Unutterably Silly Quiz

[pdurrant]

Originally Posted by Stitchawl

Ephraim is the remaining brother.
Dorcas, Liza, Alice, and Ruth are the brides.

All right. Five points to you

[orlok]

Originally Posted by pdurrant

One isn't called a sea in modern times.

Persian Gulf?

[pdurrant]

Originally Posted by orlok

Persian Gulf?

Already guessed.

The Scores are In!

Hamlet53: 11 points
orlok: 9 points
HomeInMyShoes: 8 points
Bilbo1967: 7 points
red_phoenix: 6 poiints
Stitchawl: 5 points
Shayne Parkinson: 4 points
poohbear_nc: 3 points
Bookpossum: 2 points

With only one point left to get, Hamlet53 is the winner. Over to you!

(The remaining sea was the Arabian Sea.)

Originally Posted by pdurrant

Already guessed.

The Scores are In!

Hamlet53: 11 points
orlok: 9 points
HomeInMyShoes: 8 points
Bilbo1967: 7 points
red_phoenix: 6 poiints
Stitchawl: 5 points
Shayne Parkinson: 4 points
poohbear_nc: 3 points
Bookpossum: 2 points

With only one point left to get, Hamlet53 is the winner. Over to you!

(The remaining sea was the Arabian Sea.)

Oh dear! I will spend the day on trying to come up with something that is challenging, but not to obscure or complicated in a subject area that interests me [eg mathematics].

[pdurrant]

Originally Posted by Hamlet53

Oh dear! I will spend the day on trying to come up with something that is challenging, but not to obscure or complicated in a subject area that interests me [eg mathematics].

Oh, go on... give us a bit of maths

[orlok]

Originally Posted by pdurrant

Oh, go on... give us a bit of maths

Yes! Stitchawl loves maths (math).

So by request, maths . . .


You have 25 horses and want to know which are the three fastest. Whenever you race horses, the order of finish accurately reflects the relative speeds of the horses but you can only race five at a time. What’s the minimum number of races required to determine the three fastest, and how do you do it?

To be clear you have no measure of absolute speed, that is no race times. To forestall any similar questions you are sitting in a room without a view of the track and are only handed slips of paper with the names of the horses and the relative positions of finish after each race. You may select any five horses to participate in each race.

Provide your reasoning with the answer for any credit.

[pdurrant]

Originally Posted by Hamlet53

So by request, maths . . .


You have 25 horses and want to know which are the three fastest. Whenever you race horses, the order of finish accurately reflects the relative speeds of the horses but you can only race five at a time. What’s the minimum number of races required to determine the three fastest, and how do you do it?

To be clear you have no measure of absolute speed, that is no race times. To forestall any similar questions you are sitting in a room without a view of the track and are only handed slips of paper with the names of the horses and the relative positions of finish after each race. You may select any five horses to participate in each race.

Provide your reasoning with the answer for any credit.

First, split the horses into five groups of five, call them A,B.C.D & E groups. Race each group. Call the fastest horse in each group A1, B1, C1, D1, E1; the second fastest horse A2, etc.

Take the top three horses from each group - 15 horses in all. The fastest three horses of the 25 must be in this fifteen.

Split the fifteen horses into three groups:
Alpha Group: A1,B1,C1,D1,E1;
Beta Group: A2, B2, C2, D2, E2;
Gamma Group: A3, B3, C3, D3, E3

Race each group.

The fastest horse of all is the winner of the Alpha group.

The second fastest horse is either the horse who came second in the alpha group, or the horse that came first in the beta group

The third fastest horse is either the horse that came second or third in the alpha group, or the horse that came first or second in the beta group, or the winner of the gamma group.

That is, the second and third fastest horses are two out of five horses. Run those five horses in a race, and the first and second of that race are the second and third fastest overall.

Total number of races required: 5+3+1 = 8.

(The trick is to realise that after the second lot of races, we can already see who the fastest horse is.)

Originally Posted by pdurrant

First, split the horses into five groups of five, call them A,B.C.D & E groups. Race each group. Call the fastest horse in each group A1, B1, C1, D1, E1; the second fastest horse A2, etc.

. . .

(The trick is to realise that after the second lot of races, we can already see who the fastest horse is.)

Did not go through your logic in detail because there is a solution requiring fewer than eight races.

[Bilbo1967]

Originally Posted by pdurrant

Total number of races required: 5+3+1 = 8.

To be honest, I did my best to follow it but the above just doesn't make sense. Does it?

[pdurrant]

Originally Posted by Hamlet53

Did not go through your logic in detail because there is a solution requiring fewer than eight races.

There is? Hmm….

Every horse must be in at least one race. From five separate races we can tell nothing, so the answer must be 6 or 7. I'll think more about it.

[Stitchawl]

Originally Posted by orlok

Yes! Stitchawl loves maths (math).

Having just read the question, and Paul's answer (which almost blinded me,) excuse me while I go stick needles into my eyeballs...


Stitchawl

[BaenSidhe]

Originally Posted by pdurrant

First, split the horses into five groups of five, call them A,B.C.D & E groups. Race each group. Call the fastest horse in each group A1, B1, C1, D1, E1; the second fastest horse A2, etc.

Take the top three horses from each group - 15 horses in all. The fastest three horses of the 25 must be in this fifteen.

Split the fifteen horses into three groups:
Alpha Group: A1,B1,C1,D1,E1;
Beta Group: A2, B2, C2, D2, E2;
Gamma Group: A3, B3, C3, D3, E3

Race each group.

The fastest horse of all is the winner of the Alpha group.

The second fastest horse is either the horse who came second in the alpha group, or the horse that came first in the beta group

The third fastest horse is either the horse that came second or third in the alpha group, or the horse that came first or second in the beta group, or the winner of the gamma group.

That is, the second and third fastest horses are two out of five horses. Run those five horses in a race, and the first and second of that race are the second and third fastest overall.

Total number of races required: 5+3+1 = 8.

(The trick is to realise that after the second lot of races, we can already see who the fastest horse is.)

You need 6 races to determine who is first as per Paul's logic above. Then you just need 1 more race to determine the second and third place horses, running the horses Paul has listed above to determine third place.

So Paul's logic is correct, he just added up the races incorrectly. And 5+3+1 != 8.

Actually, I see where Paul got 3 now. So my answer seems to be wrong, logic-wise, although it's probably correct from Hamlet's hint.

Anyway, 5+3+1 still doesn't equal 8.

[Bookpossum]

It seems to me you need to get a bigger racetrack so you can race more than five horses at once. You would sort it all out much more quickly that way!

[Stitchawl]

Originally Posted by Bookpossum

It seems to me you need to get a bigger racetrack so you can race more than five horses at once. You would sort it all out much more quickly that way!

Or use cars. Horses are so passé. THAT is logic!


Stitchawl