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

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?

• With replacement, so you are correct: $4$ independent events, all with probability $\frac4{52}$ to succeed. – drhab May 17 '17 at 9:55
• 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

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 $\dbinom{n+k-1}{k-1}$
• 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
• 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