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?

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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.

  • $\begingroup$ "If both buttons give the same color": what does it mean? From the previous text, it seems that there are three buttons. "Both" means that you refer to two buttons. Can you clarify? You do not refer to the "three partitions" any more. What is their role? $\endgroup$ Jun 2 at 8:08
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    $\begingroup$ For me, the video is unclear. Feynman is a great physicist but he is not famous for being clear. For what I understand, after pressing two buttons, you have 25% probabilities of getting red light, 25% of getting green light, and 50% of getting black light (bulb off). So I agree with you, it seems that Feynman is wrong. But let me add that your question is even less clear than the video. I suggest you to clarify it. The users should be able to understand without the need to watch an external video. $\endgroup$ Jun 2 at 9:21
  • $\begingroup$ @ Doriano Brogioli After watching the video again I edited the question to reflect new information and added some detail to clarify it. What Feynmann claims is a mathematical chance of ${\frac13}$. But as you said I think for a generic situation it should be a 0.5 chance. What do you think of it? $\endgroup$ Jun 2 at 12:23
  • $\begingroup$ @ Doriano Brogioli I understand Feynman used $\frac13$ as a lower limit instead of $\frac12$ to eliminate the impact of a possible statistical bias of a photon sample. As QM predicts $\frac14$ it should be sufficient for what Feynman was trying to prove. $\endgroup$ Jun 4 at 8:52

2 Answers 2


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.

  • $\begingroup$ I think I misunderstood Feynmanns three-quarter claim to be classical logic. After watching the video several times it was obvious he was referring to the statistical average obtained experimentally. He was adamant at some point in the video in reply to audience questions that it hasn't any resemblance to real particles but it is a simplified abstract model that applies to specific circumstances. But I don't understand ${\frac13}$ classical logic. Shouldn't it be ${\frac12}$? I am going to edit my question to reflect that. $\endgroup$ Jun 2 at 12:11
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    $\begingroup$ Simply check the link in my answer. It covers the reasons for the probabilities. $\endgroup$ Jun 2 at 12:20
  • $\begingroup$ I will check it out. It was what I was just beginning to read two days ago before I was bounced into these videos. So I procrastinated the article for the sake of the videos but I guess I will have to go back. I am currently watching other two videos of the series. After that, I will finish reading that. $\endgroup$ Jun 2 at 12:41
  • $\begingroup$ I finished the article. If you sum up the average probability of the table it will give you $\frac12$. I understand Feynman used $\frac13$ as a lower limit instead of $\frac12$ to eliminate the impact of a possible statistical bias of a photon sample. As QM predicts $\frac14$ it should be sufficient for what Bell was trying to prove. Many thanks for the answer. $\endgroup$ Jun 4 at 8:47

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.


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