+2489 Here, I answer a very similar question. If I choose four cards from a standard 52-card deck, with replacement, what is the probability that I will end up with all four Aces?
This solution post is based on the original:
The phrase “with replacement” is an explicit standard variation in the sampling method. In this case, it means the card (no matter what it is) is replaced after it is drawn. It’s also important to note that the deck or selection process remains randomized after the card is replaced. And, for the ultra pedantic minded (Ron), the value of the drawn card is recorded before its replacement into the deck. The records are then analyzed statistically for the probability of drawing four aces in sequence.
This question is poorly written. The primary defect is the phrase “end up with four Aces,” which is a colloquialism as used here. The capital “A” in “Aces” is nonstandard, and gives ambiguous emphases to the word aces. The word “all” is not used in this question, so the interpretation is biased toward any combination of Aces where one or more may be repeated.
Rephrased as primer statistics question:
If I choose four cards with replacement from a standard 52-card deck, what is the probability that I will select four aces (of any suit)?
A success in this experiment is four aces without regard to the suit.
\(\large \left(\dfrac{4}{52}\right)^4 = \dfrac{1}{28561}\)
Compare to the probability of selecting four aces in four different suits, where any other combination is a failure.
\(\large \dfrac{4}{52}*\dfrac{3}{52}*\dfrac{2}{52}*\dfrac{1}{52} =\dfrac{3}{913952}\)
GA
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