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Originally Posted by nffc
eww, no.
3 doors, 2 goats, 1 car.
Dood chooses which could be either a goat or a car... 1/3 it's the car, 2/3 it's the goat.
Host opens other which reveals goat. We know 1 of the three doors is the goat.
For each of the other two doors there's now a 50:50 (1/2) chance that the door he's picked is the car. There's an equal probability it's the goat.
If anything by NOT switching his chances of the car being behind the original pick has INCREASED from 1/3 to 1/2.
That make sense... (to add to what Graham said!)
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Exactly..... (if you dismiss what graham said).
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Originally Posted by Graham
Wrong.
See the example lower down the page of increasing the number of doors.
If there are 100 doors your chance of picking the right one are 1:100. If the host opens one door and it's not the right one your chances have "improved" (only slightly! But improved none the less) to 1:99
If he opens *all* the other doors bar one however (which is the equivalent of what actually happens) then the odds change from 1:100 for the first door to 99:100 for the other one!
You *may* have got lucky on that first 1:100 shot, but it's much more likely that switching will give the prize.
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Nope - your wrong - if the host has already shown you that one door is a goat then there are only two doors to pick from which represents a 50/50 chance at that point in time. Therefore it is no longer a 1/3 shot its a 50/50 shot and changing wont help you at all. Its still either the right door or the wrong door from the choice of the two.
How can the odds be more in your favor if you switched? Say at the start you picked the other door of the two that will be left after the host shows you a wrong choice your chances of it being the correct door are still the same as if you selected the other door - nothing has changed except the number of doors available - instead of 3 there are now two - therefore at that point in time the odds are 50/50.
Madness to think otherwise - that would require a third door.