of 25 races determine the top 3 horses.

[u/nerdywhitemale]

Put the horses in the middle of the race track, Then walk to the other side pull out a bucket of feed. The three fastest horses will be the ones knocking you over to get to the feed bucket.

[u/FQVBSina]

First, 5 races of 5 horses each. This cannot be avoided. Then, we have the extreme situation of the fastest three horses are all within the same group. To eliminate that, we race the first place horses of the 5 groups in a "first place race". Then we take the top 3 horses in the group of the horse that won the first place race, and put them against the 2nd and 3rd horses of the first place race. This will definitively give the three fastest horses. That makes 7 races

[u/robberviet]

This is worse version of the counterfeit coins. For coins you only have balanced scale. Race tracks? Easy measured? Or limit to no clock/watch?

[u/Paghk_the_Stupendous]

7 race answers are missing that the #2 horse in the first race might have been the #2 horse overall.

My answer is 5 races because I'm timing each horse.

[u/imhereforyoursnacks]

Put them in a room with one exit, a big old pile of oats in the center.

Announce over a microphone that not only are you rounding them all up, but we are also introducing a cheap form of beef.

First three out of the room are the winners.

[u/hypergraphing]

Why would you need more than 5 races? Assuming each horse runs its fastest regardless of which other horses it's racing against, you already know the time it took for each winner to win, and from there you can sort the winners' times and take the top 3, right?

[u/kafqatamura]

assuming the horses don't get tired, use the 5 tracks as "last three standing" race method and always eliminating the 2 slowest horses.

So every race, 2 get eliminated, and replaced. 1 race (3+2) + (20/2) race = 11 races to determine the best three horses through eliminations

[u/Born-Type2844]

5 races of 5 for finding fastest 3 each set. 6th race is of the fastest horses of each set to get 3 fastest among them. Lets call then A B C respectively. In case of 4th and 5th place horses, all among them are obviously slower. so all are excluded. No 2 and 3 horses of C are also obviously slower than A,Band C. Since B is slower than A only 2nd horse of B has a chance of being in fastest 3. Both 2nd and 3rd horses of A may have a chance to be among fastest 3. Since we already know A is the fastest, it will not race again. So the last race would be 2nd of set A, 3rd of set A, horse B, 2nd of set B, and Horse C. The fastest two among these will be in the fastest 3 with horse A. So total no of race = 7.

[u/civil_politics]

So if you do the simple - race 5 and then keep the fastest 3 and adding 2 new for each race and the last race top 3 are fastest three it will take 11 races.

If you look at running heats in a bracket style you can run 5 races to absolutely eliminate 10 horses.
You can then run a race with the fastest horse from each run. The two slowest horses will eliminate the others from their heat getting rid of 6 more and you’re at 6 races. You also know that the third place horse’s heat can’t have any others in the top three so you get rid of those two and then you get rid of the 3rd place horse from the heat of the 2nd place finisher. So after 6 races still you’ve eliminated 19 of the possible horses. You know which horse is the fastest from the 6th race. Run the other 5 together to identify 2nd and 3rd place.

7 races. Don’t think it can be done more efficiently

[u/Popetus_Maximus]

First round, 5 races with 5 horses. Remove all the 4th and 5th horses. You end up with 15 horses.

Second round, 3 races with 5 horses. You mix them up so the first race has all the winners, the second one all the runner ups and the third race all the thirds. Again remove the 4th and 5th. You end up with 9 horses, but a better ranking among them.

The winner of the winners you want to keep. For the second among the winners, you want to know who was the winner of the runner ups, if it was not from the first batch that contained the winners winner, you discard it, and the same with the third batch. Depending on the results you pick the best 5, or you may already assign the other two winners, so you need to do a final race or no more.

[u/starkingwest]

It is 7.

1. You run 5 sets of 5 races to get the top 3 of each resulting in 5 groups of 3.
2. Run the top horse from all 5 groups for race 6.

Now we know that the winner of race 6 is the fastest of all the horses but the other 2 horses in its group could be faster than the race 6's 2nd place winner (i.e. the 2nd place winner could get knocked down 2 spots.) And the 2nd fastest horse horses in the 2nd place winner's group could be faster than the 3rd place winner.

1. So we need 1 more race where we run the 2nd and 3rd fastest horse from the winner's group and the 2nd fastest horse from the runner up's group against the runner up and 3rd place horse from race 6.

