20 September 2005

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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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