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What is the probability of drawing $4$ aces from poker deck? The cards are drawn with replacement.

So my first idea was $(4/52)^4$ but my teacher said it is combination with repetition, therefore, (choose $4$ from $(52+4-1))/($choose $4$ from $(4+4-1))$. Who is correct?

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    $\begingroup$ With replacement, so you are correct: $4$ independent events, all with probability $\frac4{52}$ to succeed. $\endgroup$ – drhab May 17 '17 at 9:55
  • $\begingroup$ Thank u very much. Can u give me some arguments why is his approach wrong please? :) $\endgroup$ – maybe later May 17 '17 at 10:00
  • $\begingroup$ does his calculation get a different value? How about considering other cases, such as probability of getting Ace Spades 4 times, or probability of getting any of the 52 cards 4 times (1 in that case) - what does his method get then? $\endgroup$ – Cato May 17 '17 at 10:05
  • $\begingroup$ Yeah it has different value than mine. He said that (4/52)^4 is variance, hence, we care about the order of the cards... but we don´t care about the order so we use combination with repetition. $\endgroup$ – maybe later May 17 '17 at 10:09
  • $\begingroup$ Sorry, but I really can't follow your teacher. As argument I would simply use that $\text{his answer }\neq(\frac4{52})^4$. $\endgroup$ – drhab May 17 '17 at 10:12
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With replacement events are independent therefore $$P(A\cap B\cap C\cap D)=\left(\frac{4}{52}\right)^4$$ I think your professor got confused by combinations where we use the formula $\dbinom{n+k-1}{k-1}$

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    $\begingroup$ I don't see that, what if you had 4 people with 4 identical packs of cards? And they all drew one card, it would then surely be $(\frac{4}{52})^4$ - yet the experiment is mathematically the same as reusing the same pack with the card replaced each time $\endgroup$ – Cato May 17 '17 at 10:09
  • $\begingroup$ math.stackexchange.com/questions/474741/… $\endgroup$ – Keller May 17 '17 at 10:11
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    $\begingroup$ What do you mean "multiple cards"? We have $52$ cards and one is picked. Probability that is an ace is $\frac4{52}$. Then the card is replaced and again a card is picked... What is the probability on an ace now? $\endgroup$ – drhab May 17 '17 at 10:15
  • $\begingroup$ drawing an ace is an event with a probability. There is no link between the 4 events, the first drawing does not affect the second drawing $\endgroup$ – Cato May 17 '17 at 10:15
  • $\begingroup$ Look at this link math.stackexchange.com/questions/1756362/… $\endgroup$ – Keller May 17 '17 at 10:18

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