Riddle me this, Riddle me that!

StaggerLee said:
You are on the right track, and this is actually a very difficult probability theory problem. I am going to give it to you anyway because it's difficult.

You are actually increasing your chances to 66.66666~%. So you DOUBLE your chances by changing. The simplest way to look at this is to say that when you start, you have a 33.3333~% chance of winning and a 66.6666~% chance of losing. Once you choose one door and the host reveals a goat behind one of the doors, they are removing that loss from the equation. Now, if you change, you change an initial loss into a win, and an initial win into a loss, in other words the initial probabilities reverse, and you double your odds. So, statistically speaking, you should change.

For a more in-depth analysis of the problem, if anyone actually gives a crap or doesn't believe me, go here:
http://en.wikipedia.org/wiki/Monty_Hall_problem

And for a more technical analysis:
http://formalisedthinking.wordpress.com/2010/10/05/bayes-theorem-and-the-monty-hall-problem/

Right unstablebob, you're up.
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That was me, handing over the turn of riddle-person to unstablebob. So bob, go for it.
 
I don't know about that theory hey. Yes, initially you have a 33.3333 etc% chance of choosing the correct door, but once he opens door 3, your probability of being correct immediately goes up to 50%. If you stick with door 1, then you have 50% chance of being correct, and if you change your choice to door 2, then it's also 50% chance of being right.
 
matt said:
I don't know about that theory hey. Yes, initially you have a 33.3333 etc% chance of choosing the correct door, but once he opens door 3, your probability of being correct immediately goes up to 50%. If you stick with door 1, then you have 50% chance of being correct, and if you change your choice to door 2, then it's also 50% chance of being right.
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No guy, really, the probability doubles. It's not "just a theory" either. It's because the hosts choice is dependent on your first choice, this makes your two choices dependent. The simplest way to explain it is as I have done above. If you want to see a full mathematical proof, then you will need some background in probability theory. If you have that, then check the second link and see for yourself :)
 
The Smith family is a very wealthy family that lives in a big, circular home. One morning, Mr. Smith woke up and saw a strawberry jam stain on his new carpet. He figured out that everyone who was there that morning had a jam sandwich. By reading the following excuses, figure out who spilled the jam.
Billy Smith: “I was outside playing basketball.”
The Maid: “I was dusting the corners of the house.”
Chef: “I was starting to make lunch for later.”
Who is lying?
 
StaggerLee said:
No guy, really, the probability doubles. It's not "just a theory" either. It's because the hosts choice is dependent on your first choice, this makes your two choices dependent. The simplest way to explain it is as I have done above. If you want to see a full mathematical proof, then you will need some background in probability theory. If you have that, then check the second link and see for yourself :)
Click to expand...

Lots of things are provable mathematically, And the odds are better on your second choice than on your first choice, but the way I see it is that it's two independent events. First you have a go at 33.3% chance, then you have another go at 50% chance. Door 3 being wrong doesn't affect the 50-50 chance of door 1 or door 2. There is a whole new event of sticking with door 1 for 50% chance, or changing your choice to door 2 for 50% chance. Door 3 doesn't come into this event.

It's like if you flip a coin 10 times. The odds that it lands on heads 10 times in a row are tiny, but each flip is 50-50. The result of the previous flip has no influence on the current flip. It's entirely random.

Basically, the way I see it is that at the point when he says do you want to change your answer, door 3 is gone, and the previous odds are irrelevant because there is only the choice of two doors. The fact that there were 3 previously doesn't matter.
 
Sorry for chipping in, but here is my 2c:

matt said:
the way I see it is that it's two independent events
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It's not. You picked the door before one door was taken out of the equation.
When flipping a coin, each event is independent of the previous flip. A coin has no memory, so each flip is 50/50.
When you pick the door, however, you keep your choice into the second event when a door is opened.
Think like this: 1,000,000 doors exist and you pick one. 999,998 doors are opened. Only your door and one other door is left closed. You'll pick the other door in a second, won't you? Well, this is the same, just on a much smaller scale.
 
Mister 44 said:
Sorry for chipping in, but here is my 2c:



It's not. You picked the door before one door was taken out of the equation.
When flipping a coin, each event is independent of the previous flip. A coin has no memory, so each flip is 50/50.
When you pick the door, however, you keep your choice into the second event when a door is opened.
Think like this: 1,000,000 doors exist and you pick one. 999,998 doors are opened. Only your door and one other door is left closed. You'll pick the other door in a second, won't you? Well, this is the same, just on a much smaller scale.
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Ok, I think I get it. I suppose if you say the odds against picking the correct door were so bad in the first place that you most likely got the wrong door, and left with the choice of one that's correct and one that's incorrect, then most likely the other one is the correct one...
 
matt said:
Lots of things are provable mathematically, And the odds are better on your second choice than on your first choice, but the way I see it is that it's two independent events. First you have a go at 33.3% chance, then you have another go at 50% chance. Door 3 being wrong doesn't affect the 50-50 chance of door 1 or door 2. There is a whole new event of sticking with door 1 for 50% chance, or changing your choice to door 2 for 50% chance. Door 3 doesn't come into this event.

It's like if you flip a coin 10 times. The odds that it lands on heads 10 times in a row are tiny, but each flip is 50-50. The result of the previous flip has no influence on the current flip. It's entirely random.

Basically, the way I see it is that at the point when he says do you want to change your answer, door 3 is gone, and the previous odds are irrelevant because there is only the choice of two doors. The fact that there were 3 previously doesn't matter.
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Yes, lots of things are provable mathematically, including this. The whole point of the riddle is that it illustrates that our brains don't think probabilistically. As Mister 44 (and I) said, the two events are not independent. Flipping a coin multiple times is an example of independence, however the riddle I gave is not. The door that the host chooses DEPENDS on your choice because he will only choose a door that is incorrect. Then, you changing your decision DEPENDS on the fact that he showed you a door with a goat behind it. The events are definitely not independent. To avoid a continuation of this debate and to stay on the riddle topic, please check the (many) articles on the web (or your local library) on the Monty Hall problem.
 
unstablebob said:
The Smith family is a very wealthy family that lives in a big, circular home. One morning, Mr. Smith woke up and saw a strawberry jam stain on his new carpet. He figured out that everyone who was there that morning had a jam sandwich. By reading the following excuses, figure out who spilled the jam.
Billy Smith: “I was outside playing basketball.”
The Maid: “I was dusting the corners of the house.”
Chef: “I was starting to make lunch for later.”
Who is lying?
Click to expand...

Well, the maid is lying about the corners because it's a circular house, but I'm not sure who spilled the jam...
 
There are three desk lamps standing on a desk in room A. You, however, are in room B where you cannot see the lamps. There are three light switches in room B that control the three desk lamps in room A. All light switches are in the off position. You get fifteen minutes to think and act and then you will be brought to room A, where you have to state which switch corresponds to which lamp. How do you do it? You can only enter room A once, and you cannot go back to room B once in room A.
 
Turn on 2 lamps for 14min then switch off the one. The bulb that is lit is the one you left, the one that's hot is the one you turned off and the cold one is the switch you didn't touch.
 
Here is one from FBS 200

A truck is crossing a 2km bridge and weighs 3 tons. The bridge can only support 3 tons so everything is in balance. At the half way mark the driver stops and a bird lands on the truck.
Will the bridge collapse under the extra weight?
 
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