🚀 go-pugleaf

packer@amarna.gsfc.nasa.gov (Charles Packer) writes:

>Wait a minute, I thought I understood this a few months
>ago when it was discussed here. But now I write down what
>I understand to be the sample space and it says: don't
>bother to switch, it's a fair game:

>Assume for simplification that the prize is always behind door A.
>Probabiliy of each outcome is listed:
>I choose A, host chooses B, I stay with A 1/12 win
>I choose A, host chooses B, I switch to C 1/12
>I choose A, host chooses C, I stay with A 1/12 win
>I choose A, host chooses C, I switch to B 1/12
>I choose B, host chooses C, I stay with B 1/6
>I choose B, host chooses C, I switch to A 1/6  win
>I choose C, host chooses B, I stay with C 1/6
>I choose C, host chooses B, I switch to A 1.6  win

>Add the "win" probabilities together, and they come out
>to .5. I must have forgotten something...

Your analysis of the problem is entirely correct, and indeed
illuminating, and shows what happens if you toss a coin to decide your
strategy.  

Your interpretation of the solution to the problem is incorrect.  The
reason it pays to switch is because the host's action is forced if you
pick an incorrect door, so the host's action gives you information
about where the prize actually is which you can exploit in determining
your strategy.  Adding the probabilities of winning together above is
an answer to the problem `what if I tossed a fair coin to decide my
strategy?' (ie.  ignore the host's information in determining my
strategy).  The correct interpretation of P(win|switch) is to delete
all the "stay" lines above, and multiply the remaining probabilities
by 2 (since P(switch) has just increased by a factor of 2), and then
it becomes clear that

P(win|switch)=1/3 + 1/3,

these components coming from the situations where the host's action
was forced (ie. informative).  This is not surprising, since a-priori
there is a 2/3 chance that the host's action will tell you where the
prize is (ie. if you initially pick a loser).

Steve.
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