🚀 go-pugleaf

In article <C0Cx87.KDE@hplntx.hpl.hp.com> potamian@hpl.hp.com writes:
=>Lets generalize:
=>    a) There are N doors
=>    b) M of them contain a prize
=>    c) Monty opens K doors 
=>
=>(obvious restrictions: M<N, K <= M-N-1 -- the last one guarantees that
=>Monty will have at least k doors to open)
=>
=>If you don't switch, the probabiliy of winning is equal to the probability
=>of having selected a "correct" door in the first place, i.e.
=>
=>	P_no_switch = M/N
=>
=>--------------------------------------------------------------------------
=>Now, lets find out the probability of winning if you switch:
=>There are two possibilities:
=>
=>    a) The door you have selected contains a prize. The probability
=>       for this event is M/N.
=>       Monty opens K doors, so there are N-K-1 doors you can switch
=>       to, and M-1 of them contain a prize.  The probability of
=>       choosing one of them is (M-1)/(N-K-1).  So, the total
=>       probability of selecting the door with a prize, switching,
=>       and selecting another door with a prize is:
=>
=>       P1 = [ M/N ] * [ (M-1)/(N-K-1) ]
=>    
=>    b) The door you have selected does not contain a prize (probability
=>       is (N-M)/N.
=>       Out of the remaining N-K-1 doors, M of them contain a prize,
=>       and the probability of choosing one of them right is M/(N-K-1).
=>       So the total probability of selecting a door without a prize,
=>       and switching to another door that contains a prize, is:
=>
=>       P2 = [ (N-M)/N ] * [ M/(N-K-1) ]
=>
=>The probability of winning if you switch is then:
=>
=>
=>                                M * (N-1)
=>	P_switch = P1 + P2 = ... =  --------------
=>                                N * (N-K-1)
  
Excellent! Mathematicians always need generalization ;-)

=>An interesting thing to note is that:
=>
=>    P_switch > P_no_switch 	(for K<>0)
=>
=>So, again, it is always a good idea to switch!!!
 
Unless you want to win something behind the N-M doors, when N-M > N/2,
and K is number of doors where there *is* a price you want than it's better
not to switch. Suppose you do want to win a goat:
3 doors, 2 goats, 1 car: p(goat-no switch)=2/3 but p(goat-switch)=1/3 !!
Better not switch in that case!

=>It is obvious that *in general* the sum
=>
=>				   M * (2N-K-2)
=>	P_no_switch + P_switch = ----------------
=>				   N * (N-K-1)
=>
=>is not equal to 1, because:
=>
=>P_switch    = prob. of "switch & win"
=>P_no_switch = prob. of "don't switch & win"
=>
=>These two events are not exclusive (in general). To see why,
=>lets say that there are a zillion doors (N is very large),
=>one of them contains the prize, and Monty opens K=1 door only.

Exactly, p(win)+p(no win)=1 are the complementary events. 

Leon
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"De wereld is mijn vaderland          Email: leon@nematic.chem.ubc.ca
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