25 horses puzzle
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8 members have voted
June 10th, 2017 at 6:00:02 PM
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You have 25 horses. You need to determine the three fastest, in order. All you have is a racetrack, which can race five horses at a time. You don't have a watch, so all you know from each race is the top three finishers. Assume that each horse runs at the same speed every race.
What is the least number of races you need to run to determine the fastest three horses, in order, and how should it be done?
Free beer to the first person to present a correct answer and solution.
As always, please put the answers and solutions in spoiler tags.
The question for the poll is what is the answer?
No searching!
What is the least number of races you need to run to determine the fastest three horses, in order, and how should it be done?
Free beer to the first person to present a correct answer and solution.
As always, please put the answers and solutions in spoiler tags.
The question for the poll is what is the answer?
No searching!
Type [spoiler]Whatever you want to say[/spoiler]
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 10th, 2017 at 6:35:37 PM
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I voted 10 but miscounted. I should have said 11. Run a race of 5 and take the top 3. add two more each incremental race. I'm sure this crew will find a more efficient way.
June 10th, 2017 at 6:56:20 PM
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Quote: prozemaI voted 10 but miscounted. I should have said 11. Run a race of 5 and take the top 3. add two more each incremental race. I'm sure this crew will find a more efficient way.
Yes, you can correctly get the top 3 that way. However, there is a way to do it with fewer races.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 10th, 2017 at 7:05:34 PM
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I think 9. 5 races with each horse, eliminate the 4th and 5th horse from each race leaves 15 horses. 3 races with those 15 and then one more race with the winners of those three races
June 10th, 2017 at 7:17:17 PM
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Quote: WizardQuote: prozemaI voted 10 but miscounted. I should have said 11. Run a race of 5 and take the top 3. add two more each incremental race. I'm sure this crew will find a more efficient way.Yes, you can correctly get the top 3 that way. However, there is a way to do it with fewer races.
we could run 5 heats of 5 to get the top 15. Then run a 6th heat with the winners of each prior heat. That would get us to 6 horses left... Places 1,2,3 for the winner of heat 6... 2 and 3 for the second place finisher, and the 3rd place horse from heat 6. Run a 7th heat of 5 then an 8th heat of 4.
That would do it too I think. 8 races.
That would do it too I think. 8 races.
June 10th, 2017 at 7:23:04 PM
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Quote: prozemaQuote: WizardQuote: prozemaI voted 10 but miscounted. I should have said 11. Run a race of 5 and take the top 3. add two more each incremental race. I'm sure this crew will find a more efficient way.Yes, you can correctly get the top 3 that way. However, there is a way to do it with fewer races.we could run 5 heats of 5 to get the top 15. Then run a 6th heat with the winners of each prior heat. That would get us to 6 horses left... Places 1,2,3 for the winner of heat 6... 2 and 3 for the second place finisher, and the 3rd place horse from heat 6. Run a 7th heat of 5 then an 8th heat of 4.
That would do it too I think. 8 races.
wait... In heat 6, I already know who is first, so I just need a 7th heat of 5 to figure out 2nd and 3rd.
7 is the right answer.
7 is the right answer.
June 10th, 2017 at 7:25:48 PM
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Quote: JohnzimboI think 9. 5 races with each horse, eliminate the 4th and 5th horse from each race leaves 15 horses. 3 races with those 15 and then one more race with the winners of those three races
That method won't guarantee the fastest three horses. What if the fastest three are all in the first race? The second and third fastest would be eliminated from consideration under your method.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 10th, 2017 at 7:28:27 PM
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Quote: prozema
wait... In heat 6, I already know who is first, so I just need a 7th heat of 5 to figure out 2nd and 3rd.
7 is the right answer.
Can you elaborate on which horses would be in the 7th race. I'm not sure I understand your solution between the two posts.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 10th, 2017 at 7:30:15 PM
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Run 5 races of 5 horses each, all 25 horses are run.
Race #6 is the fastest horse from each of the first five races. Order the first 5 races from the one containing the fastest to the one containing the slowest of the 6th race
Select the 2nd and 3rd place horse from the race with the fastest horse from the first 5 races
Select the 2nd place horse from the heat with the second fastest horse from
Race those three horses with the 2nd and 3rd fastest horse from race 6.
That is race #7.
June 10th, 2017 at 7:34:42 PM
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Quote: WizardQuote: prozema
wait... In heat 6, I already know who is first, so I just need a 7th heat of 5 to figure out 2nd and 3rd.
