What is the probability that a random $5$-card hand(from a deck of $52$ cards) has:

  1. (a) exactly $3$ of the same number: that is $3$ cards of the same number but not poker or full.
  2. (b) $3$ or more of the same number?
  3. (c) At least $1$ spade, at least $1$ heart, no diamonds or clubs and the values of the spades are all greater than values of the hearts?

I think the answer for a) is $$\frac{\dbinom{13}{3}\dbinom{4}{1}\dbinom{12}{2}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{52}{5}}$$

  • 2
    $\begingroup$ For a) the number on top should be $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}$. The only significant difference from yours is the $\binom{13}{1}$ in front instead of your $\binom{13}{3}$. $\endgroup$ – André Nicolas Mar 27 '16 at 23:37
  • $\begingroup$ why? but we are choosing 3 numbers $\endgroup$ – user296169 Mar 27 '16 at 23:42
  • 1
    $\begingroup$ @Margarita: We could use your version and start by choosing three numbers, so $\binom{13}{3}$. Then from these $3$ we would choose the special number that will occur $3$ times. That can be done in $\binom{3}{1}$ ways. If we use that way of counting, then the $\binom{3}{1}$ replaces your $\binom{12}{2}$. $\endgroup$ – André Nicolas Mar 27 '16 at 23:51

We dealt with a) in a comment. Here we deal with the less standard c) by counting the number of "favourables."

There are $\binom{13}{5}$ ways of choosing the values of the cards. Arrange these values in increasing order. The values of the hearts are all less than the values of the spades, so the only issue is where the dividing line shall be between hearts and spades. This dividing line can be put in any of $4$ places, for a total of $4\binom{13}{5}$.

  1. For a): You are picking $3$ cards but $1$ number (f.e. $3$ seven's or $3$ J's etc.). So, pick firstly the number, you can do this in $\binom{13}{1}$ ways and then pick from the $4$ cards of this numbers the $3$, which you can do in $\binom{4}{3}$ ways. Next (as you correctly have it), pick $2$ more numbers from the remaining $12$, which you can do in $\binom{12}{2}$ ways and from each set of $4$ cards with the same number pick $1$, which can be done in $\binom{4}{1}\binom{4}{1}$ ways. In total (and by the product rule) this gives $$\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}\dbinom{4}{1}$$ favourable ways to get this hand.
  2. For b): You have counted the favourable ways to get exactly $3$ in a), so count now the favourable ways to get exactly $4$ of the same number, which are equal to (reasoning as above) $$\dbinom{13}{1}\dbinom{4}{4}\dbinom{12}{1}\dbinom{4}{1}$$ and sum up these two figures (in a) and b) ).
  3. For c): Any admissible hand that fulfills the given constraint must have $5$ different values on the $5$ cards (think about it). So, firstly choose $5$ numbers out of the $13$, you can do this in $\binom{13}{5}$ ways and next choose which cards will be spades and which will be hearts. Since the spades must all be greater than the hearts and since there must at least $1$ of each, you have only following $4$ choices: $HHHHS$, $HHHSS$, $HHSSS$, $HSSSS$, ($H$ for Hearts, $S$ for Spades), giving a total of $$\dbinom{13}{5}\cdot4$$
  • $\begingroup$ $$\dbinom{13}{1}\dbinom{4}{4}\dbinom{12}{1}\dbinom{4}{1}+ \dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}\dbinom{4}{1}$$ for b? $\endgroup$ – user296169 Mar 29 '16 at 19:35
  • $\begingroup$ @Margarita Yes, correct this is the numerator! The denominator is as you have it in your post $\dbinom{52}{5}$ $\endgroup$ – Jimmy R. Mar 30 '16 at 9:02

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