That’s for sure 
The question is not to give a solution that works. It’s to explain what’s wrong with the solution I gave 
Bah!
This friggin riddle had me screwed up for two days a couple years ago when someone told it to me.
You said here:
Originally, there was $30.
Each guest paid $9
3x9 = 27
The clerk kept $2
27+2=29
What happened to the other dollar?
You can’t add the $2 if you want to find the total money because the $2 is what the guest already paid in the $9.
What you should add is the money that was given back to the guest! ($1 x 3)
If you get what I mean...
Did I get it?? 
Nice Riddle, btw…great post…very tricky 
OK. Here’s a math one:
Proof that 2 equals 1:
X = Y **Given**
X^2 = XY **Multiply both sides by X**
X^2 - Y^2 = XY - Y^2 **Subtract Y^2 from both sides**
(X+Y)(X-Y) = Y(X-Y) **Factor both sides**
(X+Y) = Y **Cancel out common factors**
Y+Y = Y **Substitute in from line 1**
2Y = Y **Collect the Y's**
2 = 1 **Divide both sides by Y**
Where is the flaw?
no way…senocular is just more clever…it said so on the post
You can’t cancel the (X-Y) because X-Y is 0, given the first line. Everything after that line is meaningless, since dividing by 0 is unpredictable and crazy and mathematicians will scorn you forever if you do it. :bad:
lets get backto some probability stuff.
to be unprecise: something with goats…
the event plan of a gameshow:
there are 3 doors, behind one of them is a porsche, behind the other two doors are goats.
(1) the candidate C (who doesn’t know what is behind each door) picks one of the doors.
(2) the anchorman A (who knows what’s behind the doors) will open one of the doors he didn’t pick, and shows him the goat
(3) C can choose again
(4) C wins the price behind the door he chose
If you were the candidate - would you swap?
how likely is it that the porsche is behind the door
a) the candidate chose
b) the anchorman is offering
or another one (we were asked in a math test in school):
6 person are living together in a flat, distributing the work between each other from day to day. how likely is it that any of the roommates has to wash the dishes twice this week
100% - there are 6 people, and 7 days in a week. Someone will have to take the last day.
To the one about the goats-
I think it’s 50/50, after the anchorman shows the guy one of the goats, there’s a 50/50 chance of his door having the prize behind it, no matter how many times he switches. What happened before is irrelevant, since we know that no matter which of the 3 doors he chooses, the anchorman will always cancel out one of the other doors.
[size=1][EDIT] Am I the only one who would *want *a goat more than a Porsche?[/size]
100% - there are 6 people, and 7 days in a week. Someone will have to take the last day.
yep
I think it’s 50/50, after the anchorman shows the guy one of the goats, there’s a 50/50 chance of his door having the prize behind it, no matter how many times he switches. What happened before is irrelevant, since we know that no matter which of the 3 doors he chooses, the anchorman will always cancel out one of the other doors.
nope
Ooh okay, I’m an idiot.
First C picks a door. The chance of the prize being there is 1/3. When A reveals one of the goats, the chance of the prize being in the other door increases to 2/3. So C should logically switch.
First C picks a door. The chance of the prize being there is 1/3. When A reveals one of the goats, the chance of the prize being in the other door increases to 2/3. So C should logically switch.
yep
Here’s the basic jist of one I was asked as a 6th grade math assignment:
A man goes 1km south, 1km west, 1km north, and ends up where he started. What are his two possible locations?
“In a maze of twisty little passages, all alike”.
5 bonus points for anyone old enough to know where that is from… 
Colossal Cave! Although I’m not actually old enough. It was the only game my older cousins had until I was 13.
Awesome, Hificopymaster! The 5 points are yours.
I played that game to death 20 years ago. Almost got all the points. I think I missed 1 or 2. Never figured out what to do with the Rubiayat of Omar Khayam (I’m 100% sure that that’s misspelled).
North pole.
(South pole is not an option since you can not go south any further)
An infinite set of options is on the circle that’s approx. 1159 meters away from the south pole.
When you walk 1000 m south, you near the pole. When you walk 1000 m west, you walk in a circle around the pole (159 x 2 x Pi = 1000). Then you walk 1000 m north and end up at the same spot again.
So there’s an infinite number of points near the south pole that’ll work.
Yes Hans, that’s the other answer - sorry, I said two possible locations, I should have said two possible areas or two correct answers to the question “where is he?” or something other than what I said. Heh. See how I suck at wording?