7 is the right answer.Can you elaborate on which horses would be in the 7th race. I'm not sure I understand your solution between the two posts.
the 7th race would be between the 2nd and 3rd place horses from the original 5 heats of the winner of heat 6 plus the top two horses from the original heat of the second place winner of heat 6 plus the 3rd place finisher in heat 6. Hope that makes sense.
June 10th, 2017 at 7:48:48 PM
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Quote: Dalex64
Run 5 races of 5 horses each, all 25 horses are run.
Race #6 is the fastest horse from each of the first five races. Order the first 5 races from the one containing the fastest to the one containing the slowest of the 6th race
Select the 2nd and 3rd place horse from the race with the fastest horse from the first 5 races
Select the 2nd place horse from the heat with the second fastest horse from
Race those three horses with the 2nd and 3rd fastest horse from race 6.
That is race #7.
Well said. Looks like you got me by 3.5 minutes.
June 10th, 2017 at 8:37:19 PM
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Seven races.
I think this is the solution that has already been found, but this is how I did it.
Run five heats (a) thru (e) and assume you get horses A1-A2-A3-A4-A5 thru E1-E2-E3-E4-E5.
Sixth heat is the winners A1-B1-C1-D1-E1 (and assume A beats B beats C etc).
At this stage the winning horse is A1.
The second fastest horse is either B1 or A2.
The third fastest horse is either one of the above or A3 or B2.
So run A2 A3 B1 B2 C1 to determine second and third.
I think this is the solution that has already been found, but this is how I did it.
Run five heats (a) thru (e) and assume you get horses A1-A2-A3-A4-A5 thru E1-E2-E3-E4-E5.
Sixth heat is the winners A1-B1-C1-D1-E1 (and assume A beats B beats C etc).
At this stage the winning horse is A1.
The second fastest horse is either B1 or A2.
The third fastest horse is either one of the above or A3 or B2.
So run A2 A3 B1 B2 C1 to determine second and third.
A1 > B1 > C1
/ /
A2 B2
|
A3
It could be A1 A2 A3, A1 A2 B1, A1 B1 A2, A1 B1 B2, A1 B1 C1 - the last race determines which.
June 10th, 2017 at 8:58:16 PM
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It looks like prozema and Dalex64 are both correct. It could be argued who was first. How about I owe you both a beer? Charlie, looks like you are the third place horse.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 10th, 2017 at 9:07:23 PM
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Ha! Getting it right was prize enough, besides I live in Missouri and beer is free in Las Vegas. Thank you for the puzzle!
June 10th, 2017 at 9:15:22 PM
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Quote: WizardIt looks like prozema and Dalex64 are both correct. It could be argued who was first. How about I owe you both a beer? Charlie, looks like you are the third place horse.
Sounds great! After conceiving of so many wrong answers in these challenges, it is nice to finally get one right.
I'll actually be in Vegas in mid-September this year.
June 10th, 2017 at 9:37:50 PM
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5 races of 5 different horses.
That's 5.
Race all top 5.
That's 6 races and you've found #1.
Horses that placed 4'th and 5'th in the 6'th race are done, as well as the other horses in their races, and the 4 horses in the race where the 3'rd horse was. Also remove horses that placed 3'rd, 4'th, and 5'th from the one with horse #2. Remove horses that placed 4'th and 5'th from horse #1 race.
This should leave 5 horses remaining (the fastest horse has already been determined and no longer in action).
Take horses those 5 horses, race them, and the winner is the second fastest horse and the 2'nd placer in that race is the third fastest.
That's 7 total.
June 10th, 2017 at 10:36:20 PM
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Seeing as to how it is already answered I don't se the need to try. Before I scrolled down I was thinking a little bit about it. For some reason the Mr. Ed the Talking Horse theme song began to play in my head.
June 11th, 2017 at 2:09:49 AM
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I get 12 races, which I see is wrong. PS: boy did I come late to this!
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
June 11th, 2017 at 6:33:44 AM
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Quote: Dalex64Sounds great! After conceiving of so many wrong answers in these challenges, it is nice to finally get one right.
That was a tough one too -- congratulations!
Quote:I'll actually be in Vegas in mid-September this year.
Sounds good. Hope to see you then.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 11th, 2017 at 10:22:18 AM
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I approached it differently.
I simply killed 22 of the horses, the survivors were now my three fastest. Only ran one race and had a heck of a bbq.
Sometimes a hammer works just as well as math.
I simply killed 22 of the horses, the survivors were now my three fastest. Only ran one race and had a heck of a bbq.
Sometimes a hammer works just as well as math.
The older I get, the better I recall things that never happened
June 11th, 2017 at 10:24:13 AM
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Quote: billryanI approached it differently.
I simply killed 22 of the horses, the survivors were now my three fastest. Only ran one race and had a heck of a bbq.
