I have deduced that howdy is now trolling...
DISCLAIMER: THOUGHT INDUCING MATERIAL INSIDE
3,111 Views |
133 Replies |
Last: 16 yr ago by Red Skye
No I really am not.
Their "logic" is illogical.
Their "logic" is illogical.
Imagine if you had to pick one door out of a thousand. After you picked your door, every door except your pick and one other door were revealed to be empty. Would you switch then? I’d sure hope so.
Say you pick A. You can also assign 2/3 probability to A and B (or A and C for that matter, not just A and B). If the host removes C, what does that do to the odds? I can't explain eloquently, but I think that gets to the root of the problem. You can assign the 2/3 probability to any two of the three.
why switch ATM? because your door has beaten the odds 998 times you don't think it can beat the odds again? opening each door is a separate, unrelated event.
If you play the card game, you'll see that we're right. But you won't because you don't want to be proven wrong.
(edit) Forget it. ATMag hit the nail on the head. (/edit)
If you are unwilling to test yourself with the cards, you will be admitting to be a troll.
-------------------------------------------------------
If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.
[This message has been edited by opie03 (edited 10/18/2005 3:22p).]
If you are unwilling to test yourself with the cards, you will be admitting to be a troll.
-------------------------------------------------------
If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.
[This message has been edited by opie03 (edited 10/18/2005 3:22p).]
Guys... you arent going to convince me. Eliminating 1 automatically wrong door is pointless. There is 1 correct and 1 wrong. There lies your choice.
A + B + C
1/3 + 1/3 + 1/3 = 1 First odds
You choose A and B is revealed as not having the car. You are still sitting at 1/3 with A but the 2/3 chance that you are wrong is now all on C. So if you switch, you now have a 2/3 chance of being correct vs your original 1/3.
1/3 + 0/3 + 2/3 = 1 Second odds
1/3 + 1/3 + 1/3 = 1 First odds
You choose A and B is revealed as not having the car. You are still sitting at 1/3 with A but the 2/3 chance that you are wrong is now all on C. So if you switch, you now have a 2/3 chance of being correct vs your original 1/3.
1/3 + 0/3 + 2/3 = 1 Second odds
In order for there to be 1/3 odds. There would have to be a 1/3 chance that the first door chosen actually has the prize behind it. It seems in this example the first door is always empty. So that means theres really only 2 doors that have a chance to win hence 50%
Does this make any sense? Someone explain, I am curious.
Does this make any sense? Someone explain, I am curious.
Playing the card game and getting anything but my answer proves you arent doing it randomly. Sorry guys I know my statistics. You have faulty logic when you know 1 of 2 remaining is going to be eliminated and then switching.
If you choice the correct one off the bat, 1 of the 2 remaining is STILL ELIMINATED.
If you choice the correct one off the bat, 1 of the 2 remaining is STILL ELIMINATED.
Thanks, Hagen95. I thought just explaining it would be clearer than using equations, but that is exactly correct.
dan... that is exactly what I am saying.
howdy,
in response to your two door theory. If they "tested" the door not chosen and then asked if you wanted to switch then obviously you would get it correct 100% of the time. the whole point is the "test"
it is very counterintuitive, but it makes some sense.
in response to your two door theory. If they "tested" the door not chosen and then asked if you wanted to switch then obviously you would get it correct 100% of the time. the whole point is the "test"
it is very counterintuitive, but it makes some sense.
PA = 1/3
PnotA = 2/3
its that simple.
PnotA = 2/3
its that simple.
quote:
A + B + C
1/3 + 1/3 + 1/3 = 1 First odds
You choose A and B is revealed as not having the car. You are still sitting at 1/3 with A but the 2/3 chance that you are wrong is now all on C. So if you switch, you now have a 2/3 chance of being correct vs your original 1/3.
1/3 + 0/3 + 2/3 = 1 Second odds
No because during the first choice it is
A + B + C
1/3 + 1/3 + 1/3 = 1
The fact that one of the two not chosen is elimated does NOT make make the other jump to 2/3s. That is asinine.
quote:That's what your intuition tells you, but it is wrong. Try the card game.
In order for there to be 1/3 odds. There would have to be a 1/3 chance that the first door chosen actually has the prize behind it. It seems in this example the first door is always empty. So that means theres really only 2 doors that have a chance to win hence 50%
But the first door that the "host" chooses doesnt really have 1/3 odds, because it never wins...
Right?
So then the 1/3 + 1/3 + 1/3 thing doesnt hold up.
Right?
So then the 1/3 + 1/3 + 1/3 thing doesnt hold up.
The best explanation was done by aTmAg.
