




Posts: 856 Joined: 18 Mar 2005 Last Visit: 22 Oct 2019
 


Posted: Tue 09 Jan, 2018 


Considering this further: No, I was wrong, and moogle's solution was closer than mine. It is 7 races.
As before, in the first round, each group of horses runs a race, making 5 races in the second round.
Then, in the next round, there is just one race, with the firstplace winner from each group from the first round. As before, the firstplace winner from this race is firstplace overall, and need not run another race.
In the final round, both the secondplace and thirdplace winners from the firstround group that the overall first place ran in need to run, because all of the top 3 might have ended up in that group.
The second place horse from the second round, and the runner up to that horse from the second round also need to compete in the final round. The lower places don't need to run, because the thirdplace horse is slower than the secondplace horse from the first round, which is slower than the firstplace horse from that round, which is slower than the firstplace horse from the second round. Thus, we know there are 3 horses that are faster than those horses.
Likewise, the the thirdplace horse from the second round is, at best, thirdplace overall, so none of the runnerups which that horse beat in the second round need to compete in the final round.
Therefore, in the final round, the 5 horses are the second and thirdplace horses from the second round, the second and thirdplace horses that the overall firstplace horse outran, and the runnerup from the firstround to the horse that came in secondplace in the second round. The top 2 horses from the final round are second and thirdplace overall.
So, there are 5 races in the first round, one in the second round, and one in the final round, for 7 races in all.








