Can we find a quantum analogue of the Bertrand paradox?

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jk22
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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 ?
 
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jk22 said:
Suppose the presentator opens door 3
What does that mean quantum mechanically?

jk22 said:
the intermediate state is then a mixture

Cos a|100>+sin a|010>
Where do the cos and sin come from?
 
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 ?
 
jk22 said:
It probably means that now we have the knowledge its 1 in door 1 or 2 hence the first two superposed states.
I don't understand how you can do that. What measurement are you doing?

jk22 said:
The cos and sin are unknown coefficient that will be found afterwards for the presentator to choose where to put the 1 ?
Again, how do you do that? What kind of manipulation are you making on the quantum state?
 
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