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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?

Jonas Dahlbæk
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maybe later
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    With replacement, so you are correct: $4$ independent events, all with probability $\frac4{52}$ to succeed. – drhab May 17 '17 at 09:55
  • Thank u very much. Can u give me some arguments why is his approach wrong please? :) – maybe later May 17 '17 at 10:00
  • 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? – Cato May 17 '17 at 10:05
  • 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. – maybe later May 17 '17 at 10:09
  • Sorry, but I really can't follow your teacher. As argument I would simply use that $\text{his answer }\neq(\frac4{52})^4$. – drhab May 17 '17 at 10:12
  • it depends if you want the probability, or the number of different sets of successful outcomes (one of which is ace of spades 4 times and one of which is each of the aces appearing), critically though, 4 aces of spades is less likely than each of the aces separately, so a probability calculation has to weight this, making it a tougher approach – Cato May 17 '17 at 10:29
  • If your notation means the teacher's answer is ${55\choose4}\over{7\choose4}$, that is obviously wrong, since a probability cannot be greater than $1$. Are you sure this is what the teacher said? – Barry Cipra May 17 '17 at 10:30
  • Sorry the other way around. choose 4 from 7 is numerator. – maybe later May 17 '17 at 11:14

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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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Keller
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    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 – Cato May 17 '17 at 10:09
  • https://math.stackexchange.com/questions/474741/formula-for-combinations-with-replacement – Keller May 17 '17 at 10:11
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    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? – drhab May 17 '17 at 10:15
  • 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 – Cato May 17 '17 at 10:15
  • Look at this link https://math.stackexchange.com/questions/1756362/combination-with-replacement-why-is-the-formula-not-nk-n?noredirect=1&lq=1 – Keller May 17 '17 at 10:18
  • that approach seems to count the number of successful outcomes in terms of which aces were drawn (ignoring order). That's ok, but some combinations can occur in more ways than others, if that is allowed for, it can then yield the answer of the simpler approach (but seems wrong to me so far) – Cato May 17 '17 at 10:32
  • I think your right that case is with combination but in this case it's probability and events are independent $P(N)=P(A).P(B).P(C).P(D)$ – Keller May 17 '17 at 11:28