Hint for Problem 5

It is impossible.

Since among X, Y, Z, there is bound to be at least two zeros or two ones, at least one of the events {X=Y}, {X=Z}, {Y=Z} is certain to happen. Then since the probability of the union is less than or equal to the sum of the probabilities,

1 ≤ P(X = Y) + P(X = Z) + P(Y = Z).

Or is it?

Bell's analysis of the Einstein-Podolsky-Rosen experiment in quantum mechanics shows that things are not quite so simple. Here is a probabilist's view of Bell's paradox
  1. There is an actual physical experiment we may perform that will produce any of six (dependent) random variables, X, Y, Z, X', Y', Z', each taking values 0 or 1. This experiment may be replicated independently any number of times.
  2. We cannot observe more than two of the variables at the same time. In fact, we can only observe one primed variable and one unprimed variable.
  3. When we observe X and X' together, they always have the same value, both zero or both one. Probabilistically, this means P(X = X') = 1. Similarly, P(Y = Y') = 1 and P(Z = Z') = 1. (Using this, we may reduce the problem to X, Y, Z, by identifying X with X', Y with Y', and Z with Z'.)
  4. When X and Y' (= Y) are observed together, we find that they are the same about 1/4 of the time, and different about 3/4 of the time. So we may presume P(X = Y) = 1/4 approximately. Similarly, P(X = Z) = 1/4 approximately, and P(Y = Z) = 1/4 approximately.
  5. But we must have 1 < = P(X = Y) + P(X = Z) + P(Y = Z), because at least one of the three events {X = Y}, {X = Z}, {Y = Z} must occur. (This is a special case of Bell's inequality.) So we have

    1 ≤ P(X = Y) + P(X = Z) + P(Y = Z) = 3/4 approximately?????!

COMMENT: This paradox does not depend on the assumption that the experiments are independent and identically distributed.

In fact, we may imagine a magician assigning his own personal choices of six zeros and ones and placing them in sealed envelopes labeled X, Y, Z, X', Y',and Z'. The envelopes are placed on a table in front of a scientist, who must then select one primed envelope and one unprimed envelope. Then the magician retains and destroys the four unselected envelopes, and the scientist opens the two selected envelopes.

It turns out that if a primed and an unprimed envelope of the same letter were selected, then the two numbers they contain will always be the same. However, if a primed and unprimed envelope of different letters were selected then the numbers they contain would be the same only about 1/4 of the time.

How can the magician do it? He seems to be able to guess, with success greater than possible by chance, whether or not the scientist is going to choose to observe the same lettered envelopes. He must have ESP! What does the Amazing Randi say to all this?

Addendum 10/28/19

There is a lesser known variation of the EPR paradox known as the "GHZ experiment". Using the notation above, this may be described as follows. Thanks to Marc Ferguson for pointing this out.
  1. There is an actual physical experiment we may perform that will produce any of six (dependent) random variables, X, Y, Z, X', Y', Z', each taking values 0 or 1.
  2. We cannot observe more than three of the variables at the same time. In fact, we can only observe either the primed or the unprimed variable for each of X, Y and Z.
  3. The sum of three observations that include exactly one primed variable is always odd. This means that (X'+Y+Z) mod 2 = 1, (X+Y'+Z) mod 2 = 1, and (X+Y+Z') mod 2 =1.
  4. Summing the equations in 3., we find (X'+Y'+Z'+2X+2Y+2Z) = 1 mod 2, which implies (X'+Y'+Z') mod 2 = 1.
  5. Yet, observations of (X'+Y'+Z') are always even.
Note that there are no probabilities involved here. This time, the magician is not guessing better than chance. He can predict what the scientist will choose every time.

A fuller exposition of the above including a description of the experiment is part of Sidney Coleman's lecture, "Quantum Mechanics In Your Face", found on YouTube.