The chance of light being red after 2nd button press if first button gives red light (Richard Feynman: Quantum Mechanical View of Reality)

Question

This question is related to a Richard P. Feynmann lecture on "Quantum Mechanical View of Reality"

There are 3 buttons L, M and R. Each switch is connected to a bulb that can randomly light up as either red or green. The light has a probability of 0.5 lighting up as either Red or Green per each button press (Red= 0.5, Green = 0.5). If both button presses give the same colour (for example if pressing the first button press gives red light and if pressing the second switch also gives red light bulb will continue to be red), the light will be that colour but if the buttons give the opposite colour then the light will be black. For the purpose of clarification that I need for this example, we will be only pressing one button after the other. The third switch is used in the example in the video because three switches or buttons are needed for what Feynmann trying to show (experimental details invalidate EPR assumptions of local realism).

So what are the chances light can give red after two buttons are pressed if the first button press gives red colour?

The answer seems to me it is a 0.5 or 50% chance of the bulb lighting up red because of the four chances RR, RG, GR, and GG we have eliminated GG and GR by pressing the first button which gives red light then only leaving RR and RG. So the second button has a 50% chance of causing red so the bulb has a mathematical chance of being 50% red or black. So why does Feymann say several times through the video that the bulb has one-third ( ${\frac13}$ )of a chance of being red upon pressing the second button?

Here is where the related example begins in the video: https://youtu.be/ZcpwnozMh2U?t=1091

This may be more of a maths question but I thought since it regards an explanation related to a lecture by Feynmann regarding quantum entanglement, Bell's inequalities and EPR I thought maybe it is not pure mathematics so probably the answer depends on the context of physics related.

Answer 1

You are completely missing the point of Feynman covering this problem, and it is not really your fault. He is trying to abstract away the details of the actual experiment and that made it even more confusing than before.

The point he is trying to get you to understand, is that the experiment is showing that if you try to get an answer using classical logic and probability arguments, then you will get an answer that is in disagreement with experiment. This is the crux of his argument. The only way to get an agreement with experiment, is to abandon classical logic and use quantum logic.

He is actually talking about Bell's theorem. The 3 buttons are the three $120^\circ$ angle polarisations that you can choose to use. After you measure one polarisation, if you immediately measure the same polarisation, then you get the same result. You can measure a different polarisation, you get the $\frac14$ same colour vs $\frac34$ different colour in experiment, but as you can see in the website, it should always be more than $\frac13$ agreement if you use classical logic.

That is, what we now understand about quantum theory is that you can reduce Bell's theorem into a box-colouring problem, and realise that there is no way to colour the boxes in the way that classical logic says it should be possible to. If you understand that, then the experiments are really telling you that you have no choice but to give up using classical logic on quantum problems.

Answer 2

One simple way to see the 1/3 vs 2/3 distribution is to imagine that the internal state of a classical box is one of six possible states with equal probability. The states being RGR, RRR, RGG, GRR, GGG, GRG, each occurring with probability 1/6

This set of states has the property that any particular symbol (left, center or right) will be R half of the time, and G half of the time. But the first two symbols agree only 1/3 of the time.

All I have done is to let the first and last values be arbitrary, but when the first and last value disagree I force the middle value to match the third value. And then I assigned an equal probability to the resulting symbols.

I recently saw the beginning of the first of the Feynman videos in question, and I don't know much about 120-degree polarization, so I took Feynman's analogy at face value to mean there was a simple way to build a classical box that gave you 1/3 (not 1/2), and I thought about it for a moment and found a way. It is interesting that you cannot get to 1/4 by any hidden variable approach, and I suppose that can be seen by enumerating all such possible models and seeing that 1/3 is the minimum. There is probably a clever way to see it without having to list all the possible models, unless I am somehow missing the point of how "hidden variable" approaches that fail would work in the case of these red/green box setups.