🚀 go-pugleaf

On 2007-03-23, Eraser <Eraser@Pencil.Box> wrote:
>
>
> Something like this was used to explain quantum entanglement 
> in a book by Julian Brown called, Quest for the Quantum Computer:
>
> A single black box has:
>    an on-off switch, 
>    a dial with six positions, 
>    one red and one green light, 
>    a counter which counts how many times the box was turned on. 
> To operate the box, randomly select any dial position 
> and then switch the box "on". Note which light turns on, 
> red or green. Then switch the box "off". The color of 
> the light that gets turned on is guarranteed to always be 
> completely random. There is a 50:50 chance for either 
> the red or green light coming on, regardless of 
> whether the dial's position was changed or not.
>
> Now whenever two such black boxes have been switched "on" 
> the same number of times, they will have the same counter number.
> Then, they will both show the same color light if their dial 
> positions are the same. But whenever the dial on one of the 
> two boxes is in a position opposite to the position of the 
> dial of  the other box, the color of the two box's lights 
> will be different (one will be red and the other will be green).
> And whenever the dial on one of the two boxes is in 
> a position adjacent to the dial position of the other box, 
> the color of the two box's lights will be the same on average
> 3/4ths of the time.
>
> Two such black boxes can be said to be 'entangled'.
>
> Can they be physically manufactured without the 
> ability to send information between them?  
>
>
>


It's not possible. 

Given that A implies B, C, or D, then P(A) <= P(B) + P(C) + P(D).

Chance (Light 1 != Light 4) <= Chance (Light 1 != Light 2)
                                 + Chance (Light 2 != Light 3)
                                 + Chance (Light 3 != Light 4)

Which implies

1  <= 1/4 + 1/4 + 1/4 == 3/4

This reasoning shows that two adjacent positions will give the same 
result more than 2/3 of the time.

It is possible to achieve 2/3 by having the device give green on three 
adjacent numbers, with each choice being used 1/6 of the time.

It is possible to achieve 0 by using three alternating positions, each 
1/2 the time.

It is possible to achieve any value in between by an average of those 
two methods.