[jason1990]

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What does it mean to "[deal a real number]"? I would assume it means choose one uniformly and at random. That, in my mind, means you have a probability measure and each real number is assigned the same nonzero mass.

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But you proved that it cannot mean this. So what can it mean? We have to have a probability measure. And we certainly want that measure to have the property that if X is the chosen number, then P(a<X<b) depends only on b-a.

Here's my attempt at a rigorous formulation of the problem.
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Let X, Y, U, V be independent, uniformly distributed random variables, where "uniform" is meant in the standard probabilistic sense. Let

h:[0,1]x[0,1] --> {0,1} and
g:{0,1}x[0,1] --> {0,1}

be Borel-measurable. Define

S_A = h(X,Y) and S_B = g(h(X,Y),Y).

Define

C_A = U*S_A + X*(1 - S_A) and
C_B = V*S_B + Y*(1 - S_B).

Define

E = E(h,g) = 50P(C_A > C_B) - 50P(C_A < C_B).

Find max_h(min_g(E)) and min_g(max_h(E)) and find functions h and g such that E(h,g) attains these values.
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Interpretations:
X,Y - original cards dealt to A and B, respectively
U,V - new cards dealt, if they choose to switch
h,g - strategies
S_* - 1 if * switches, 0 otherwise
C_* - *'s card after deciding whether to switch
E - A's expectation
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P.S. You can take X,Y,U,V as the four coordinate projections of [0,1]^4 onto [0,1]. Then C_A and C_B are functions from [0,1]^4 to R and, for example, P(C_A > C_B) is just

m({x:C_A(x) > C_B(x)}),

where m is Lebesgue measure on R^4.