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I guess I have the answer to it
I ran through some combinations and figured out the following so far:

If B1 is in

  • 2, B2 hasn’t been in 1
  • 3, D hasn’t been in 1
  • 4, DT hasn’t been in 1
  • 5, RT hasn’t been in 1
  • 6, S hasn’t been in 1

If B2 is in

  • 1, B1 hasn’t been in 2
  • 3, D hasn’t been in 2
  • 4, DT hasn’t been in 2
  • 5, RT hasn’t been in 2
  • 6, S hasn’t been in 2

I will do the rest later and edit it here, but if we continue this pattern, we will probably get the following, but need to be proven:

If D is in

  • 1, B1 hasn’t been in 3
  • 2, B2 hasn’t been in 3
  • 4, DT hasn’t been in 3
  • 5, RT hasn’t been in 3
  • 6, S hasn’t been in 3

If DT is in

  • 1, B1 hasn’t been in 4
  • 2, B2 hasn’t been in 4
  • 3, D hasn’t been in 4
  • 5, RT hasn’t been in 4
  • 6, S hasn’t been in 4

If RT is in

  • 1, B1 hasn’t been in 5
  • 2, B2 hasn’t been in 5
  • 3, D hasn’t been in 5
  • 4, DT hasn’t been in 5
  • 6, S hasn’t been in 5

If S is in

  • 1, B1 hasn’t been in 6
  • 2, B2 hasn’t been in 6
  • 3, D hasn’t been in 6
  • 4, DT hasn’t been in 6
  • 5, RT hasn’t been in 6

One example to make it easier to follow:
If you opened Door 1 and found the Special,

grafik

you wont find Boss 1 in 6

grafik

This changes the probabilities again:
If you start with 4 or 5 (lowest probability for being a trap) and find no trap
i.e. open 4 and find B1:

grafik

You will know that there is no DT in 1
So Door 1 has now just a 25% chance of being a trap (RT in this case)

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