Notes to Supplement: Rubinstein's Proof

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Notes

1. If this does not look immediately obvious, consider that either

E = [T₂ = t] = my (Lizzi's) t^th confirmation was lost,

<em>F = [T₂ = t] = my t^th confirmation was received and Joanna's t^th confirmation was lost

must occur, and that

₁(T₁ = t | E) =

₁(T₁ = t | F) = 1 because Lizzi can see her own computer screen, so we can apply Bayes' Theorem as follows:

₁(E | T₁ = t) = ₁(T₁ = t | E) ₁(E)
₁(T₁ = t | E) ₁(E) + ₁(T₁ = t | F) ₁(F)

= ₁(E)
-----------
₁(E) + ₁(F)

=
-------
+ (1)

First published: August 27, 2001
Content last modified: August 27, 2001