A riddle to curve the boredom!

Originally posted by SLH
Which way would your bro tell me to go if i wanted to go home?

Choose the other way.
Click to expand...

bah, easy!
 
30 secs to read through the rose stuff, 5 secs to work it out (once i rolled it)

Does this mean i am really stupid?
 
You only need one trip and you have to flip two switches. Flip one switch on, wait about 5 minutes, then turn it off. Flip one of the other switches on and walk upstairs. If the lightbulb is on, you know that the second switch you flipped opperates it. If the lightbulb is off and warm, you know that the first switch you flipped operates it. If the lightbulb is both off and cool, you know that switch you didn't flip operates it. If the lightbulb happens to be florescent, you're making another trip buddy.
 
the brother one is easy, ask the people what there brother would say is the correct answaer then go the oppersite. piss.
 
Ok, here's a good math question.

There's a game show in which a contestant is shown three doors. Behind one of the doors is a prize, and behind the other two are nothing. The contestant is asked to pick a door. Once the contestant has picked a door, without opening any of them, one of two other doors containing nothing is removed. The contestant is then asked to pick between the remaining two doors.

Now, after you've picked the first door, which strategy makes more sense mathematically? Stick with your first choice, change your mind and go with the other door, or does it not matter?
 
Choose another door.

When you made your first choice you had a 1 in 3 chance, now you have a 1 in 2 chance.


Don't ask me to prove the other way of thinking is wrong though.... i.e. you'd still have a 1 in 2 chance because there are 2 doors one of which has a prize, so i doesn't matter which of the 2 you choose (could choose the currently selected door).
 
Yah, I guess in order to get the question right, you'd have to prove it. Otherwise it's pretty much a guess.
 
Originally posted by iamironsam
Ok, here's a good math question.

There's a game show in which a contestant is shown three doors. Behind one of the doors is a prize, and behind the other two are nothing. The contestant is asked to pick a door. Once the contestant has picked a door, without opening any of them, one of two other doors containing nothing is removed. The contestant is then asked to pick between the remaining two doors.

Now, after you've picked the first door, which strategy makes more sense mathematically? Stick with your first choice, change your mind and go with the other door, or does it not matter?
Click to expand...

Go with the other door as it has a higher probability of containing the prize. It's easier to understand when it's exagerated. (There's a million doors, you choose one... the host then removes all the doors containing nothing except yours and another one... yours still has 1:1000000 probablity while the other becomes a 1:2 probablity. Obviously, it's better to switch.
 
Originally posted by Letters
Go with the other door as it has a higher probability of containing the prize. It's easier to understand when it's exagerated. (There's a million doors, you choose one... the host then removes all the doors containing nothing except yours and another one... yours still has 1:1000000 probablity while the other becomes a 1:2 probablity. Obviously, it's better to switch.
Click to expand...

I don't think that works! If there were a million doors and now there is only 2 left, then you've got bugger all chance of the one you picked still being there. You'd have to pick a new one!

:)
 
Your door has JUST as much chance of having it as the other one... 50/50

They could have removed the first door before you picked..same outcome....
 
Originally posted by oldi1knoby
Your door has JUST as much chance of having it as the other one... 50/50

They could have removed the first door before you picked..same outcome....
Click to expand...

I agree.

The "do you want to change your mind?" question is equally the same as "choose one of the remaining doors", of which there are now two that have an equal chance of containing the prize. The chance is 50/50.
 
Originally posted by MrD
I don't think that works! If there were a million doors and now there is only 2 left, then you've got bugger all chance of the one you picked still being there. You'd have to pick a new one!

:)
Click to expand...
No, you misunderstand... the one you choose isn't gotten rid of... hello? Did you read the riddle?
 
No, this same thing was in another forum and some guy linked to a page with a detail descritpion why it was more sensible to change, i think it was somewhere around 90% better probability of winning when changing door.
 
I think you have a better chance when you change doors:
The probability of your picking the door with the prize for the first choice is 1/3.
The probability of picking the door without a prize is 2/3.
Therefore by the time you're ready to pick the door on the second time, you're more likely to have picked a door without a prize for your first choice so you'd be better off switching doors.
 
