how is the last calculation of this question?
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Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’) .
Thus, P(S)=61=p, P(F)=65=q
P(A wins in first throw)=P(S)=p
P(A wins in third throw)=P(FFS)=qqp
P(A wins in fifth throw)=P(FFFFS)=qqqqp
So, P(A wins)=p+qqp+qqqqp+…
=p(1+q2+q4+…)
=1−q2p=1−362561=116
P(B wins)=1–P(A wins)
P(B wins) =1−116=115
So, P(A wins)=116 , P(B wins)=115
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