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Q&A

How does $P(C > D \mid C = 2) \neq P(C > D \mid C \neq 2)$ prove that B > C depends on C > D?

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I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C \neq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below, colored in blue. How do these two probabilities prove that B > C DEPENDS ON C > D?

  1. Consider four nonstandard dice (the Efron dice), whose sides are labeled as follows (the 6 sides on each die are equally likely).

A: 4; 4; 4; 4; 0; 0
B: 3; 3; 3; 3; 3; 3
C: 6; 6; 2; 2; 2; 2
D: 5; 5; 5; 1; 1; 1

These four dice are each rolled once. Let A be the result for die A, B be the result for die B, etc.

(a) Find P(A > B), P(B > C), P(C > D), and P(D > A).

(b) Is the event A > B independent of the event B > C? Is the event B > C independent of the event C > D? Explain.

Solution:

Image alt text

Blitzstein, Introduction to Probability (2019 2 edn), Ch 2, Exercise 30, p 88. pp 13-14 in the publicly downloadable PDF of curbed solutions.

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As it states and you can derive, $B > C$ is logically equivalent to $C = 2$ in this scenario. Thus th... (1 comment)

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