Posts: 687 Joined: 28 Jun 2012 Last Visit: 22 Oct 2019
LD count: 11.8 estimate
 

horse race part 2 πππ 
Posted: Tue 06 Feb, 2018 


From initial analysis, I would say it's not guaranteed to be possible to find the three fastest without knowing which is fastest, since you might get lucky (or unlucky) and race the three fastest in one of the initial races. However, the odds of that would be 231/53130 or approx 0.43% if I managed to get my probabilities right.. (22C2 over 25C5?)
Anyway, there's still a higher probability that it will be possible, so here's how I would propose to proceed:
Click here to see the hidden message (It might contain spoilers)
Round 1: Same as before, group the horses into 5 groups of 5, and have the preliminary races, numbered 1 to 5. Immediately, the 4th and 5th place finishers of each race are eliminated. 15 horses are left.
For easier reference, let us call each horse as (n,r) where n is the race number that the horse was grouped into in Round 1 and r is the rank of the horse in that race. n is from 1 to 5, and r is from 1 to 3.
For example, horse (4,1) was in race 4 of the preliminaries, and came in first place.
Round 2: Group the remaining horses into 3 matchups of 5 such that no five horses in the same group had previously raced against each other in the preliminaries. To even things up, I would split them in this pattern:
Race 1: (1,1) vs (2,2) vs (3,3) vs (4,2) vs (5,1)
Race 2: (1,2) vs (2,3) vs (3,1) vs (4,1) vs (5,3)
Race 3: (1,3) vs (2,1) vs (3,2) vs (4,3) vs (5,2)
From the results of the 2nd round, I think a maximum of two more rounds would be needed depending on the results. Odds can be determined accordingly based on whether the three finalists had raced against each other in the previous two rounds, and perhaps based on the performance of the other horses as well, however that might get more complicated.
Results of Round 2:
Note: Any horse that comes third is automatically out of the running.
Proof: Since each race has a maximum of two first placers from the previous round, at least one of the top three finishers in any given race has at least one horse faster that raced in another match. Therefore at least three horses are faster than the third place finisher of any given race.
With only 6 horses left in the running, at least two have raced against each other in the previous round, therefore we would be able to determine the three fastest with at most one final round. However, to lessen the possibility of knowing which of the horses is the fastest, we should minimize the number of horses that race against each other in Round 3, and thus a fourth round may also be needed.
I made a spreadsheet to illustrate my solution, though it's not fully done yet, and only based on scenarios. Basically I numbered the horses from 1 to 25 with 1 being the fastest, and randomly grouped them for Round 1. Orange colored cells are the horse numbers. White colored cells refer to their respective rank from Round 1 races. Press F9 or edit any cell to generate another scenario.
Example scenario 1:
Click here to see the hidden message (It might contain spoilers)
Round 2 results (top 2 of each matchup):
Race 1: (5,1) and (1,1)  (horses 2 and 4)
Race 2: (5,3) and (1,2)  (horses 5 and 6)
Race 3: (2,1) and (5,2)  (horses 1 and 3)
Solution:
(1,2) is out, because (1,1), (5,1), (5,3) and (5,2) are all faster.
(5,3) is out because (5,2), (5,1) and (2,1) are all faster.
Round 3: (1,1) vs (5,2) > whoever wins is in the top 3 with (2,1) and (5,1). Either (5,1) or (2,1) is the fastest, but the winner won't be known until after the finals π
Example scenario 2:
Click here to see the hidden message (It might contain spoilers)
Round 2 results (top 2 of each matchup):
Race 1: (5,1) and (1,1)  (horses 1 and 2)
Race 2: (4,1) and (3,1)  (horses 3 and 4)
Race 3: (2,1) and (3,2)  (horses 5 and 7)
Solution:
(3,2) is out, because (3,1), (4,1) and (2,1) are all faster.
Round 3: (1,1) vs (3,1) vs (2,1)
Results: (1,1) beats (3,1) beats (2,1) > (4,1), (1,1) and (5,1) are in the top 3. Either (5,1) or (4,1) is the fastest, but the winner won't be known until after the finals π
Example scenario 3:
Click here to see the hidden message (It might contain spoilers)
Round 2 results (top 2 of each matchup):
Race 1: (5,1) and (1,1)  (horses 1 and 6)
Race 2: (4,1) and (3,1)  (horses 3 and 5)
Race 3: (5,2) and (2,1)  (horses 2 and 4)
Solution:
All horses are still in the running.
Round 3: (1,1) vs (3,1) vs (5,2)
Results: (5,2) beats (3,1) beats (1,1) > (1,1) is out, because (3,1), (5,2) and (5,1) are faster.
(3,1) is out because (4,1), (5,2) and (5,1) are all faster.
(5,2) and (5,1) are in the top 3. We will need a Round 4 to determine if (4,1) or (2,1) is our last contender.
Alternate Round 3: (1,1) vs (3,1) vs (2,1)
Results: (2,1) beats (3,1) beats (1,1) > (1,1) and (3,1) are out, because (2,1), (5,2) and (5,1) are faster.
Same result, (5,2) and (5,1) are in the top 3. We will need a Round 4 to determine if (4,1) or (2,1) is our last contender.
Round 4: (4,1) vs (2,1)
If (4,1) wins, then we have our top 3, and either (4,1) or (5,1) is the fastest but the winner won't be known until after the finals π
However, if (2,1) wins, then we already know that (5,1) is the fastest, followed by (5,2) then (2,1)

EDIT v2 of spreadsheet here (with odds)
Click here to see the hidden message (It might contain spoilers)
So, finally managed to work out the program until Round 3 (Edited the terminology a bit so that Round 4 is now called Round 3 race 2), and realized my original computation of Top 3 meeting in the first round was off. It should be .43%*5, or roughly 2.2%, still small enough.
However, it seems that the probability of successfully determining the top 3 without knowing which is the fastest is only about 75% using the method above, and only about 1011% that it will be an evenly anyone'sgame 3way race. This still could be improved, as I think there's still a bug with the rules on how to determine the Round 3 match ups, and it doesn't always give the best solution (for example, by wasting a race by choosing 2 horses whose rankings against each other are already known).
Anyway, I've identified five cases based on known WinLoss data after Round 3. Two of the cases end up as failures (Fastest horse is already known at the end of Round 3). Corresponding theoretical and simulated probabilities are shown below.
Edit: Added Sim2 to address the bug of wasting Round 3 match ups. Even threeway race odds has increased to about 14% in the simulation.


