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$8$ cards are drawn without replacement from an ordinary deck. Find the probability of obtaining exactly $3$ aces, or exactly $3$ kings, or both.

The answer is $\frac{2~\binom 4 3 \binom {48} 5 - \binom 4 3 \binom 4 3\binom {44} 2 }{ \binom {52} 8 }$

I am trying hard to understand the reasoning behind this answer but still don't fully get it

Can someone explain please. Thank you very much.

please read $\binom m n$ as "from m pick n"


Things I believe I understand:

$\binom {52} 8\,\leftarrow$ obvious. # of possible different 8 card hands

$2~\binom 4 3 \binom {48} 5\,\leftarrow$ the act of taking 3 figures of the same rank (be it aces or kings) and multiplying it by the way in which the 5 remaining cards can be chosen.

Thanks again.

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    $\begingroup$ No thank you. I will read it as '$m$ choose $n$' :) $\endgroup$ – Em. Feb 26 '16 at 0:09
  • $\begingroup$ @probablyme why you didn't suggest an answer? $\endgroup$ – Prefijo Sustantivo Feb 26 '16 at 21:42
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It's the principle of inclusion and exclusion at work.

$$\mathsf P(A\cup K) = \mathsf P(A)+\mathsf P(K)-\mathsf P(A\cap K)$$

$\mathsf P(A)~=~{\binom 4 3\binom {48} 5}\big/{\binom {52} 8}$ is the probability of obtaining 3 aces and 5 non-aces.   Some of those 5 non-aces may be kings.

$\mathsf P(K)~=~{\binom 4 3\binom {48} 5}\big/{\binom {52} 8}$ is the probability of obtaining 3 kings and 5 non-kings.   Some of those 5 non-kings may be aces.

$\mathsf P(A\cap K)~=~{\binom 4 3\binom 4 3\binom {44} 2}\big/{\binom {52} 8}$ is the probability of obtaining 3 aces, 3 kings, and 2 cards of other faces.   However we've counted this event among both the above.   So to avoid over counting we subtract one occurrence.

$$\mathsf P(A\cup K) = \dfrac{\binom 4 3\binom {48} 5 + \binom 4 3\binom {48} 5- \binom 4 3\binom 4 3\binom {44} 2}{\binom {52} 8}$$

That is all.

$\Box$

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The number of hands of 8 cards with exactly 3 kings is:

  • a group of 3 kings taken from 4 possible kings i.e. $\binom{4}{3}=4$ multiplied by
  • a group of 5 cards that are not kings taken from $52-4=48$ possible cards i.e. $\binom{48}{5}$

The number of hands of 8 cards with exactly 3 aces is the same that the number of cards with exactly 3 kings.

But in any case we counted the number of both, 3 kings and 3 aces, so the number of hands with 3 kings and 3 aces are counted twice so we must subtract one time this quantity (this is a case of inclusion-exclusion principle).

The number of hands with exactly 3 kings and 3 aces is $\binom{4}{3}^2\binom{44}{2}$.

And all possible hands of 8 cards are $\binom{52}{8}$, so the probability will be:

$$\frac{2\binom{4}{3}\binom{48}{5}-\binom{4}{3}^2\binom{44}{2}}{\binom{52}{8}}$$

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