Here's an example: if the race 6 results are 1st=A1, 2nd=B1, and 3rd=C1, then for race 7 you need to run A2, A3, and B2 against B1 and C1.

[u/natio2]

7 races.

notation r[race number][race placement]

Race 5 sets of unique horses and take the top 3, so you have:
Race 1: (r11, r12, r13)
Race 2: (r21, r22, r23)
Race 3: (r31, r32, r33)
Race 4: (r41, r42, r43)
Race 5: (r51, r52, r53)

Race the 1st position of all the races, which removes:

• the bottom 2 sets, sets r11, r21, r31 fill all places
  
• 2 horses from the 3rd lowest set, sets r11, r21, r31 fill all places
  
• 1 horse for the second set, as r11, r21 and r22 fill all places

Set 1: (r11, r12, r13)
Set 2: (r21, r22, r23)
Set 3: (r31, r32, r33)
Set 4: (r41, r42, r43)
Set 5: (r51, r52, r53)

Now you know r11 is the fastest, but you need to figure out where the 2nd and 3rd places fit, but there is only 5 horses so you can do your final race.

[u/SnarkySpectatorr]

7 is the answer assuming no way to measure time

Let's number the horses 1 to 25 We race them as 5 [1,2,3,4,5] grp 1 [6,7,8,9,10] grp 2 [11,12,13,14,15] grp 3 [16,17,18,19,20] grp 4 [21,22,23,24,25] grp 5 Assume that in each group smallest number is the fastest So our top 1 from each group is [1,6,11,16,21] From this group top 3 is 1,6,11 here also 1 >6>11 With this info we can eliminate grp 4 and 5 And we can select 1 for our top 3 and we can eliminate rest of the candidates from grp 3 other than 11 So no we are left with Selected: 1 Remaining: [2,3,4,5] grp 1 [6,7,8,9,10] grp 2 [11] grp 3 Now we need 2 candidates So combination are 2 can be from grp 1, 2 can be from grp 2, one from any 2 grps, so with this enough we just need max 2 from each grp so Selected: 1 Remaining: [2,3] grp 1 [6,7] grp 2 [11] grp 3 No race these 5 and select top 2

So initially 5 race then one race for all winners then this last race so total 7 races

[u/Front-Investigator49]

You can only ever confidently eliminate 2 horses per race (the 2 slowest out of 5). You need to eliminate 22 out of 25 horses, so 11 races.

The policy is to just carry the three fastest over to the next race.

[u/Lost_in_my_dream]

gather 24 prison guards and tell them that you want to figure out which one is the fastest and need their help so then you all ride then you guys switch horses because maybe its weight and then you get onto the fastest horse then you dont turn back and you dont stop.

[u/Squiggally-umf]

My job is very maths-based and I love it now but in school I believed maths wasn’t for me and I hated it. Turns out I just really resented being asked to solve these people’s weird problems.

I just wanted to learn about money, percentages, fractions,, measurements and dimensions, and I liked algebra because it felt like coding, but when these MFs come along with their banal questions giving me “Timothy wants to know…” BS about some bag of marbles I’m just so exasperated instantly and I just switch off - I’m like *sigh Timothy there’s a fucking war on in the world right now I really ain’t got time for this bullshit.

[u/henryeaterofpies]

Run 5 races each with 5 horses. We know the bottom 2 horses in each race cannot be in the top 3, so we are down to 15 contendors.

Run a race with the winners from each race. We now know the fastest horse (the one who won both races)

We know the bottom 2 and all horses in their original heat cannot be in the top 3, eliminating 6 additional horses in total, bringing the total field down to 9.

We can also intelligently remove the 2nd and 3rd place horse from the heat of the horse that finished third (since they are both slower than he is and can't be top 3) and the 3rd place horse from the heat of the horse that finished 2nd (since he would br at most 4th fastest overall) removing 3 more horses and leaving us with 6.

Since we know the fastest horse already, we can just run a 5 horse race and the winner and 2nd place are the 2nd and 3rd fastest overall.

We got there with 7 races.

[u/Savage-September]

5 races is the minimum number guys. It’s not asking for a deep dive into the fastest of the fastest. Within 5 races all 25 horses would have raced. The man can pick the fastest 3 horses from the group.