Five cards are randomly dealt from a standard deck of cards. Show the probability distribution of the number of cards that are either face cards or aces.

You can create a table or not.


closed as off-topic by JMoravitz, Graham Kemp, Saad, cansomeonehelpmeout, Xander Henderson Jun 3 '18 at 12:36

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    $\begingroup$ This should be a straightforward application of counting principles or an even more straightforward application of the hypergeometric distribution. Where exactly did you get stuck in your own personal attempts at solving this problem? $\endgroup$ – JMoravitz Jun 3 '18 at 4:08
  • $\begingroup$ This question was given to me. It asks to put this into a table. I'm not used to doing this type of question when dealing with cards that request face or aces. Do you have a solution to this problem. $\endgroup$ – Chris Milano Jun 3 '18 at 4:45
  • $\begingroup$ Yes. The question is whether I consider it worth giving you the answer directly. I would much rather see you put a bit of effort into solving the problem yourself because otherwise I am not confident that you will have learned anything. Lets break down the definitions... How many cards in the deck are facecards or aces? (remember, face cards are jacks, queens, and kings) $\endgroup$ – JMoravitz Jun 3 '18 at 4:48
  • $\begingroup$ I have put a lot of effort into this question (several hours at that) which Is why i'm asking for help. There are 12 face cards and 4 aces in a deck. 16 in a deck of 52 cards. $\endgroup$ – Chris Milano Jun 3 '18 at 4:50
  • $\begingroup$ Next leading question: How many cards are neither face nor ace? Okay... so, lets look at a specific entry of the table... let us count how many hands contain exactly one ace or face. How many ways are there to choose the one specific ace or face that appears? How many ways are there to choose the remaining four nonace-nonface cards? Out of how many ways to have chosen five cards ignoring the question of how many face/ace cards there are? $\endgroup$ – JMoravitz Jun 3 '18 at 4:51

Here is such a table for a related but different question of instead counting how many Spades ($\spadesuit$) are in our hand of five cards.

Let $X$ represent the random variable counting the number of spades in our hand when drawing five random cards from a well-shuffled standard deck of $52$ cards without replacement.

Note: there are $13$ spades and there are $39$ non-spade cards.

$\begin{array}{|c|c|c|}\hline k&Pr(X=k)&\text{approximation}\\\hline 0&\dfrac{\binom{13}{0}\binom{39}{5}}{\binom{52}{5}}&0.222 \\\hline 1&\dfrac{\binom{13}{1}\binom{39}{4}}{\binom{52}{5}}&0.411 \\\hline 2&\dfrac{\binom{13}{2}\binom{39}{3}}{\binom{52}{5}}&0.274 \\\hline 3&\dfrac{\binom{13}{3}\binom{39}{2}}{\binom{52}{5}}&0.082 \\\hline 4&\dfrac{\binom{13}{4}\binom{39}{1}}{\binom{52}{5}}&0.011 \\\hline 5&\dfrac{\binom{13}{5}\binom{39}{0}}{\binom{52}{5}}&0.0005\\\hline\end{array}$

Note: Here $\binom{n}{r}$ represents the binomial coefficient "n choose r", the number of ways of selecting $r$ objects from a set of $n$ distinct objects. It is not division. Instead $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ so for example $\binom{7}{3}=\frac{7!}{3!4!}=35$

The idea behind where this formula comes is that there are $\binom{52}{5}$ equally likely five-card hands. When counting the number of hands with $k$ spades, we choose which $k$ spades those happen to be in $\binom{13}{k}$ number of ways. Multiplying that by the $\binom{39}{5-k}$ number of ways of choosing the remaining $5-k$ non-spade cards gives us the total number of hands with $k$ spades as $\binom{13}{k}\binom{39}{5-k}$. Taking the ratio then gives us the probability of it occurring.

This example and yours both follow the hypergeometric distribution. There are only very minor changes to be made to find the answer to your exact question compared to this one.

  • $\begingroup$ How would I implement the question of the probability distribution of "either face cards or aces" into this table. $\endgroup$ – Chris Milano Jun 3 '18 at 5:23
  • $\begingroup$ In the above table we were interested in $13$ spades. In the table you want to make, instead we are interested in $16$ face-or-ace cards... In the above table there were $39$ non-spade cards. In the table you want to make there are......... _______ $\endgroup$ – JMoravitz Jun 3 '18 at 5:25

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