That's how I was thinking. I think Glenn's approach is
more intuitive but there are a lot of nonintuitive results in
probability. I first applied the coin toss because I thought
these were independent events and then the way Glenn described it
I could see how it may be nonindependent events but have no idea
how to calculate . Just the odds would be somewhere between my guess of 50 and
Glenn's 99.
So it doesn't matter that it is the same person guessing with the
two doors its still independent events? I can still kind of
see Glenn's point that person does seem to have a little more
info then if a 2nd person came out (who didn't see the first odds)
and picked, that definitely is 50-50.
Erin
>Ah, OK, I see the original puzzle. Jonathon is right: with the new
>information you now know that the good door is one of the remaining two.
>While the odds HAVE changed, from 1-in-3 to 1-in-2, switching your choice
>will neither increase nor decrease the odds of picking the right one.
>Amazing that anyone would think it might...
>
>Lawry
>
>
>> -----Original Message-----
>> From: owner-moq_discuss@venus.co.uk
>> [mailto:owner-moq_discuss@venus.co.uk]On Behalf Of Jonathan B. Marder
>> Sent: Thursday, July 11, 2002 8:25 AM
>> To: moq_discuss@moq.org
>> Subject: RE: MD Let's Make a Deal
>>
>>
>> Hi Rick, Erin, Glenn,
>>
>> I've been in a discussion on this before. The fact that Monty provides
>> new information DOES change the statistics. You now know that door #1 is
>> worthless, so the stats are as they would be if you knew that in the
>> first place - it's 50/50 on door #2 vs. #3.
>>
>> The interesting thing about the previous debate I saw on this was that
>> nobody could agree on the solution . . . until someone ran a simulation
>> and confirmed empirically that the chance of guessing right is indeed
>> 50%. So much for theory!
>>
>> Jonathan
>>
>>
>> ---Original message from Rick [VALENCE] ---
>>
>> Here's the Setup:
>> Imagine you're a contestant on Let's Make a Deal. Monty Hall calls
>> you up to the stage and explains the game to you. He tells you there
>> are three doors. Door #1. Door #2. And Door #3. Behind two of the
>> doors (he won't tell you which of course) are worthless gag prizes, but
>> behind the third is a valuable prize (for this group, we'll imagine it's
>> an unlimited one-on-one Q&A with Robert Pirsig as he sails you down the
>> Hudson River on the Arête).
>> Monty asks you choose a door... you pick door #2. Monty says, "Well
>> it's a good thing you didn't pick door#1." Door#1 opens and you see one
>> of the gag prizes revealed (let's say... a goat in a wheelbarrow).
>> Now you're down to your chosen door (door#2) and the remaining door
>> (door#3). Monty says, "I'll give you $100 to switch to door#3.... I'll
>> give to $200...$300...etc...etc."
>>
>> Here's the Question:
>> Does switching doors improve your odds of winning?
>>
>> Here's the Possibilities:
>> 1. Switching won't help. It's a 50/50 chance. Door #2 or Door #3.
>> Switching don't mean diddley.
>>
>> 2. Switching will help. You started with 3 doors. 2 bad and 1 good.
>> Odds are, you picked a bad one to begin with. So odds are, if you
>> switch, you're switching to a good one.
>>
>>
>>
>> Can anyone crack this one for me? Does switching improve the odds?
>>
>>
>> thanks,
>> rick
>>
>>
>>
>> MOQ.ORG - http://www.moq.org
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>>
>> To unsubscribe from moq_discuss follow the instructions at:
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>>
>
>
>
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>
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