# What is the probability one card from each suit will be represented when 5 cards are dealt?

If each suit is represented then we will have two cards from one suit and one card each from the remaining suits. So I am counting the ways this can happen like so -

${52 \choose 1}$ ways of selecting the first card.

${12 \choose 1}$ ways of selecting a card from the same suit as the first.

${13 \choose 1}$ ways of selecting a card for each of the 3 remaining suits.

So I have -

$${52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}$$

Then I have the sample space as $52 P 5$ as we are counting with regard to order on the numerator so we need to do the same for the denominator.

So I have $$\frac{{52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}}{52P5}$$

Does that look correct?

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Are you sampling with or without replacement? –  ncmathsadist Oct 28 '12 at 22:22

When $5$ cards are dealt, the reasonable assumption is that they are dealt from the same deck, so the number of hands is $\dbinom{52}{5}$. All these hands are equally likely.

Now let us count how many hands have at least one card from each suit.

The suit that we have $2$ of can be chosen in $\dbinom{4}{1}$ ways. For each of these ways, the actual $2$ cards can be chosen in $\dbinom{13}{2}$ ways. For each of these choices, there are $\dbinom{13}{1}$ ways to choose $1$ card from the highest ranking remaining suit, and so on, for a total of $$\binom{4}{1}\binom{13}{2}\binom{13}{1}\binom{13}{1}\binom{13}{1}.$$ To find the probability, divide.

Now we can look at an entirely different problem, in which after drawing a card and writing down what it is, we replace the card in the deck and shuffle before drawing again. That is a somewhat unreasonable interpretation of the problem.

The natural denominator is then $52^5$, since there are $52^5$ sequences of cards, all equally likely. Now we need to count the number of "favourables."

The suit we will have $2$ of can as before be chosen in $\dbinom{4}{1}$ ways. For each of these ways, the location of the draws at which we got this suit can be chosen in $\dbinom{5}{2}$ ways. For each way of specifying the location, the two slots can be filled in $13^2$ ways. Once this is done, there are $3$ empty slots. The first one can be filled in $39$ ways. For each such way, the next empty slot can be filled in $26$ ways. And for each of these, the remaining empty slot can be filled in $13$ ways.

There are other ways of organizing the count. For example, onece we are left with $3$ empty slots, there are $3!$ ways of deciding the order of the suits in thse empty slots. And once this is done, the slots can be filled in $(13)(13)(13)$ ways.

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...need to rethink my way of counting it with regard to order. –  Jim_CS Oct 28 '12 at 22:29
@Jim_CS: I have also written out the calculation if we take the to my mind much less natural assumption that we are sampling with replacement. –  André Nicolas Oct 28 '12 at 22:37
@Jim_CS: Your denominator has changed, so now you are not allowing five Jacks of Spades. Counting the favourables should be done in a manner very close to my replacement solution. The only difference is that for filling the places where we have the doubled suit, we need $(13)(12)$ instead of $13^2$. –  André Nicolas Oct 28 '12 at 22:50