Probability of drawing no aces with replacement of cards?

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If you choose n cards with replacement (randomly choose a card, note what it is in a list, replace it in the deck, and repeat), then what is the probability that the list of cards contains no aces?

I know that without replacement, the probability would be

Choose(48, n)/Choose (52, n)

I'm not sure what to do when there's replacement. Any help?

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1 Answer

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The probability that the first card is not an ace is $P(A_1) = 48/52$. The probability that the second card is not an ace is $P(A_2) = 48/52$, and since this is an independent event to the first draw (thanks to drawing with replacement), the probability that neither card is an ace is $P(A_1\ and\ A_2) = P(A_1)P(A_2) = (48/52)^2$.

More generally, the probability that no card in $n$ draws is an ace is $P(A_1\ and\ A_2\ and\ A_3\ and \ldots\ and\ A_n) = (48/52)^n$.

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