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Drawing cards both with and without replacement and finding probability of drawing Ace of Spades. What would that be for the two specific cases below?

1) Randomly draw cards from a standard deck of cards WITHOUT replacement. What is the probability that the last card drawn is the Ace of Spade?

2) Randomly draw cards from a standard deck of cards WITH replacement. What is the probability that the 40th card drawn is Ace of Spade?


marked as duplicate by JMoravitz, YuiTo Cheng, drhab, Paul Frost, воитель Jul 6 at 15:07

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  • $\begingroup$ Any thoughts? If this is too hard, maybe try a smaller deck, so you can just list all the cases. Start with a two card deck, then move up from there until you get a sense of the problem. $\endgroup$ – lulu Jul 6 at 13:09
  • $\begingroup$ These are both $\frac{1}{52}$. With and without do not matter here. If you choose to subject yourself to the torture of performing conditional probability calculations to confirm the result... be my guest, but you will still get $\frac{1}{52}$. $\endgroup$ – JMoravitz Jul 6 at 13:10
  • $\begingroup$ I do not understand why 1/52 is the answer to the first question. Would the number of cards drawn be a factor in calculating the probability? For example, wouldn't the probability of the last card being Ace of Spades different between 10 cards being drawn and 14 cards being drawn? $\endgroup$ – Randy Giedrycz Jul 6 at 13:45

For question one, each of the cards are equally likely to start out at the bottom. Thus, the probability of the ace of spades being drawn last is $1/52$. For question two, if we want the 40th draw to be the firsr time the ace of spades is drawn, this is just an application of the geometric distribution. We need to draw other cards 39 times and the ace of spades once, so the probability is $(51/52)^{39}*1/52$. If we just want the 40th draw to be the ace of spades without regard to whether it appeared before, then the probability is again $1/52$

  • 1
    $\begingroup$ The answer you give of $(\frac{51}{52})^{39}\cdot\frac{1}{52}$ is the correct answer to a totally different problem. That is the probability that among the first $39$ draws, none of them are the ace of spades and the $40$'th draw is the ace of spades. This is a different question to the one where we simply ask for the $40$'th draw to be the ace of spades and we don't care whether or not the ace of spades appeared before then. $\endgroup$ – JMoravitz Jul 6 at 13:13

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