The worlds first riddle!

When rolling 2 six-sided dice, there are 36 possible combinations. Out of these 36 combinations, there is only one way that a twelve (2 sixes) can appear.

You are offered an even-money bet (let's say for $10) that if you can throw a twelve in 21 throws, you win $10. If not, you lose $10. Also, you may continue to make bets for as long as you'd like.

You reason that a twelve is 1 out of the 36 dice combinations, therefore your chances are 50-50 after 18 rolls and with 21 tries you are going to come out a winner more often than not. Is this reasoning correct

[size=8]DICE
The probability of rolling at least one 12 in 21 rolls is 1-(35/36)^21 = 0.44655

Bad bet to take. You need 25 rolls to barely turn the odds in your favor.[/size]

Mark:


DRIFT SPEED
1 nm/s



BEFORE YOU
The future


DICE
The probability of rolling at least one 12 in 21 rolls is 1-(35/36)^21 = 0.44655

Bad bet to take. You need 25 rolls to barely turn the odds in your favor.


In fact in 1952, a gambler named "Fat the Butch", agreed to this bet (except at $1,000 for 21 rolls). After losing $49,000 he realized that something had to be wrong with his reasoning.





From the integers 1, 2, 3, 4, 5, 6, 7, 8, and 9. Can you find SIX different integers, call them A, B, C, D, E, and F such that:

A x B x C = Which is the same as D x E x F =

1 x 8 x 9 = 72
3 x 4 x 6 = 72

but you still can't get my riddles.

r u talking to me, darkgirl?

Thoh:

1 x 8 x 9 = 72
3 x 4 x 6 = 72

Good thinking partner.


Or
A = 2
B = 3
C = 6
D = 1
E = 4
F = 9

2 x 3 x 6 = 36
1 x 4 x 9 = 36

(I doubt if there are any more)



I note there was no post by Mark, this has only happened twice before, once when he was sick, and the other when he had a meeting with the President. I have checked the Presidents diary; get well soon dude.


Darkgirl wrote, "you still can't get my riddles"

Whilst your contribution to the forum is to be highly applauded, Thoh and others have done a pretty good job in answering the problems. I would only attempt them if I did not already know the answer.

I did indeed prepare an answer:

a) Take away the letter S, and you make it (s)even

b) If you assume that "them" refers to the children and the mother, it means that she has 3 girls and 4 boys

c) "half of them are boys" 3. So are the other half 3. + 1 girl


However, as usual Mark beat me to it.

Keep up the good work, it's a dirty job, but someone has to do it. If there were no questions, there could be no answers.




Riddle for eight year olds.
Damn, this is annoying:

Given the eight digits 1, 1, 2, 2, 3, 3, 4, and 4, can you come up with an eight digit number using all these eight digits

Such that the:

1's are separated by 1 digit,
2's are separated by 2 digits,
3's are separated by 3 digits,
4's are separated by 4 digits.

23421314

If I am correct, does this mean I am relegated to hanging out with the 8 year olds from now on?

[size=8]41312432 or its reverse[/size]

Try..are you insulting me?

Can we leave this till I get home?

"them" refers to the children. i thought you of all people would know that. it looks like stupidity has rubbed on to you.

Them was refering to the children. besides, i wasn't even talking about that one. you need to read more.

In area code 530 it's the phone number for Ron Hilton from the Church Of The Nazarene. In area code 607 it's the phone number for DeWitt Mall. Dunno what it is in area code 404.

Stormwatch:

23421314

Yup! That is what I have.



Mark:

41312432 or its reverse


I am sorry, you have already reversed the reverse, if we reverse it again we will be back to where we started.



Newbie rote, "…it looks like stupidity has rubbed on to you."


It is true, Stupidity and I are old friends.


Seasoned Member Adrian, who I have not seen for several years, but nevertheless overjoyed at his return, wrote:

"In area code 530 it's the phone number for Ron Hilton from the Church Of The Nazarene. In area code 607 it's the phone number for DeWitt Mall. Dunno what it is in area code 404."

Keep taking the pills m8, see you in 2008. He is from OZ so please don't report him to the authorities.



Today I am giving away squares…
You are each given a free square, each of whose sides measure x inches.
If the corners of the square are cut off so that a regular octagon remains, how long is each side of the resulting octagon in terms of x

Well!

x
-----------
1+sqrt(2)

Thoh:

x
-----------
1+sqrt(2)


That is a very interesting answer, can it be so simple?

How about:

The four corner triangles are isosceles right triangles.
Let each of the eight sides of the octagon be y.
Then AB = y.
So, AC = y divided by the square root of 2.
So each side of the square would equal:
x = y + (y\/2) / 2 + (y\/2) / 2
x = y + y\/2

solving for y:
y = x / (1 + \/2)
Rationalizing the denominator gives you
y = (\/2 - 1) x or .414x

Or am I way out here?



I know Mark is missing because…

Mark went shopping for his wife's birthday party.
At the first stop, he spent half of his total amount of money for food.
At the second stop, he spent half of what was left for paper plates and napkins.
At the next stop, he spent 1/4 of what was left for table favors.
At the last stop, he spent half of what was left for a birthday cake and the candles to go on it.
Mark returned with only $6.

How much did he start out with

[size=8]SHOPPING
$64[/size]

Mark:

SHOPPING
$64



Let x = total amount of money Mark begins with.
He spends x/2 dollars on food, leaving him with x/2 dollars.
He spends x/4 dollars on paper plates, leaving him with x/4 dollars.
He spends x/16 dollars on party favors, leaving him with 3x/16 dollars.
He spends 3x/32 dollars on cake and candles, leaving him with 3x/32 dollars.

So set 3x/32 = 6 and solve for x
3x = 192, so x = $64.00




What is the next number in this fiendishly difficult sequence?

4, 16, 37, 58, 89, 145, 42, 20,