29 August 2005
20 September 2005
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Which Door has the Cadillac
puzzle:
Given that a valued prize (a Cadillac automobile) is hidden
behind one of three
doors. You must select correct door to win the prize, however
once you make your initial guess a knowing game host reveals one of the
empty doors. At this point you are given the opportunity to
change your initial guess in favor of the only remaining door.
What should you do, and what is the probability of obtaining the
prize given this choice?
answer:
Clearly you begin with the simple odds of one in three, however the
knowing host's act of revealing a dud choice may lead you astray
into considering the resulting situation a game of simple odds (2 options 1 choice)
and an incorrect answer. Consider
instead, that the
probability of all the states any system sums to 1. As the
original probability is 1/3
goes to your initial guess the other 2/3 must evenly
distributed to the remaining two doors ... that is until the
knowing revelation of a dud door. Following this, the
probability redistribution of that 2/3 must be re-accounted and it may
only transfer to only the unselected door. You should select this
so far untouched door as it now
holds a 2/3 chance of being the prize door.
discussion:
If this all seems unintuitive consider that mathematical genius Paul
Erdos was among the fooled (bio. The
Man Who
Loved Only Numbers, ref. Decision
Sciences). Quantum Chemistry is entirely based on implication
of state with respect to probability. Forbid a state (as in
expose a empty door) and instantaneously the probability re-expresses
itself to the now permissable states. Why does the probability
not equally allocate itself amoung your pick and the remaing unknown
door, resulting in a 1/2-1/2 split? Here is an observation that
would forbid this to occur. If the puzzle were to end without you
getting any more knowlege then you would expect to win on average 1 in
3 times. Likewise it would be identical, if you were given a
chance but decided never to change your mind. You initial choice
is and always will be until and unless
you have and take an opportunity to change 1 in n. Where n is the total number of
doors. Let n =
1,000,000 (= 1M). You would still expect to win 1 in n times. Play random games
exposing the unchoosen doors, expose 23,345-120,000 if you didn't
happen to pick a number there or even if you did. Pick odd
numbers if you picked even, expose doors as multiples and or factors of
the number you choose. Expose by any method you choose it will
not change the game and the underlying probability. Its your
lucky day when given a chance to revise your guess when the knowing
game host is compelled to inject knowlege into the system. From
the perspective of
this person who is compelled to expose a door there are two significant
cases. Case 1: [1 in 3 probability] the game host
gets lucky and has the opportunity to expose one of two losing
doors. Case 2, [ 2 in 3 probaility] you have started
out with a losing choice and the game host must
consequently route out the other loser, leaving the winner.
If you still are uncovinced or just as a fun exercise download my CadillacProgram.cpp
examine that it follows the rules and
run
a simulation of
it for yourself. For those without a C++ compiler here's a peek
at the results.
$ time
./CadillacProblem.exe 1073741824
of 1073741824 trials
Strategy "keep first choice" wins : 357915152
(33.333446)
Strategy "Switch door" wins :
715826672 (66.666554)
----------- ----------
total : 1073741824 (100.000000)
real 11m46.656s
user 0m0.015s
sys 0m0.000s
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