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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?

0 answers · posted 2y ago by DNB‭ · edited 2y ago by DNB‭

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#2: Post edited by $user avatar$ DNB‭ · 2021-12-24T07:00:33Z (over 2 years ago)

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~~I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C~~
~~eq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below. How do these two probabilities prove that B > C DEPENDS ON C > D?~~
>32. 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](https://math.codidact.com/uploads/fzxk7HMTvrBLDuo6o9LHfkA6)
Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 30, p 88. pp 13-14 in the publicly downloadable PDF of curbed solutions.

I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C
eq 2) = P(C > D \mid C = 6) = 1}$. 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?
>32. 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](https://math.codidact.com/uploads/fzxk7HMTvrBLDuo6o9LHfkA6)
Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 30, p 88. pp 13-14 in the publicly downloadable PDF of curbed solutions.

#1: Initial revision by $user avatar$ DNB‭ · 2021-12-24T07:00:13Z (over 2 years ago)

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

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. How do these two probabilities prove that B  > C DEPENDS ON  C > D?

>32. 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](https://math.codidact.com/uploads/fzxk7HMTvrBLDuo6o9LHfkA6)



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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