- #1
jk22
- 729
- 24
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 ?
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 ?