Monty Hall problem - SP edition

 chart listing out all the cases

If you count, you'll see that if you stay, there are two cases where you'll get a pineapple, and 1 case where you'll get a chinchilla, but if you switch, there are two cases where you'll get a chinchilla, and 1 case where you'll get a pineapple.

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

C is struken since it was eliminated by the host, revealing it contained a . This makes the probabilty 50% chance of winning if you stay and 50% chance of winning if you switch.

You can't strike C because if you had chosen it, the host would have revealed what is behind door B.

edit: Ok, I re-read your post and have a clarification to make.

Yes, in the case you stated, the probability is 50%, but under the assumption that initially door B is always picked and you have the option of staying or switching, in which case you're only really picking between door A and door B (door C is irrelevant in this case). What you should do is repeat your steps assuming you pick door A first, and again assuming you pick door C first. Individually you'll still get 50% for each case, but when you combine them you'll find that you have a 2/3 chance of picking the chinchilla if you switch.