Sometimes a hammer works just as well as math.
In that case, a race wasn't necessary.
DUHHIIIIIIIII HEARD THAT!
June 11th, 2017 at 10:26:21 AM
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Quote: billryanI approached it differently.
I simply killed 22 of the horses, the survivors were now my three fastest. Only ran one race and had a heck of a bbq.
Sometimes a hammer works just as well as math.
Glue and hotdogs are +EV?
June 11th, 2017 at 10:27:54 AM
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22 dead horses is like six weeks of free Carl Jr's.
The older I get, the better I recall things that never happened
June 11th, 2017 at 10:31:52 AM
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This reminds me of the episode of All in the Family where Archie eats horse meat.
DUHHIIIIIIIII HEARD THAT!
June 11th, 2017 at 1:07:24 PM
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I googled it and still too slow to win a beer.
Here's a graphical explantaion
http://puzzles.nigelcoldwell.co.uk/fiftynine.htm
Here's a graphical explantaion
http://puzzles.nigelcoldwell.co.uk/fiftynine.htm
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
June 11th, 2017 at 1:33:28 PM
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Quote: OnceDearI googled it and still too slow to win a beer.
Here's a graphical explantaion
http://puzzles.nigelcoldwell.co.uk/fiftynine.htm
That's a good site. I think that will give me more math problem ammunition for a while.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 11th, 2017 at 1:54:29 PM
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I was asked this exact question as part of the interview for an entry level casino analyst (bean counter) position by one of the major Strip companies.
Other questions included:
-I give you three identical brown paper bags. One contains only apples, one has only oranges, and one has both apples and oranges. I intentionally mislabeled each bag (ie the bag with only apples is definitely not labeled as "apples only"). You can reach into one bag and pull out one item. After that you must properly label each bag. Which bag do you pull from and why?
-You have 9 cue balls but one of them weighs slightly less than the rest. To be able to determine which is lighter I give you a two-sided scale. What is the least number of scale uses needed to identify the light cue ball with 100% success?
-You have a pistol and sit at a table with two other armed men. Your weapon has a 1/3 chance of successfully firing when you pull the trigger. The man to your right's gun successfully fires 2/3 of the time and the other man's weapon always fires with success. Everyone in the room knows the individual success probabilities for the firearms and they all know that the other men in the room know this as well. You will be first to shoot your gun, then the man to your right has his turn to shoot (if he is still alive), then the other man (if he is still alive). What strategy gives you the greatest chance of survival?
I have no beers to give but folks might find it interesting to see how some of these dreadful bean counters are selected.
Other questions included:
-I give you three identical brown paper bags. One contains only apples, one has only oranges, and one has both apples and oranges. I intentionally mislabeled each bag (ie the bag with only apples is definitely not labeled as "apples only"). You can reach into one bag and pull out one item. After that you must properly label each bag. Which bag do you pull from and why?
-You have 9 cue balls but one of them weighs slightly less than the rest. To be able to determine which is lighter I give you a two-sided scale. What is the least number of scale uses needed to identify the light cue ball with 100% success?
-You have a pistol and sit at a table with two other armed men. Your weapon has a 1/3 chance of successfully firing when you pull the trigger. The man to your right's gun successfully fires 2/3 of the time and the other man's weapon always fires with success. Everyone in the room knows the individual success probabilities for the firearms and they all know that the other men in the room know this as well. You will be first to shoot your gun, then the man to your right has his turn to shoot (if he is still alive), then the other man (if he is still alive). What strategy gives you the greatest chance of survival?
I have no beers to give but folks might find it interesting to see how some of these dreadful bean counters are selected.
Views are my own...
June 11th, 2017 at 2:07:36 PM
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Quote: TheoHuxtable
-You have a pistol and sit at a table with two other armed men. Your weapon has a 1/3 chance of successfully firing when you pull the trigger. The man to your right's gun successfully fires 2/3 of the time and the other man's weapon always fires with success. Everyone in the room knows the individual success probabilities for the firearms and they all know that the other men in the room know this as well. You will be first to shoot your gun, then the man to your right has his turn to shoot (if he is still alive), then the other man (if he is still alive). What strategy gives you the greatest chance of survival?
This is a common paradox used as an example of a perfectly competitive market failure used in Economic study.
June 11th, 2017 at 2:52:04 PM
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My gun has a 1/3 chance of success, the other 2/3, the last, 3/3. If I shoot the 2/3 and succeed, surely the sure thing will kill me if I succeed. We will both fire at the sure thing, if I succeed, the 2/3 will have no other option than to try and take me out, and if the sure thing survives, he will go after the 2/3 first because he is a greater threat. Being the 1/3, I will never be the first to die.