Forget about it in terms of 3 doors... Think of it terms of 100 or 1000 doors. If you pick one and all of the doors minus the one you picked and one other door are revealed to be empty, you would be crazy not to switch to the other door. The odds are overwhelmingly in your favor if you switch your choice.
Try the card game and track what your results. You will win twice as often by switching. Trust me.
Forget about it in terms of 3 doors... Think of it terms of 100 or 1000 doors. If you pick one and all of the doors minus the one you picked and one other door are revealed to be empty, you would be crazy not to switch to the other door. The odds are overwhelmingly in your favor if you switch your choice.
Try the card game and track what your results. You will win twice as often by switching. Trust me.
quote:But entirely correct.
The fact that one of the two not chosen is elimated does NOT make make the other jump to 2/3s. That is asinine.
Dan, the host will get it right 1/3 of the time. Point being that the other door will be correct the other 2/3 of the time.
Ok, im about to do the card thing. Do you have to know which one to take out on the first one.
What if that one wins?
What if that one wins?
quote:
That is asinine.
Mark that for later.
I can get you the mathematical proof if you really want it. This has been discussed for over 30 years. Do you think you're discovering some new truth here or something?
I give up. This is a no win situation, because I can not explain it any simpler.
I can probably find a mathematic proof that shows it is wrong, since this has been debated for 30 years appearenty.
A + B + C = 1
1/3 + 1/3 + 1/3 = 1
If B = 0/3, then A + C = 1
You're assuming C = 2/3 because that's how you choose to group them. If you initially say there is 2/3 chance that the right answer is A or B, you pick A, and B is removed, the odds of A then become 2/3!
1/3 + 1/3 + 1/3 = 1
If B = 0/3, then A + C = 1
You're assuming C = 2/3 because that's how you choose to group them. If you initially say there is 2/3 chance that the right answer is A or B, you pick A, and B is removed, the odds of A then become 2/3!
I will make my last point the quote that proves us all wrong...
quote:
Now, a clever man would put the poison into his own goblet, because he would know that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the wine in front of you. But you must have known I was not a great fool, you would have counted on it, so I can clearly not choose the wine in front of me.
Howdy:
If you had 100 doors and you chose one... you would agree that you have 1/100 chance of being correct right? That means there is 99/100 chance that one of the other doors is the right one.
Revealing the 98 other doors as empty does not change the odds! You still have a 1/100 chance of having chosen the correct door the first time. While the other door has a 99/100 chance of being the right one.
If you had 100 doors and you chose one... you would agree that you have 1/100 chance of being correct right? That means there is 99/100 chance that one of the other doors is the right one.
Revealing the 98 other doors as empty does not change the odds! You still have a 1/100 chance of having chosen the correct door the first time. While the other door has a 99/100 chance of being the right one.
OK i did the cards 15 times(i know its not much of a sample)
Host won 5 times
when i switched i won 4 times and lost once
when i didnt switch i didnt win 4 times and won 1 time
SO i guess this is kinda proving the theory right, although i still dont understand it.
Host won 5 times
when i switched i won 4 times and lost once
when i didnt switch i didnt win 4 times and won 1 time
SO i guess this is kinda proving the theory right, although i still dont understand it.
When you always switch, you only lose when you had the car picked with your first choice, in which case switching gets you off the car.
This happens 1/3 of the time, the other 2/3 of the time you win.
This happens 1/3 of the time, the other 2/3 of the time you win.
i dont see what the doors has to do with it?
What if the 63rd door has it behind it
What if the 63rd door has it behind it
Pwned.... but 1/3 of the time the one that is revealed first will win.
Right?
Right?
The 1/1000 example really hits it home. I believed it at first, but I didn't "get it" until then. It makes sense now.
Well if we're using Vizzini as our argument base.
Howdy: INCONCEIVABLE.
Texags: You keep using that word. I do not think it means what you think it means.
Howdy: INCONCEIVABLE.
Texags: You keep using that word. I do not think it means what you think it means.
quote:
If you had 100 doors and you chose one... you would agree that you have 1/100 chance of being correct right?
What if your choice was the right one? then when there are two cards left, yours has a 100/100 chance of being the winner and the other card has a 0/100 chance of winning, why would you switch?
It seems we're assuming our first choice is always wrong.
Featured Stories
See All
Potential of a 300-yard three-peat would be a Texas A&M first
by Olin Buchanan
46:01
10h ago
5.8k
6:32
6h ago
1.1k
12:32
11h ago
3.6k
DeangeloVickers
Dan Campbell has to sell his home due to fan harassment
in Billy Liucci's TexAgs Premium
147