As many have prefaced this so far, "I wasn't going to chime in, but..."
(now you've CHANGED THE RULES drastically enough I feel I *must* make a
statement...)
> -----Original Message-----
> From: Gavin Scott [mailto:[log in to unmask]]
>
> Ok, I know this is cruel, but I'm going to do it anyway
> because Stan just
> came in and made *me* figure it out.
>
> We've all pretty much got the basic Monty question down now, so
> let's move on to a slightly more complicated version (which I
> suppose we
> should call the Full Monty).
>
> Three contestants, three doors. Each contestant picks a door.
>
> Monty opens one (guaranteed to be non-winning) door, and
> eliminates the
> player who chose that door.
>
> Now there are two contestants left who each have an unopened door.
>
> Each appears to be in the same position as the contestant in
> the original
> problem. Each has chosen a door and seen one of the
> remaining doors opened.
[KEY word here is "APPEARS"]
> If both are attentive HP3000-L readers, shouldn't they *both*
> think that
> they should switch to the *other's* door in order to increase
> their chances
> of wining from 1/3 to 2/3?
>
> Should they trade doors or not?
Like all good answers before, "it depends..."
IF: the THIRD player was eliminated, player 1 should switch and player 2
should stay
LIKEWISE: if the SECOND player was eliminated, p1 should switch and p3
should stay
HOWEVER: if P1 was eliminated, there is no advantage in switching. [err, I
think...]
Remember, the KEY word was that each player APPEARS to be "in the same
position", but that's not really true. ONLY the first player to pick is "in
the same position" -- remember, each player has to pick a different door
[this is a NEW CONSTRAINT that is ADDED by this variation], so players #2 &
#3 don't get the same odds as player 1 [more on this later]
At the outset, it CAN BE SHOWN that each player has a 1-in-3 chance of being
correct:
* P1 picks a door -- 1-in-3 he's right; players 2 & 3 now pick from doors
that COLLECTIVELY have a 2 in 3 chance of being correct.
* P2 picks a door. Since he cannot choose the door that P1 chose [due to
that pesky NEW CONSTRAINT], he has 1-in-2 chance of picking what still might
be "the winning door", and that "winning door" will be found 2/3rds of the
time; 50% of 2/3rds is [amazingly] 1-in-3.
* P3 has no choice -- he is given the final door and STILL has the "same"
1-in-3 as everyone else. [because players 1 & 2 ALSO COLLECTIVELY have a
2/3rds chance of being correct "so far"]
Now, Mr. Hall INTRODUCES NEW INFORMATION which CHANGES THE STATE OF THE
SYSTEM
suppose Mr. H eliminates P3 (who had no choice in the matter), where do we
stand now?
-- P1 chose 1 from THREE unknowns [1/3 of the
3-out-of-3-potentially-winning doors]
-- P2 chose 1 from TWO unknowns [1/2 of the REMAINING 2-of-3 doors, or
1/3]
-- P3 chose 1 from ONE unknown [ALL of the final 1-of-3 doors, again
1/3] and was eliminated. This gives us NEW INFORMATION about P2's choice:
P2 chose the "good" half of the 2-in-3, which for all practical purposes is
now "all" of the 2-in-3 chance that the second and third doors COLLECTIVELY
had of being "the correct door". P1 switches provided he can convince P2 to
switch [who, if being the "astute" 3000-L reader he is, has read THIS
message & refuses to do so... ;)]
Should Mr. H eliminate P2, then P3 should hold firm for the above reason.
[this time p3 was "forced" to take the "good" half of the (collective)
2-in-3 chance]
Finally, we get to the case of P1 being eliminated. this NEW INFORMATION
tells us something quite interesting:
-- P2 chose 1 from TWO possibilities, which we NOW KNOW have a 100%
chance of COLLECTIVELY containing the winning door, not 2/3'rds as before.
With this NEW INFORMATION, we NOW KNOW P2 has a 1-in-2 chance
-- likewise, P3 had the other 1-in-2 of what is NOW 100% guaranteed to be
behind either the second or third door chosen.
It is this final case where it makes no difference to switch.
Well, while writing this I see 5 new entries on the topic came in, so I
might as well hit >>send<< now...
Tom
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