I considered a quantum version of the problem(adsbygoogle = window.adsbygoogle || []).push({});

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 ?

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# I Quantum Monty Hall problem

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