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    if we have a crooked coin, with P(H)=0.55 and P(T)=0.45. Then what is the difference between finding E(X) and E(Y) . Given X is the number of coin tosses until 2 successive heads (HH) occur and Y is the number of coin tosses until a head is immediately followed by a tail (HT)
    does a crooked coin effect the method of finding E(Y) different from finding E(X) due to the chance of getting heads is more likely than getting tails ?
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    (Original post by vanessa_tram)
    if we have a crooked coin, with P(H)=0.55 and P(T)=0.45. Then what is the difference between finding E(X) and E(Y) . Given X is the number of coin tosses until 2 successive heads (HH) occur and Y is the number of coin tosses until a head is immediately followed by a tail (HT)
    does a crooked coin effect the method of finding E(Y) different from finding E(X) due to the chance of getting heads is more likely than getting tails ?
    Well, what have you tried? The obvious way to get started for E(X) is to condition on the first toss. If you let E(X|H) be the number of remaining throws needed after throwing a head on the first toss (with corresponding E(X|T)), then you observe that E(X) = p(1+E(X|H)) + q(1 + E(X|T)) = p(1+E(X|H)) + q(1 + E(X)), as throwing a tail on the first toss leaves you back where you started. Now you work on E(X|H) by conditioning on the second toss and so on and so on.

    How might this analysis differ for the rv Y?
 
 
 
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