Conquer Club

lgoasklucyl wrote:The one who looks like the dog's been at his head.

If there's only one barbershop in town, the two barbers must cut eachothers hair. Since the other barber's hair looks spiffy, the one with the mess on his head must have cut his hair, and is therefore the correct choice.

Correct! Do you have a riddle to share?

Looks like I'm going to have to post one again...

Using the logic of the first 4 examples, decode the final word:

40-1-19-13 = FONT
2-81-90-7-11 = TENSE
16-1-19-9-8-3 = SONNET
20-1-4-5-8-87 = TOFFEE
14-1-1-10-98-1-20-18 = ?

No one? It's not that hard... cmon people...

I looked briefly, but didn't put too much thought into it.

If I get a few I'll see what I can muster up

slowreactor wrote:Looks like I'm going to have to post one again...

Using the logic of the first 4 examples, decode the final word:

40-1-19-13 = FONT
2-81-90-7-11 = TENSE
16-1-19-9-8-3 = SONNET
20-1-4-5-8-87 = TOFFEE
14-1-1-10-98-1-20-18 = ?

FOOTNOTE

each number corresponds with the first letter of its name

Blinkadyblink wrote:

slowreactor wrote:Looks like I'm going to have to post one again...

Using the logic of the first 4 examples, decode the final word:

40-1-19-13 = FONT
2-81-90-7-11 = TENSE
16-1-19-9-8-3 = SONNET
20-1-4-5-8-87 = TOFFEE
14-1-1-10-98-1-20-18 = ?

FOOTNOTE

each number corresponds with the first letter of its name

The rules say that the person who answers the current riddle must now post a new one. Do you have one to share?

A standard deck of playing cards is emptied into a hat. A man pulls a card out of the hat, looks at it, and puts it back in the hat. Then, once again, he pulls a card out of the hat, looks at it, and puts it back. If one of the two cards was a spade, what is the probability that the other one also was.

1/4?

Gozar wrote:1/4?

Nope, that would be too easy.

1/16?

If the jokers were included, the chances are 13/54.
If not, since the first card was returned (and thus could be picked again) the chances are 1 in 4.

Mr. Squirrel wrote:1/16?

No.

jonesthecurl wrote:If the jokers were included, the chances are 13/54.
If not, since the first card was returned (and thus could be picked again) the chances are 1 in 4.

Jokers are not included (there are 52 cards, 13 of each suit,) and, yes, the first card could be picked again, but it's not 1 in 4.

Remember, one of the cards is a spade, not necessarily the first one.

ah - I misread it.
I'll think further.

Looks like a 1 in 7.


Possibilities are:
Spade/Spade
Spade/not
Spade/not
Spade/not
not/Spade
not/Spade
not/Spade.

jonesthecurl wrote:Looks like a 1 in 7.


Possibilities are:
Spade/Spade
Spade/not
Spade/not
Spade/not
not/Spade
not/Spade
not/Spade.

Correct, but restated a little easier to follow...

For the experiment, there are 4 cases.
Case A: spade/spade Probability 1/4 x 1/4 = 1/16
Case B: spade/not spade Probability 1/4 x 3/4 = 3/16
Case C: not-spade/spade Probability 3/4 x 1/4 = 3/16
Case D: not-spade/not-spade Probability 3/4 x 3/4 = 9/16

The question is "If one of the two cards was a spade, what is the probability that the other one also was?" If you rephrase the question, "what is the probability that both cards are spades, given that at least one card is a spade?" you have a classic question of conditional probability and the answer is (1/16)/(1/16+3/16+3/16)= 1/7.

Note since the draws are independant, a nitpicker could argue that that the original question is better answered by the 1/4 answer. Although the question doesn't specify whether the first draw or the second is the given spade, it does specify that you are looking at the probability that the OTHER draw is spade- thus you are looking at either Case A and Case B or at Case A and Case C, but either way your conditional probability is back to 1/4.

