Recently I attended a workshop on probability, I was asked the following question but I did not find a way to solve it. A discussion on this question can help to learn something new.

We were told that, We are guests in a game show and close to win a great fortune. The quiz master asks us to choose one of three (closed) doors. She explains that behind one of them awaits you a million Euros. Once you fixed your choice the quiz mastress opens one of the other doors and shows you that this was only a goat. She gives you a final chance: you may either retain your door or switch to the remaining closed one.

(i) Say door 3 is opened. How can we Calculate the conditional probability that our door is the winning one given that the door 3 is a fail, and its complement.

(ii) How to Calculate the unconditional probability that our door is the winning one, and its complement.

Thank you in advance.

  • 4
    $\begingroup$ This is a famous, and much discussed problem in probability. If you search using the name 'Monty Hall problem' you can find discussion of the problem here on CV and on math.SE and vastly more on it elsewhere. It's readily solved by considering the appropriate probability tree, or if you're even reasonably adept at manipulating conditional probability, you can do the calculations directly (NB: the host knows where the prize is & always reveals a goat). If neither of those yield insight, simulation is a very useful tool for exploring the problem $\endgroup$ – Glen_b -Reinstate Monica May 11 '14 at 23:59
  • $\begingroup$ Consider, without loss of generality, that the valuable prize is behind door 1, and that contestant initially chooses door 1, 2 or 3. Consider what happens under each choice (what does the host do?) and how the two strategies perform in each of the three cases. $\endgroup$ – Glen_b -Reinstate Monica May 12 '14 at 0:12
  • $\begingroup$ Here's a paper from the american statistician that may aid your understanding of the nuances of both the unconditional and conditional probability calculation. (google.com/…) $\endgroup$ – jsk May 12 '14 at 2:22