You're correct, I made a serious error in the above calculations. Here are the corrected results:

Prior probability for situation A, namely both trials result in red doors: .25; Prior probability for situation B, namely one red and one blue: .50; Prior probability for situation C, namely both trials result in blue doors: .25; Prior probability for me getting a blue door: .745; Prior probability for me getting a red door: .255; Prior probability of the other trial getting red: .745; Prior probability of the other trial getting blue: .255;

Then probability of situation A, given I have a red door = (Pr(A)/Pr(red)) x P(red given A). Pr(red given A)=1, so the result is pr(A given red) = .25/.255 = .9803921...

So the probability that I will be told red, given I have red, is not 1/3, but over 98% (namely the same value above)! And so the probability that I will be told blue, given I have red, is of course .01960784, namely the probability of situation B given that I have a red door.

So using Bayes' theorem with the corrected values, the probability me having a red door, given that I am told the other resulted in red = (pr being red/ pr other red) x pr (told red given red) = (.255/.745) x .9803921... = .33557... or approximately 1/3.

You can work out the corresponding calculation (probability of being blue given told red) by starting with the probability of situation C given I have a blue door, and then deriving the probability of B given I have a blue, and you will see that it matches this one (i.e. it will be approximately 2/3.)