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2 votes
$\frac{1}{6}*[\: (1-\frac{\binom{48}{1}}{\binom{52}{1}}) + (1-\frac{\binom{48}{2}}{\binom{52}{2}}) +(1-\frac{\binom{48}{3}}{\binom{52}{3}}) + (1-\frac{\binom{48}{4}}{\binom{52}{4}}) + (1-\frac{\binom{48}{5}}{\binom{52}{5}}) +(1-\frac{\binom{48}{6}}{\binom{52}{6}})]$

Explanation: P(Ace)=P($1\: on \: dice$)*P($\frac{Ace}{1\: on \: dice}$) + P($2\: on \: dice$)*P($\frac{Ace}{2\: on \: dice}$) + ….. + P($6\: on \: dice$)*P($\frac{Ace}{6\: on \: dice}$)
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0 votes
The answer should be

P(Ace) =  P(1on dice)*P(1Ace from deck) + P(2 on dice)*P(2 Ace from deck) + P(3 on dice)*P(3 Ace from deck) + P(4 on dice)*P(4 Ace from deck)

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