Monty Hall problem - SP edition

I don't see why it doesn't split into a 50-50 chance. Couldn't you say either door now has a 2/3 chance of having a ? Also, You could possibly say that you knew your door had a 1/3 chance of not having a and now, since a door was eliminated, that your door has a 2/3 chance of not having a . The numbers can be manipulated too much; just choose a door.

EDIT: Another thing I can point out, you can say the other door had 1/3 chance of not getting the chinchilla. Therefore, once a door is eliminated, using that logic, you could effectivly say the other door now has a 2/3 chance of not getting the . The numbers can go anyway you want them to, as I see it.

Door A: Good
Door B: Bad
Door C: Bad

You choose 1 door and can either switch or stay. That yields 6 possibilities

Choose A + Stay = Win
Choose A + Switch = Lose
Choose B + Stay = Lose
Choose B + Switch = Win
Choose C + Stay = Lose
Choose C + Switch = Win

As you can see, switching gives you 2/3 winning while staying gives you 1/3

Also can be thought of variable change.