Conquer Club

Foxglove wrote:

Big Whiskey wrote:I don't know if this riddle has been used here yet but here we go:

A father and his son were on there way to the store when they were suddenly struck by another car.The father wasn't hurt all that bad,so he was taken to the local hospital.The son on the other-hand was seriously injured and had to be life flighted to a hospital 100 miles away.When the doctor walked in the e.r he said"I can not operate on this boy for he is my son."Who is the doctor?

His mother.

Oh wait...are women allowed to be doctors?!

Note: this solution is not possible. Unless the mother had a sex-change.

jonesthecurl wrote:

Foxglove wrote:

Big Whiskey wrote:I don't know if this riddle has been used here yet but here we go:

A father and his son were on there way to the store when they were suddenly struck by another car.The father wasn't hurt all that bad,so he was taken to the local hospital.The son on the other-hand was seriously injured and had to be life flighted to a hospital 100 miles away.When the doctor walked in the e.r he said"I can not operate on this boy for he is my son."Who is the doctor?

His mother.

Oh wait...are women allowed to be doctors?!

Note: this solution is not possible. Unless the mother had a sex-change.

true so the answer should be the father in the car wreck was an odoptive father and the docter was a birth father.

Incidentally, the correct answer in the beer-barrel one was to drink enough of the beer to make a fair distribution possible.
One of the perks of being a lawyer.

jonesthecurl wrote:Incidentally, the correct answer in the beer-barrel one was to drink enough of the beer to make a fair distribution possible.
One of the perks of being a lawyer.

Plausible, but I think it more likely that the lawyer would take a portion of the beer, resell it at a price which bankrupts the brewery, then sweep up the operation through foreclosure.

ender516 wrote:

jonesthecurl wrote:Incidentally, the correct answer in the beer-barrel one was to drink enough of the beer to make a fair distribution possible.
One of the perks of being a lawyer.

Plausible, but I think it more likely that the lawyer would take a portion of the beer, resell it at a price which bankrupts the brewery, then sweep up the operation through foreclosure.

My way's more fun...and maybe why I'm not a lawyer.

Since no one has posted a puzzle in a while, and I think I got the right answer to "How many times can you subtract 5 from 135?", I will throw a new log on the fire:

I have four horses that I wish to transfer from point A to point B. Being of different ages and states of health, these animals can make the trip in 1, 2, 4, and 6 hours, respectively (regardless of whether I am riding one or not). For reasons known only to the Great Puzzler, I can only transfer a maximum of two horses at a time (say I ride one and lead the other). When travelling with two horses, naturally we are forced to go at the slower speed of the two. How long will it take to transfer all four horses from A to B?

ender516 wrote:Since no one has posted a puzzle in a while, and I think I got the right answer to "How many times can you subtract 5 from 135?", I will throw a new log on the fire:

I have four horses that I wish to transfer from point A to point B. Being of different ages and states of health, these animals can make the trip in 1, 2, 4, and 6 hours, respectively (regardless of whether I am riding one or not). For reasons known only to the Great Puzzler, I can only transfer a maximum of two horses at a time (say I ride one and lead the other). When travelling with two horses, naturally we are forced to go at the slower speed of the two. How long will it take to transfer all four horses from A to B?

14 hours?

Let's call the horses Gallop, Canter, Trot, and Walk (in order of fastest to slowest).

Leg 1, A to B: Gallop and Walk (6 hours)
Leg 2, B to A: Gallop (1 hour)
Leg 3, A to B: Gallop and Trot (4 hours)
Leg 4, B to A: Gallop (1 hour)
Leg 5, A to B: Gallop and Canter (2 hours)

Poor Gallop is going to have a heart attack after running for 14 straight hours, though.

I get 13

outward 1,2 = 2
return 1 = 1
outward 4,6 = 6
return 2 = 2
outward 1,2 = 2

=13

nippersean wrote:I get 13

outward 1,2 = 2
return 1 = 1
outward 4,6 = 6
return 2 = 2
outward 1,2 = 2

=13

I believe you have the right answer, [player]nippersean[/player]. Good try, [player]Bones2484[/player]. I took the same approach the first time I tried this puzzle. I guess the "Aha" comes when you realize that you can come out ahead by taking the two fastest together and the two slowest together. I should admit that I lifted this puzzle from the Nintendo DS game "Professor Layton and the Diabolical Box", second in a series which started with "Professor Layton and the Curious Village", both of which I would recommend to puzzle fans who already own a DS. I'm not sure it would be worth buying the unit just to play these games, but there is a third game promised, and there are lots of other puzzle games available, so it's a good bet overall for puzzle fanatics who need a portable fix.

Oooohhh! sneaky!

Haha, then my turn I believe.

What is the expected number of times to throw a die until all the numbers come up at least once?

This is based on a fair 6 sided die. Not some dodgy intesity cube that conspires against you.

Good luck!

Yay, I loved this in High School :p

First dice you throw is always a number you haven't had before.
So thats 1/1
Then you just go from 1/1, 1/2...to 1/6.
=2.45
At last you multiply the number by 6 (because you have 6 numbers).

EDIT:
forgot to post the answer
2.45x6= 14.7

So, you have to throw atleast 14.7= 15 times

frankiebee wrote:Yay, I loved this in High School :p

First dice you throw is always a number you haven't had before.
So thats 1/1
Then you just go from 1/1, 1/2...to 1/6.
=2.45
At last you multiply the number by 6 (because you have 6 numbers).

EDIT:
forgot to post the answer
2.45x6= 14.7

So, you have to throw atleast 14.7= 15 times

On the button Mr Bee. Nice roll and your turn