🚀 go-pugleaf

Leon ter Beek (leon@unixg.ubc.ca) writes:

>I've made up another problem, similar to the Monty Hall problem. The
>similarity is very obvious, but it does give it a different twist,
>though.
>
>Disclaimer: First, I have been reading this group for just two days, so
>excuse me if a similar extension has already been posted.
>Second, I don't claim this to be a smart problem (what do you expect ;-)).
>
>There are five doors, behind two of them a car, and behind each of the
>other three doors a goat. You pick a door and the quizmaster (who knows
>what's behind the doors) opens a door with a goat behind it.
>1) Should you change doors?
>2) Why is the sum of the probability of changing and not changing not
>equal to one?

MONTY NEVER DIES!!!!!
LONG LIVE MONTY!!!!!!

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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)

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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

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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)

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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!!!

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For N=5, M=2, K=1 we get:

    P_no_switch = 2/5
    P_switch    = 8/15

For N=3, M=1, K=1 we have the original puzzle, and we get:
    
    P_no_switch = 1/3
    P_switch    = 2/3

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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.

Then
P_no_switch = 1/N = a very small number.
P_switch = a slightly bigger, but still very small number.
	   (because K<<N, the doors that Monty opens don't give
	   you too much information)

Obviously their sum is less than 1.

In the case of the original puzzle (N=3,M=K=1), it so happens that these
two events are exclusive:

P_switch    = prob. of "switch & win" = prob. of "first door has no prize"
P_no_switch = prob. of "don't switch & win" = prob. of "first door has prize"

The events "first door has no prize" and "first door has prize" are
obviously exclusive. This also happens when M=1, K=N-2.
However, I suspect that these are not the only values of N,M,K that
satisfy the equation "P_no_switch + P_switch = 1"...
Has anybody a better idea????

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If we really want to generalize this problem, we must consider non
uniform distribution for the probabilities of a door containing a
prize, and of Monty choosing the empty doors.

I have solved this problem too, but there is not enough room in this post to
write the proof :-) :-) :-)

Spyros Potamianos
potamian@hpl.hp.com