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You are dealt five cards from a standard deck find probability of getting $1$ ace, $2$ aces, $3$ aces, $4$ aces


Attempt

There are $\binom{ 52}{ 5}$ possible outcomes.

$p_1$ is probability of being dealt one ace

$$ p_1=\frac{4}{\binom{ 52}{ 5} }$$

$p_2$ is probability of being dealt one ace

$$ p_2=\frac{\binom{5}{2} \binom{50}{3} }{\binom{ 52}{ 5} }$$

$p_3$ is probability of being dealt one ace

$$ p_3=\frac{\binom{5}{3} \binom{49}{2} }{\binom{ 52}{ 5} }$$

$p_4$ is probability of being dealt one ace

$$p_4=\frac{\binom{5}{4} \binom{48}{1} }{\binom{ 52}{ 5} }$$

$p_5$ is probability of being dealt one ace

$$ p_5=\frac{1 }{\binom{ 52}{ 5} }$$


have a follow up question needing correct $p$'s

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  • $\begingroup$ Your question isn't clear. What is $p_i$? $\endgroup$ – AnotherJohnDoe Oct 11 '17 at 16:53
  • $\begingroup$ probability of getting $i$ aces $\endgroup$ – Tiger Blood Oct 11 '17 at 16:54
  • $\begingroup$ Ok, consider $p_i$ - the probability of getting $exactly$ i aces. That means you need to choose $i$ from $4$, and the rest, that is, $5-i$, from what? $\endgroup$ – AnotherJohnDoe Oct 11 '17 at 16:57
  • $\begingroup$ 52 cards in total. $\endgroup$ – Tiger Blood Oct 11 '17 at 16:59
  • $\begingroup$ Ok, so what you have done is for $p_i$, chosen $i$ aces from $4$, and have chosen the remaining $5-i$ from $52-i$. Do you see what the problem is? $\endgroup$ – AnotherJohnDoe Oct 11 '17 at 17:02
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Count the number of ways to pull there required number of aces times the number of ways you can pull the required number of non-aces. There are 48 non-aces in the deck.

0 aces. $\frac {{48\choose 5}}{52\choose 5}$

1 ace $\frac {{4\choose 1}{48\choose 4}}{52\choose 5}$

2 aces $\frac {{4\choose 2}{48\choose 3}}{52\choose 5}$

etc.

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We use counting techniques to calculate the probability. We first count how many ways we can have exactly a hand of $5$ cards with exactly $i$ aces and $5-i$ non-ace cards. To do this, approach via multiplication principle.

  • Pick which $i$ aces are used in your hand. This can be accomplished in $\binom{4}{i}$ ways.
  • Pick which $5-i$ non-ace cards are used in your hand. This can be accomplished in $\binom{48}{5-i}$ ways.

The notation $\binom{n}{r}$ above represents the binomial coefficient. Notice that for the second step, after having $i$ aces in our hand that implies we will need an additional $5-i$ cards to make our handsize total $5$. Further notice that we choose our additional $5-i$ cards specifically from the non-ace cards in the deck. If we chose from all $52-i$ cards left in the deck, we might accidentally get additional aces which would make it so there weren't actually $i$ aces in our hand in the first place.

Now, for the probabilities, divide by the total number of ways you can select a hand of five cards from a standard deck, i.e. $\binom{52}{5}$. We have then $$Pr(X=i)=\frac{\binom{4}{i}\binom{48}{5-i}}{\binom{52}{5}}$$

As a sanity check, you should be able to confirm that $Pr(X=0)+Pr(X=1)+Pr(X=2)+Pr(X=3)+Pr(X=4)=1$, as expected since every hand will have one of exactly 0, 1, 2, 3, or 4 aces.

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