hw1_solutions

# hw1_solutions - EE 649 Pattern Recognition – Spring 2008...

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Unformatted text preview: EE 649 Pattern Recognition – Spring 2008 Homework 1 - Solutions 1. Let us define the events: A = { The door first opened by the contestant has the prize } B = { The last unopened door has the prize } There are three possibilities: the contestant should never switch, always switch, or it does not matter, if P ( A ) > P ( B ), P ( A ) < P ( B ), or P ( A ) = P ( B ), respectively. It is obvious that P ( A ) = 1 3 . As for P ( B ), we have: P ( B ) = P ( B | A ) P ( A ) + P ( B | A c ) P ( A c ) Clearly, P ( B | A ) = 0 and P ( B | A c ) = 1 (the latter is so because the host is forced to open the remaining donkey door), so that P ( B ) = 0 + 1 .P ( A c ) = 1- P ( A ) = 1- 1 3 = 2 3 Therefore, P ( B ) > P ( A ), and the contestant should always switch. The solution was easy because of the appropriate definition of the events. The solution can be much more complicated if the events are defined differently — for example, take a look at the solution in the Wikipedia entry on this problem: http://en.wikipedia.org/wiki/Monty Hall problem Note also that the answer would be different if the host was dishonest. For example, if the host would only open a door and offer a switch if the contestant picked the correct door first, then the probability of winning by switching would be zero. Note: This problem is sometimes called the “Monty Hall Problem” (Monty Hall was a popular TV game show host). It is a “paradox” in the sense that many people intuitively expect that, since there are two options available after the host opens a door, one should have P ( A ) = P ( B ) = 1 2 , so it should not matter whether the contestant switches or not. On the other hand, other people feel that the host may be trying to mislead the con- testant, who thus should never switch doors (this is of course not allowed in the above formulation of the problem). These wrong perceptions do not stand up to the proba- bilistic analysis of the problem. This problem is completely equivalent to another (and older) paradox called “The Three Prisoners Problem,” proposed by Martin Gardner, in which there are three prisoners, one of which is going to be executed and the rest will be...
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hw1_solutions - EE 649 Pattern Recognition – Spring 2008...

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