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Each scenario has a 1 / 6 probability. The original three prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1 / 6 probability of occurring. However, the warden in the original case cannot reveal the fate of a pardoned prisoner.
The host always reveals a goat and always offers a switch. If and only if he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p. [38] [34] If the host opens the rightmost ( P=1/3 + q/3 ) door, switching wins with probability 1/(1+q).
If you wish to assume that after door 3 has been opened the conditional probability that the car is behind door 1 remains the same 1/3 as the unconditional probability, then trivially the conditional probability that it is behind door 2 is equal to 1 - 1/3 = 2/3. So, essentially, you want to assume what actually has to be proven.
Berkson's paradox, also known as Berkson's bias, collider bias, or Berkson's fallacy, is a result in conditional probability and statistics which is often found to be counterintuitive, and hence a veridical paradox. It is a complicating factor arising in statistical tests of proportions.
Then the unconditional probability that = is 3/6 = 1/2 (since there are six possible rolls of the dice, of which three are even), whereas the probability that = conditional on = is 1/3 (since there are three possible prime number rolls—2, 3, and 5—of which one is even).
Referring to either the figure or the tree, in the cases the case that the host opens Door 3, and the car is behind Door 2 with has probability 1/3 but the case that the host opens Door 3, and the car is behind Door 1 with has probability 1/6. The sum of these probabilities is 1/2, meaning as the host opens Door 3 only 1/2 half of the time.