I am a robot.
June 11th, 2017 at 3:08:18 PM
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Quote: onenickelmiracleMy gun has a 1/3 chance of success, the other 2/3, the last, 3/3. If I shoot the 2/3 and succeed, surely the sure thing will kill me if I succeed. We will both fire at the sure thing, if I succeed, the 2/3 will have no other option than to try and take me out, and if the sure thing survives, he will go after the 2/3 first because he is a greater threat. Being the 1/3, I will never be the first to die.
It's kind of a trick question as the unstated third option of "firing at the ground" is the best strategy. Shooting at the ground is the best strategy for all the men - this is what make's the paradox prozema states above.
If you shoot at the ground then the man to your right can either shoot at you (which guarantee's his death 2/3 of the time) or shoot at the other man (which results in 2/3*1/3=2/9 chance of his death on the next round) so he will also shoot at the ground. If the last man chooses to shoot directly at you then the last man now has to face a 2/3 chance of death when the man to your right shoots and if he shoots at the other man he will have to face a 1/3 chance of death from your next shot.
After seeing that the strategy chosen by the first two men poses no threat, the last man will also shoot at the ground so as to maintain this peaceful state with zero chance of death. So you all end up sitting at this table shooting your infinite rounds of ammo into the dirt until the end of time.
However, if the situation absolutely required each man to take aim at someone else at the table then I agree that your strategy is best.
Views are my own...
June 11th, 2017 at 8:56:10 PM
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Yes it's fun. I liked the one about two eggs ...Quote: WizardThat's a good site. I think that will give me more math problem ammunition for a while.
Quote:60. There is a building of 100 floors
-If an egg drops from the Nth floor or above it will break.
-If it's dropped from any floor below, it will not break.
You're given 2 eggs.
How do you find N in the minimum number of drops?
June 12th, 2017 at 1:15:59 AM
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Quote: TheoHuxtableI was asked this exact question as part of the interview for an entry level casino analyst (bean counter) position by one of the major Strip companies.
Other questions included:
-I give you three identical brown paper bags. One contains only apples, one has only oranges, and one has both apples and oranges. I intentionally mislabeled each bag (ie the bag with only apples is definitely not labeled as "apples only"). You can reach into one bag and pull out one item. After that you must properly label each bag. Which bag do you pull from and why?
-You have 9 cue balls but one of them weighs slightly less than the rest. To be able to determine which is lighter I give you a two-sided scale. What is the least number of scale uses needed to identify the light cue ball with 100% success?
-You have a pistol and sit at a table with two other armed men. Your weapon has a 1/3 chance of successfully firing when you pull the trigger. The man to your right's gun successfully fires 2/3 of the time and the other man's weapon always fires with success. Everyone in the room knows the individual success probabilities for the firearms and they all know that the other men in the room know this as well. You will be first to shoot your gun, then the man to your right has his turn to shoot (if he is still alive), then the other man (if he is still alive). What strategy gives you the greatest chance of survival?
I have no beers to give but folks might find it interesting to see how some of these dreadful bean counters are selected.
The cue ball can be isolated in 2 weighs.
1. Pick six balls.and weigh them, three on each side.
A. If the scale does not balance, you
2. Keep the 3 balls from the higher side, empty the scale, and place any 2 of the balls in it, one each side.
A. If the scale does not balance, the higher side holds the lighter cue ball
B. If the scale balances, the ball you did not weigh is the lighter ine.
1B. If the scale balances, you keep the balls not weighed and empty the scales. Place 2 unweighed balls in the scales, one each side. Proceed as in 2 above.
1. Pick six balls.and weigh them, three on each side.
A. If the scale does not balance, you
2. Keep the 3 balls from the higher side, empty the scale, and place any 2 of the balls in it, one each side.
A. If the scale does not balance, the higher side holds the lighter cue ball
B. If the scale balances, the ball you did not weigh is the lighter ine.
1B. If the scale balances, you keep the balls not weighed and empty the scales. Place 2 unweighed balls in the scales, one each side. Proceed as in 2 above.
If the House lost every hand, they wouldn't deal the game.
June 12th, 2017 at 1:24:59 AM
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3 mislabeled bags. You must reach into the mixed orange and apples bag. Say it is an orange. That bag must be oranges only.
That leaves 2 incorrectly labeled bags, oranges only and apples only. Since the apples only label is wrong, the apples only must be in the bag marked oranges only. That leaves the apples only bag to hold apples and oranges.
The answer is reversed if you pull an apple from the mixed bag.
If the House lost every hand, they wouldn't deal the game.
