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I Quantum Monty Hall problem

  1. Sep 28, 2016 #1
    I considered a quantum version of the problem

    There is one winning position so the initial state is |100>+|010>+|001> divided by sqrt3

    Suppose the presentator opens door 3 the intermediate state is then a mixture

    Cos a|100>+sin a|010>

    We suppose finally the player chooses door 2 hence the end state were |010>

    Going through those steps we can compute the probabilities pi*pf=(cos a+sin a)^2/3*sin^2 a

    We find the extremas to be .06 up to .48

    How to interprete those probabilities ? Does it mean that the game can be won only 48% of the time and hence it would be a lucrative game for the presentator ?
  2. jcsd
  3. Sep 28, 2016 #2


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    Staff: Mentor

    What does that mean quantum mechanically?

    Where do the cos and sin come from?
  4. Sep 28, 2016 #3
    It probably means that now we have the knowledge its 1 in door 1 or 2 hence the first two superposed states.

    The cos and sin are unknown coefficient that will be found afterwards for the presentator to choose where to put the 1 ?
  5. Sep 28, 2016 #4


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    Staff: Mentor

    I don't understand how you can do that. What measurement are you doing?

    Again, how do you do that? What kind of manipulation are you making on the quantum state?
  6. Sep 28, 2016 #5


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    Gold Member

  7. Sep 29, 2016 #6


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    Science Advisor

    Last edited: Sep 29, 2016
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