It doesnt matter, you had a 1/3 chance the first time, anot matter what door you picked, one with nothing is removed. Wouldnt make any sense to switch because it could still be either. Tests have shown that your first instinct is usually right, and not to chnage your mind tihnking about it too much.
 
the second time you pick, it could still be either.
however, is the door that you're at before you make your second choice (which is 50/50 and doesn't count in) more likely to be one with the prize or without the prize?
it's more likely to be the one without the prize since initially there were more without prizes than with prizes, so to get the one with the prize you should pick the other door.

if you set out the possible combinations of choices like so (and allow door a to hold the prize):
if you pick A first, then for 2nd choice you have either
A(stay) or the wrong one(switch) (it doesn't matter B or C; whichever is left is still wrong)
If you pick B first, then for 2nd choice you have
B(stay) or A(switch)
If you pick C first, then for 2nd choice you have
C(stay) or A(switch)
So, if you decide to switch doors for the 2nd choice:
the probability of picking the wrong one is 1/3 (which is the chance that you picked A first)
the probability of picking the right one is 2/3 (which is the chance that you picked either B or C first)
 
Its just a 50/50 shot, if you picked the winner first or one of the worng doors, one of them wrongs is taken away. No matter what. From there you got a 50/50 shot, you can stick with your instinct or change adn hope. Maybe im just dumb, but I think your logic is flawed dfc, you got a 50/50 not 2/3
 
Originally posted by The Terminator
Its just a 50/50 shot, if you picked the winner first or one of the worng doors, one of them wrongs is taken away. No matter what. From there you got a 50/50 shot, you can stick with your instinct or change adn hope. Maybe im just dumb, but I think your logic is flawed dfc, you got a 50/50 not 2/3
Click to expand...
He's right... why can't people just accept it?! Haha...
 
It doesn't matter which door you pick because no one has heard of this door-picking show. With ratings so low, the prize wouldn't be more than $10 american cash money anyways.
 
Hi guys, this is my first post but anyway...
I made a simple program to test this and I found that it is better to change, with 1000 trials it came out as 755 wins and 245 losses if u change every time.
 
Originally posted by Letters
No, you misunderstand... the one you choose isn't gotten rid of... hello? Did you read the riddle?
Click to expand...

oops :eek:

Yeah in that case change your mind, though I can't think why ...
 
Originally posted by C4_Explosive
Hi guys, this is my first post but anyway...
I made a simple program to test this and I found that it is better to change, with 1000 trials it came out as 755 wins and 245 losses if u change every time.
Click to expand...

which is about twice as many wins as losses.
when you're standing there wondering what to do for your second pick and the prize happens to be in door A, you could think that there are three possibilities:
1) you picked A first, therefore you lose if you switch
2) you picked B first, therefore you win if you switch
3) you picked C first, therefore you win if you switch
so there's two chances that you win if you switch and one that you lose.

edit: oops your program didn't give twice as many more wins, more like three times
 
Originally posted by dfc05
edit: oops your program didn't give twice as many more wins, more like three times
Click to expand...

Yeah, thats because i had 4 doors in my program, and two get eliminated after the first choice. The correct door and the first choice is selected randomly.
 
I dunno, it was just easier that way. Well not much difference but who cares.
 
Hows about ressurecting this thread eh!

Heres a new riddle for you to chew over.

1.

I know a word of letters three.

Add two, and fewer there will be.

What is the word?
 
I've got another math puzzle if anyone's interested:

You have in your possession two pieces of string. Let's say that each is a couple of feet long, but it doesn't really matter. And they can both be different lengths, it doesn't really matter either. And they're burnable, like the fuse that they use to, you know, light dynamite. Now you could light either end of either sting, and it would burn. In fact, if you lit one end of a string, it would burn in exactly an hour.

Here's the wrinkle: the strings do not burn at a constant rate. For example, the string might burn for two minutes and then go crazy and burn like mad and then slow down, and da-da-da. You don't know what rate the string's burning at any specific time. All you know is that in an hour's time, the whole sting is burned. It's not linear. And not predictable.

So, the question very simply is with a Zippo light and these two strings, how would you measure 15 minutes of time?
 
do you just get a light and the strings ???? or do we get a watch or something to add in ???
 
I'll make an exception just for you crushenator 500, you can use a watch.
 
Pretty easy, since one string burns in an hour, you just cut it into 4 evenly large pieces and light one of them, or if they both burn togheter an hour then cut one in half and burn it.
 
that doesn't work maxi, b/c the burning rate of any given section is unknown, one of the quarters could burn in 5 sec and another could burn in 45 mins.. all four together have to burn in a total of an hour, but you know nothing about the burning time of any given quarter.

i would try to make some sort of rudimentary sundial with the lighter and making a circle out of the rope, marking arclength off. if you know your latitude (and it's a clear day), you should be able to mark of a 48th (ish) of the arc of the string to be approx. 15 mins. but i think there must be a better way.
 
Not sure if this would work, but do you cut both into 4 pieces (making 8 in total). Light all at once, and when 4 finish burning, then that's roughly 15 mins.
 
how about 30 minutes; does that count, lol.
cause then you just light both ends of the same string and when the fires coming from either way meet up is half an hour... but that's not the question, hmm. *goes back to thinking*

Edit: ok i think i got it... you light both ends of one string and only one end of the second string at the same time. now when the fires from both sides on string A meet is 30 minutes. so string B has also burned for 30 minutes. so at 30 minutes you light the other end of string B. when the fires on string B meet is 15 minutes.
 
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