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  #1  
Old 06-10-2005, 03:53 AM
Scotch78 Scotch78 is offline
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Default Three Doors

There are three doors (A, B and C) in front of you. One, and only one, contains a prize. You choose a door, say A, and an empty door, say C, is revealed. Some girl is insisting to me that she took a stats course that told her B is now a 2-1 favorite over A. I'm telling her she's fool of it, but she keeps insisting her textbook proved it. Of course, she doesn't have the proof anymore. Anyone?

Scott
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  #2  
Old 06-10-2005, 05:24 AM
mantasm mantasm is offline
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Default Re: Three Doors

SHe's talking about the monty hall problem, try googling that
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  #3  
Old 06-10-2005, 07:28 AM
Scotch78 Scotch78 is offline
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Default Re: Three Doors

Ok, so from what I found, the answer really does appear to be 33-67. So, there must be a fallacy somewhere in the 50-50 argument. Has it been identified yet?

Scott
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  #4  
Old 06-10-2005, 08:03 AM
Arnfinn Madsen Arnfinn Madsen is offline
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Default Re: Three Doors

[ QUOTE ]
Ok, so from what I found, the answer really does appear to be 33-67. So, there must be a fallacy somewhere in the 50-50 argument. Has it been identified yet?

Scott

[/ QUOTE ]

2/3 of all instances you choose a door which does not contain the price. In those 2/3-instances the door not opened has a 100% of containing the price. In the 1/3-instances that you choose the correct door, the other door not opened has 0% chance of containing the price. Thus she is right (2/3 vs 1/3 is 2 vs 1).

Tell her you have never met such a smart girl and invite her for dinner [img]/images/graemlins/cool.gif[/img].
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  #5  
Old 06-10-2005, 08:46 AM
kyro kyro is offline
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Default Re: Three Doors

Welcome to the Probability section. I can see this is your first time here.
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  #6  
Old 06-10-2005, 10:57 AM
mostsmooth mostsmooth is offline
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Default Re: Three Doors

[ QUOTE ]
Welcome to the Probability section. I can see this is your first time here.

[/ QUOTE ]
welcome to earth would have fit better. i didnt think there was a person alive that hasnt heard this one. should we let him in on the 3 guys and the bellboy and the missing $1? [img]/images/graemlins/cool.gif[/img]
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  #7  
Old 06-10-2005, 12:00 PM
BruceZ BruceZ is offline
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Default Re: Three Doors

[ QUOTE ]
There are three doors (A, B and C) in front of you. One, and only one, contains a prize. You choose a door, say A, and an empty door, say C, is revealed. Some girl is insisting to me that she took a stats course that told her B is now a 2-1 favorite over A. I'm telling her she's fool of it, but she keeps insisting her textbook proved it. Of course, she doesn't have the proof anymore. Anyone?

Scott

[/ QUOTE ]

The way you phrased the problem, the answer could be either 1/2 or 2/3. It depends on whether or not the empty door was chosen randomly from all 3 doors so that the prize door could have been revealed, or if only an empty door could have been chosen. The later is the way the game is supposed to be played, and then the answer is 2/3.
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  #8  
Old 06-10-2005, 12:35 PM
kyro kyro is offline
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Default Re: Three Doors

[ QUOTE ]
[ QUOTE ]
Welcome to the Probability section. I can see this is your first time here.

[/ QUOTE ]
welcome to earth would have fit better. i didnt think there was a person alive that hasnt heard this one. should we let him in on the 3 guys and the bellboy and the missing $1? [img]/images/graemlins/cool.gif[/img]

[/ QUOTE ]

Only if he tells us the odds of two people flopping a flush draw.
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  #9  
Old 06-10-2005, 01:24 PM
Scotch78 Scotch78 is offline
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Default Re: Three Doors

[ QUOTE ]
Only if he tells us the odds of two people flopping a flush draw.

[/ QUOTE ]

Assuming there are two people dealt two of the same suit as each other . . . with four of the suit seen and 48 unseen cards, I'm going to say it's ((9/48 + 9/47) * 8/46) - (9/48 * 8/47 * 7/46) = 6.1%.

And my guess for two people sharing four of the same suit is 12/51 * ((11/50 * 10/49) + ((13-m)/(52-n) + ((13-m1)/(52-n2) + . . .) where the m series represents the number of said suit previously dealt and the n series represents the number of total cards dealt, with ((n/2) - 1) = x terms in the parenthetical where x equals the number of players. I'm sure there's a formulaic way to write all that, but I'll let the mathematicians in here deal with that; I'm just a philosopher.

Now, what's this other brain teaser?

Scott
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  #10  
Old 06-10-2005, 01:28 PM
kyro kyro is offline
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Default Re: Three Doors

Touche...

3 guys stay at a hotel room. It comes to $30. The guys give the manager $10 each. Later, the manager realizes he overcharged the 3 men and sends his bellboy to the room to return $5. The bellboy, (who is a dishonest fellow), only gives $3 back to the men and pockets the other $2.

So the men each paid $9 for the room, or $27 altogher. The bellboy kept $2. $27 + $2 = $29.

Where'd the missing dollar go?
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