Wiki - The Monty Hall problem
 

http://en.wikipedia.org/wiki/Monty_Hall_problem

Okay people - debate - i say they are talking buttkiss because after the host opens the door and shows that there is a goat behind it there is a 50/50 chance of the player winning at this point - so therefore switching wont increase the odds of winning.

Right? :p:

Re: Wiki - The Monty Hall problem
 

Well I say you still have / had the same chance 1 in 3, just beacuse you now know 1 door is a goat it doesn't change ... does it ??

Re: Wiki - The Monty Hall problem
 

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.

Re: Wiki - The Monty Hall problem
 

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!)

Re: Wiki - The Monty Hall problem
 

Exactly..... (if you dismiss what graham said). :D
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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.

Re: Wiki - The Monty Hall problem
 

OK so let's consider the implications of combining the two options- which up until now, I intentionally hadn't.

The "rules" dictate that when combining the probs they are multiplied... so:
- pick one door, there's a 1/3 chance of the car (which after all is what you want, unless you happen to be from the south-west :D) let's call the door you chose door 1.
- host opens one of the goats leaving you a 1/2 chance of the car being in either. let's call this opened door 2.

viz:
- we stick with door 1 then we have a 1/3 x 1/2 (= 1/6) chance of the original guess being right.
- we change to door 3 then we have a 2/3 x 1/2 (= 2/6) chance of being right.

and:
- from choosing either door after the opening the chances are the same for either door. And for either door, they are increased anyway from the original 3-door problem.
- from following through the guess then changing then we have an increased chance if we switch.

It's counterintuitive. Their logic / stats are correct but their explanation is a tad flawed.

And all that neglects the psychological effect of shall i switch or not!
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Excuse my pedantry but it's 1:99.

Re: Wiki - The Monty Hall problem
 

Ah :PP: you knew what I meant! :grind: ;)