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I have a deck of 24 cards, 3 of which are aces. I want to figure out my chances of drawing at least one ace based on the number of cards I draw. I'm pretty sure if I draw one card my chance is 12.5% I'm not sure how to figure out what my chances are if I draw 6 or 8, or how to achieve a 60% chance of getting an ace.

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  • $\begingroup$ are you replacing the cards after you draw them? $\endgroup$ – lulu Jun 24 '16 at 1:25
  • $\begingroup$ No, a single hand of size N. $\endgroup$ – aslum Jun 24 '16 at 1:27
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    $\begingroup$ Hint: in problems like these it is easier to work from the compliment, that is, it's easier to compute the probability that you don't draw any aces. If $n=3$, say, then the probability of drawing no aces is $q_3=\frac {21}{24}\times\frac {20}{23}\times \frac {19}{22}\sim .657$. The probability you want is just $p_n=1-q_n$. Can you finish from here? $\endgroup$ – lulu Jun 24 '16 at 1:30
  • $\begingroup$ I'm pretty sure that's what I'm looking for @Lulu, if you post it as an answer I'll accept it. $\endgroup$ – aslum Jun 24 '16 at 2:30
  • $\begingroup$ The title is misleading. You're not actually optimising anything. $\endgroup$ – joriki Jun 24 '16 at 8:55
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I think @lulu's solution is too obvious and not interesting, so let's try to solve it in a different way.


The probability that you get an ace on your first card is $\frac 3{24}$. For $n=1$, it stops here and this is your answer. $$n=1 \implies p_1=\frac{3}{24}$$

The probability that you fail and need to keep drawing is $\frac{21}{24}$. The probability that you get an ace on your second card is $\frac{3}{23}$. For $n=2$, it stops here: $$n=2 \implies p_2=\frac{3}{24}+\frac{21}{24}\cdot \frac{3}{23}$$

The probability that you fail and need to keep drawing is $\frac{20}{23}$. The probability that you get an ace on your third card is $\frac{3}{22}$. For $n=3$, it stops here: $$n=3 \implies p_3=\frac{3}{24}+\frac{21}{24}\left(\frac{3}{23}+\frac{20}{23}\cdot \frac{3}{22}\right)$$

The probability that you fail and need to keep drawing is $\frac{19}{23}$. The probability that you get an ace on your third card is $\frac{3}{21}$. For $n=3$, it stops here: $$n=4 \implies p_4=\frac{3}{24}+\frac{21}{24}\left(\frac{3}{23}+\frac{20}{23}\left(\frac{3}{22}+\frac{19}{22}\cdot\frac{3}{21}\right)\right)$$

Do you see the pattern now? What is the probability for $n=5$ and $n=6$? Is it easier to get a formula using this method or @lulu's method?

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  • $\begingroup$ Lulu's in my opinion, but it is nice to point out that there is more than one approach. $\endgroup$ – JMoravitz Jun 24 '16 at 2:15
  • $\begingroup$ Maybe I'm not understanding completely, but I think @lulu has it, because the choice of how many cards to draw would be made before drawing any of them, rather drawing one card and then drawing again if you didn't draw an Ace. $\endgroup$ – aslum Jun 24 '16 at 2:30
  • $\begingroup$ If I am not mistaken then this is the same as @lulu's method...you are just calculating the probability it at each step, and his method directly gives the probability at each step. $\endgroup$ – User Not Found Jun 24 '16 at 2:32
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    $\begingroup$ @aslum This method yields the same answer as lulu's, it's just that I wanted to point out an alternate method. It doesn't matter if you draw all of the cards at once or if you draw them consecutively because that doesn't affect the probability. In this problem, we can avoid this entirely with lulu's method, but in other probability problems, this makes the problem so much easier, so it's important to know this method. $\endgroup$ – Noble Mushtak Jun 24 '16 at 3:20
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    $\begingroup$ Also, it doesn't matter if you stop drawing cards after you get an ace or you keep drawing. Once you get an ace, you have satisfied the condition, so whatever cards you get after don't matter. That's also another important trick that comes up sometimes. $\endgroup$ – Noble Mushtak Jun 24 '16 at 3:22
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To elaborate (a little) on the comment, and to phrase it slightly differently, let's suppose you had a deck with $C$ cards containing $A$ aces, from which you make a hand of $n$ cards. We let $q_n$ denote the probability that your hand fails to contain an ace, and let $p_n=1-q_n$ denote the probability that it contains at least one ace. We'll compute $q_n$.

Method I: The probability that the first card you draw is not an ace is $\frac {C-A}C$. Given that the first card is not an ace, the probability that the second is also not an ace is then $\frac {C-A-1}{C-1}$. Continuing in this way we get $$q_n=\frac {C-A}C \times \frac {C-A-1}{C-1} \times \cdots \times \frac {C-A-n+1}{C-n+1}$$

Method II. We note that there are $\binom Cn$ ways to choose a hand of $n$ cards (with no constraints). Similarly, there are $\binom {C-A}n$ ways to choose a hand of $n$ cards without getting an ace. Thus we have $$q_n=\frac {\binom {C-A}n}{\binom Cn}=\frac {(C-A)!(C-n)!}{(C-A-n)!(C)!}=\frac {(C-A)(C-A-1)\cdots(C-A-n+1)}{C(C-1)\cdots(C-n+1)}$$

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