RIDDLE GAME!!!

Nathan · Post by **Nathan** » 24 Mar 2005 16:56

ok then:

What 7 letter word becomes longer when the third letter is removed?

ted · Post by **ted** » 24 Mar 2005 17:02

lounger

EDIT: [hard one]
You are a winning contestant on a game show with the choice of three doors. Behind one door is a Prize and behind the others is nothing. You choose door one and the show host opens door two, revealing nothing. You are then offered the chance to swich your choice to Door three. Should you stick with your original choice or switch?

Iain · Post by **Iain** » 24 Mar 2005 19:26

AH HA! I know the answer to this. This is one of my fav's. I'll wait to let people who don't know the answer toy with it and then i'll step in.

C-Fan · Post by **C-Fan** » 24 Mar 2005 22:42

ted wrote:lounger

EDIT: [hard one]
You are a winning contestant on a game show with the choice of three doors. Behind one door is a Prize and behind the others is nothing. You choose door one and the show host opens door two, revealing nothing. You are then offered the chance to swich your choice to Door three. Should you stick with your original choice or switch?

Stick to the original. If you are a winning contestant, you must have picked the right door.

ted · Post by **ted** » 25 Mar 2005 13:11

mattkain · Post by **mattkain** » 26 Mar 2005 16:11

Statistically speaking, you would always want to switch doors. This is the famous Monty Hall riddle that I remember seeing in a combinatorics class. The solution can be a bit tricky to explain and even trickier to wrap your brain around. There's actually been quite a bit of debate surrounding this riddle and many computers models have been devised to prove the validity of the answer. Anyway, let me try to explain.

Let's say we pick Door 1. Through simple probability theory we know that when we choose Door 1 we have a 1/3 chance of getting the prize. This does not change. No matter what, there is always a 1/3 chance that Door 1 is the winner. So if we stick with door 1 (or our original choice) we have a 1/3 chance of getting the prize regardless of what the host shows us is behind the other doors.

However, if we change doors, we have a 2/3 probability chance of winning. This is because by switching doors we essentially are being allowed to choose 2 doors for the price of 1: the new door and the door that the host showed us. Here's another way to think of it. Picture the 3 outcomes and remember that each has a 1/3 probability of occurring:

Outcome 1: Door 1: Prize | Door 2: Lose | Door 3: Lose
Outcome 2: Door 1: Lose | Door 2: Prize | Door 3: Lose
Outcome 3: Door 1: Lose | Door 2: Lose | Door 3: Prize

Again, choose Door 1 and we can ONLY win in Outcome 1 which has a 1/3 chance of occuring. However, if we switch doors, we win in both Outcome 2 and Outcome 3. In other words, we win 2/3 of the times. This is because in both of these cases we chose Door 1 which had no prize behind it and then switch to the winning door. In Outcome 2, we choose Door 1, the host reveals that there is nothing behind Door 3, we switch to Door 2, and we win. In Outcome 3, we choose Door 1, the host reveals that there is nothing behind Door 2, we switch to Door 3, and we win. By switching we win 2/3 of the time.

Get it? Definitely tricky. Anyway, I don't know any riddles, just Math puzzles from my short tenure as a Math major at Uni. Ted, you can go again.

Tripp · Post by **Tripp** » 26 Mar 2005 16:11

Well if Ken is wrong, then I'm going to say you have to switch. It was a 50/50 chance.

mattkain · Post by **mattkain** » 26 Mar 2005 16:13

Josh, check out my solution above.

Tripp · Post by **Tripp** » 26 Mar 2005 16:17

Haha. Argh. You posted like thirty seconds before me.

Iain · Post by **Iain** » 26 Mar 2005 18:25

http://www.grand-illusions.com/simulator/montysim.htm

delakong · Post by **delakong** » 27 Mar 2005 08:18

Wow, that was hard. I did it ten times on my own and got the car half the time, but when I used the 100 time thing it was better to change all the time. Too bad if you actually had the chance to do it in real life you'd only have one try.

ted · Post by **ted** » 28 Mar 2005 19:01

Yarrgh.

Consider a pair of non-zero numbers x and y such that x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.

What's wrong with this?

mattkain · Post by **mattkain** » 28 Mar 2005 19:55

ted wrote:Consider a pair of non-zero numbers x and y such that x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.

What's wrong with this?

If x = y then:
x-y = 0

Therefore, the step where you said, "Dividing by (x-y)..." is the same as saying "Dividing by 0..." Division by zero is illegal and hence any subsequent assumptions or statements are fallacious.

MK

You can go again, Ted.

ted · Post by **ted** » 29 Mar 2005 08:51

Yarrrgh. Get this one and I'll give you a scholarship

One hundred bugs are placed on a meter-long stick. Each bug is traveling either to the left or the right with constant speed 1 meter/minute. When two bugs meet they reverse directions and when a bug reaches an end of the stick, it falls off.

What is the longest amount of time that you would need to wait to guarantee that the stick has no more bugs?

slapdash21 · Post by **slapdash21** » 29 Mar 2005 13:10

30 seconds?

edit-scratch that, i read that as the shortest time. need to think more...

Iain · Post by **Iain** » 29 Mar 2005 17:38

59.28 seconds?

if i'm right i'll tell what i did

EDIT: without rounding it would be 59.4 seconds

Colin · Post by **Colin** » 29 Mar 2005 18:16

One minute.

forget about how the bugs collide and bounce, just realise that it's the exact same result as if they were to pass through each other. When you realise this, it's clear that the bug's can't be on the stick longer than a minute.

I'll post one shortly.

Colin · Post by **Colin** » 29 Mar 2005 18:29

This is a pretty famous one, but it's real cute. Sorry if this thread has gotten too mathy.

A census taker knocks on a door, and asks the woman inside how many children she has and how old they are.

"I have three daughters, their ages are whole numbers, and the product of their ages is 36," says the mother.

"That's not enough information," responds the census taker.

"I'd tell you the sum of their ages, but you'd still be stumped."

"I wish you'd tell me something more."

"Okay, my oldest daughter Annie likes dogs."

"Thank you maam!"

What are the ages of the three daughters?

peanutbuterboy · Post by **peanutbuterboy** » 29 Mar 2005 19:27

I hate math...

delakong · Post by **delakong** » 30 Mar 2005 05:56

Okay, I'm still a little drunk but isn't that not enough information? I mean, all three daughters could be 12 or any combination of whole numbers that add up to 36.

8,10,18
6,12,18
other combinations like this

I don't know many mathish riddles but I still think there's not enough info.