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Old 12-07-2001, 03:18 PM   #131
Campino
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Join Date: November 12, 2001
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Why has Barry asked this question???
This is mental torture [img]smile.gif[/img]
Is there a mathsteacher on this forum,PLEASE????
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Old 12-07-2001, 03:24 PM   #132
Sir Kenyth
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quote:
Originally posted by Campino:
Why has Barry asked this question???
This is mental torture [img]smile.gif[/img]
Is there a mathsteacher on this forum,PLEASE????



Once you discover what a variable is, it's no longer a variable and you must re-write the problem to take that into account.
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Old 12-07-2001, 05:15 PM   #133
Talthyr Malkaviel
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Join Date: August 31, 2001
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Yeah, Sir Kenyth is right, it can't be 33% each on the last remaining doors because you know one of them, it is no longer taken into account, if you know for sure it's definitely not behind that one door, it's got to be an even spread between the 2 which are left.
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Old 12-08-2001, 03:11 AM   #134
LennonCook
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Join Date: November 10, 2001
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quote:
Originally posted by Campino:
Why has Barry asked this question???
This is mental torture [img]smile.gif[/img]
Is there a mathsteacher on this forum,PLEASE????



Well... im not a maths teacher, but i am thinking of becoming one... and i have read many many books on stats, it is my main area of concern. So i know what im talking abut when i say things like mutually exclusive.
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Old 12-08-2001, 05:11 AM   #135
Barry the Sprout
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Join Date: October 19, 2001
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quote:
Originally posted by Campino:
Why has Barry asked this question???
This is mental torture [img]smile.gif[/img]
Is there a mathsteacher on this forum,PLEASE????



Bwhahahahahahaha!!!!!! You shall all fall before my incomprehensible maths problems!!!

Seriosly guys. This discussion took place via newspaper letter pages between extremely good academic mathematicians. They argued for ages but the 33% 66% view is now widely held. It is not 50/50 trust me!
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Old 12-08-2001, 05:43 AM   #136
250
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quote:
Originally posted by Barry the Sprout:
Applause for Legolas I feel. Nice one mate!

OK, heres a probability one for you. You are playing a game show and the host asks you to pick your prize. There are three doors you can pick and one of them has a prize behind it. The other two have chickens (which, I admit, would be a prize to some people but not to most).

He asks you to guess so you pick one. He then opens one of the other doors to reveal a chicken.

He asks you if you are still sure.

On the basis of pure probability should you stick or change?



joint little late, but this is actually something I studied in Calc III

the answer is should change, because it doubles the chances of getting a prize

pick randomly is: 1/3
switch is: 2/3

so always switch
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Old 12-08-2001, 06:06 AM   #137
SixOfSpades
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When all else fails, run a simulator.

So I did. My little Pascal program puts a washer/dryer behind one random door, and a goat and a pair of socks behind the other 2 doors. It then picks a random door, and kicks out one of the 2 losing doors.

Out of 1000 trials where the computer stayed put with its first choice, it won the prize 351 times.

Out of 1000 trials where it decided to switch to the remaining, unopened door, it got the washer/dryer 657 times.

The results are counter-intuitive, I know. That's what makes it a PUZZLE! It does make sense, though, when you look at it right: Let's say you pick Door A. Now, there's a 33% chance the prize is behind Door A, and a 66% chance it's behind (Door B OR Door C). These numbers will NOT change. When the gameshow host opens Door C to reveal a broken Rubik's Cube, the chance of Door B OR Door C being the winner is STILL 66%.....except that Door C has been eliminated, leaving that 66% sitting behind Door B.
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Old 12-08-2001, 09:02 PM   #138
LennonCook
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The results WILL change- because no matter what happens ur left with a right door and a wrong door.
At the first Choice- 1/3 Chance of Getting It right, 2/3 Getting it wrong.
Second Choice- U MUST BE LEFT WITH A RIGHT AND A WRONG DOOR. The argument for u changing is that once u have made a choice the probablility of it staying right remains the same EVEN IF THE CONIDTIONS ARE CHANGED. You do not have any more chance with one than u do With the Other. IF THE CHANCES WERE TH SAME BETWEEN CHOICES, THE CONIDITIONS WOULD HAVE TO CHANGE ASWELL. I will explain this in more statistical ways in a little while when i can think of how to word it.
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Old 12-08-2001, 09:19 PM   #139
Dresdan
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Join Date: November 18, 2001
Location: Bay City, MI. , U.S.A.
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quote:
The results are counter-intuitive, I know. That's what makes it a PUZZLE! It does make sense, though, when you look at it right: Let's say you pick Door A. Now, there's a 33% chance the prize is behind Door A, and a 66% chance it's behind (Door B OR Door C). These numbers will NOT change. When the gameshow host opens Door C to reveal a broken Rubik's Cube, the chance of Door B OR Door C being the winner is STILL 66%.....except that Door C has been eliminated, leaving that 66% sitting behind Door B.


The way I see it is that there is more than one math problem here.
In the first problem you have three doors meaning that you have a 33.3..% chance of choosing correctly. If you fail to do so you are wrong. So the new the problem is that you have two doors to choose from which means that you have a 50% chance of choosing correctly.
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Old 12-08-2001, 09:52 PM   #140
SixOfSpades
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The results say you have a 66.6% chance of winning the prize if you switch your guess. End of story.
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