`In the following article of 2004, Stefan Krauss and Silke Atmaca don't use the problem formulation which is here called the ``standard problem`` but the Parade version with a remarkable supplement:  Wie man Schülern Einsicht in schwierige stochastische Probleme vermitteln kann; Unterrichtswissenschaft 32(1)/2004, S. 38-57  The supplement reads Ich zeige Ihnen mal was (Now I'll show you something) and is placed just before the host opens another door with a goat. The whole problem set is that of Gero von Randow in the German newspaper DIE ZEIT, July 19, 1991, which made the problem known in Germany. The supplement is not compatible with the ``standard problem``, although the authors claim in a footnote that these intended rules are implicitly assumed by almost all experimental subjects. This is a blank claim, and - an important point - answers that it doesn't matter if switching or not can in no way be compared with thousands of letters written to authors who presented the problem without the crucial rule together with the 2/3 solution and attemts to prove it. I don't know ``storms of protest`` following an article with the correctly formulated problem and corresponding explanations; and with that we even reached the topic going on here: If the contestant has chosen door 1, he will win the car by switching in two of three cases: If the host opens door 2, he will open door 3, and if the host opens door 3, he will open door 2. So he will win the car, if it is behind door 2 or door 3. This is completely true, and at the beginning of the game the probability for the ``switcher`` to win is 2/3. (There may be fine additions which extend this true statement to a proof of the fact that the 2/3-probability holds just before the second choice. But I think that they all are a little too abstract for a ``simple solution``.) Seamlessly we may write that a mathematical proof for the 2/3-solution just before the second choice (which is asked for) reads as follows (``Bayes light``): The host opens door d with probability 1 if the player has chosen a goat, and with probability 1/2 if the player has chosen the car. Therefore the final chance for the switching player is (1/3 * 1)/(1/3 * 1 + 1/3 * 1/2) = 2/3 . We also may add some lines with ``cases`` to improve understanding of this short calculation.    `