Related riddle- Monty Hall paradox.
You are on the game show, Let's Make A Deal. The host shows you three doors. Behind one of the doors is an expensive convertible, but behind the other two are various farm animals. You are asked to pick one of the doors, and you win whatever is behind it. You pick your door. Befort the host opens the door you picked, he opens one of the doors you did not pick, revealing an incontinent goat. (Ewg!) He then offers to trade you, your door and the contents it would reveal for the other unopened door and its contents. Assuming that you would rather have a convertible then a farm animal, should you make the trade, reject the trade, or does it matter?

jonesthecurl wrote:Looks like a 1 in 7.


Possibilities are:
Spade/Spade
Spade/not
Spade/not
Spade/not
not/Spade
not/Spade
not/Spade.

We have a winner!

Blinkadyblink wrote:

jonesthecurl wrote:Looks like a 1 in 7.


Possibilities are:
Spade/Spade
Spade/not
Spade/not
Spade/not
not/Spade
not/Spade
not/Spade.

We have a winner!

OK this one's a riddle that goes back to Roman times at least, I first encountered it while studying some Anglo-Saxon stuff:

What has two legs, one eye, and a hundred heads?

Switch. The probability that your original door hides the car is still 1/3, but the probability that the car is behind the other two doors is 2/3. You are in essence getting to trade your door for both of the others, increasing your chances of winning to 2/3.

Another way to put it:

3 possibilities -
1. You pick the car. Monty shows you goat A. You switch to goat B and lose.
2. You pick goat A. Monty shows you goat B. You switch to the car and win.
3. You pick goat B. Monty shows you goat A. You switch to the car and win.

72o wrote:Switch. The probability that your original door hides the car is still 1/3, but the probability that the car is behind the other two doors is 2/3. You are in essence getting to trade your door for both of the others, increasing your chances of winning to 2/3.

Another way to put it:

3 possibilities -
1. You pick the car. Monty shows you goat A. You switch to goat B and lose.
2. You pick goat A. Monty shows you goat B. You switch to the car and win.
3. You pick goat B. Monty shows you goat A. You switch to the car and win.

This one was done to death (mine, mainly) a coupla months ago.
What convinced me of the correct answer was using three coins, two heads and one tail, and doing the swap/don't swap thing until I had to admit I'd been wrong all along. I'm sure someone will happily direct to the thread on which I was proved so embarassingly wrong-headed. I'm still hurtin' too much to look it up again.

jonesthecurl wrote:OK this one's a riddle that goes back to Roman times at least, I first encountered it while studying some Anglo-Saxon stuff:

What has two legs, one eye, and a hundred heads?

Anyone?

jonesthecurl wrote:

jonesthecurl wrote:OK this one's a riddle that goes back to Roman times at least, I first encountered it while studying some Anglo-Saxon stuff:

What has two legs, one eye, and a hundred heads?

Anyone?

A pirate with a chest of cabbages? ("Arr, Mateys, shiver me timbers and bury me's cole slaw!") Not a frickin' clue...

you're not a million miles away from the answer.

YoursFalsey wrote:

jonesthecurl wrote:

jonesthecurl wrote:OK this one's a riddle that goes back to Roman times at least, I first encountered it while studying some Anglo-Saxon stuff:

What has two legs, one eye, and a hundred heads?

Anyone?

A pirate with a chest of cabbages? ("Arr, Mateys, shiver me timbers and bury me's cole slaw!") Not a frickin' clue...

That is an awesome answer.

Hmmm...this has me stumped.

Actually, I'm gonna give you the pirate as "correct".

The answer is "A one-eyed garlic seller".

Your go

OK, Three identical triplets, Andy, Bob, and Chad.
Due to childhood trauma- their father was a sadistic professor of logic- one of them always lies, one of them always tells the truth, and one of them alternates true statements and false statements. Unfortunately, at the moment you cannot remember which is which. Suppose you are looking for Bob, because he owes you money, and run into two of the triplets. What is the minimum number of yes/no questions you must ask to guarentee you can determine which triplet (either one of the two, or the one who's not there) is Bob?