Comments on: Morris: A Day at the Races http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/ The Math Factor Podcast Site Fri, 08 Aug 2014 12:52:06 +0000 hourly 1 https://wordpress.org/?v=4.9.25 By: sohbet http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-1067 Wed, 06 Feb 2013 15:38:46 +0000 http://mathfactor.uark.edu/?p=1296#comment-1067 three fastest from the first race compete in the second. The three fastest from the second compete in the third and so on, in 11 races you have the three fastest and the 22 other horses cycle in behind

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By: Brock Hauser http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-907 Fri, 02 Dec 2011 20:51:01 +0000 http://mathfactor.uark.edu/?p=1296#comment-907 11 races, the three fastest from the first race compete in the second. The three fastest from the second compete in the third and so on, in 11 races you have the three fastest and the 22 other horses cycle in behind in the others.

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By: Stephen Morris http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-874 Tue, 12 Jul 2011 19:35:14 +0000 http://mathfactor.uark.edu/?p=1296#comment-874 This is my proof that the solution is minimal.  Try to find a proof yourself first.

[spoiler] With six races we can determine the fastest horse (lets call it A) but we cannot guarantee finding the second fastest (lets call it B).  Suppose that we succeed in finding A. 24 horses must lose a race (that is finish second or lower) so that we know they aren’t the fastest.  With six races there are only 24 opportunities to do this, so each horse loses precisely one race.  This is the pigeon-hole principle.  It is possible that the first race includes A but does not include B, and that some other horse, say C, finishes second.  C has lost to A, so we know it isn’t the fastest, but it does not lose any other race so we will never find out that it is slower than B. We cannot rule out C as the second fastest horse and so we cannot identify that B is the second fastest. [/spoiler]

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By: Sue VanHattum http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-869 Tue, 05 Jul 2011 15:25:50 +0000 http://mathfactor.uark.edu/?p=1296#comment-869 Just testing the spoiler tag.
[spoiler]
And I thought it was 9. I thought I had to have all the second place horses run against each other, and all the 3rds. Then I ended up with the same last race. Now your solutions (BP and RMJ) make total sense to me.
[/spoiler]

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By: rmjarvis http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-867 Tue, 05 Jul 2011 06:59:30 +0000 http://mathfactor.uark.edu/?p=1296#comment-867 Interesting.  Fewer races than I’d guessed right off the bat.

[spoiler]
You only need 7 races.

First, race 5 heats of 5 horses each.

Then race the 5 winners against each other.

Let’s call the 5 heat winners A1, B1, C1, D1, E1 in the order that they finished the 6th race.

Then, A1 is necessarily the fastest horse, since every other horse has either lost to her or to a horse who’s lost to her.  So we just need to determine the 2nd and 3rd place horses.

The contenders are:

A2, A3 (the 2nd and 3rd place horses in A1’s original heat)
B1, B2
C1

A2 and B1 are the only possible horses for second place, and the other one of these two plus the other three listed above could be the third fastest.

So, race these 5 in the 7th race to determine who is 2nd and 3rd.
[/spoiler]

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By: Blaise Pascal http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-866 Tue, 05 Jul 2011 02:53:08 +0000 http://mathfactor.uark.edu/?p=1296#comment-866 (SPOILER — probably wrong, but a decent attempt anyway) [spoiler] I can do it in 7 races, 5 heats of 5 and 2 runoffs.   The first 5 races yield 5 horses which could be 1st, 2nd, or 3rd, 5 horses which could be 2nd or 3rd, and 5 horses which could be 3rd (and 10 horses which are definitely not 1st, 2nd, or 3rd). For the 6th race, run all the winners of the 5 heats.  Let’s call the first three finishers Able, Baker, and Charlie.  Able is the fastest horse over all. We can’t tell if the 2nd pace finisher in Able’s heat or Baker is 2nd fastest yet.  Every other horse has either lost to one of these two contenders for 2nd place, or lost to a horse that lost to one of them, so can’t be 2nd fastest.  Likewise, Charlie, or the horse that came in 2nd to Baker, or the horse that came in 3rd to Able, are all possible contenders for 3rd.  Every other horse is definitely slower than at least one of those three. So the 7th race is between the 2nd and 3rd finisher in Able’s heat, the 1st and 2nd finisher in Baker’s heat, and the 1st place finisher in Charlie’s heat.  The winner of that race is the 2nd fastest, and the 2nd place finisher is 3rd fastest.[/spoiler]

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By: Stephen Morris http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-865 Tue, 05 Jul 2011 01:16:49 +0000 http://mathfactor.uark.edu/?p=1296#comment-865 Sue (and everyone),

Yes please post your answers here.  However use the spoiler tag if you think you have the actual answer.  No need if it’s just thinking it through.

Spoiler tags are [ spoiler ] my brilliant answer [ /spoiler ]

Remove the spaces in the tags, I had to include them or the system would have treated it as a proper use of spoiler tags.

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By: Sue VanHattum http://mathfactor.uark.edu/2011/07/morris-a-day-at-the-races/comment-page-1/#comment-864 Tue, 05 Jul 2011 01:07:52 +0000 http://mathfactor.uark.edu/?p=1296#comment-864 Nice one! Do you want answers here?

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