What is the probability that a five-card poker hand contains all 4 suits?

I know that my answer is wrong but I don't know why it is wrong.

What is the probability that a five-card poker hand contains all 4 suits?

What I did was I this:

$$\frac{C(4,1)\,C(13,1)\,C(3,1)\,C(13,1)\,C(2,1)\,C(13,1)\,C(1,1)\,C(13,1)\,C(4,4)\,C(48,1)}{C(52,5)}.$$

My logic for $C(4,1)$ was to choose one of the four suits, then the $C(13,1)$ was to pick one of the cards of that suit, and similarly for $C(3,1)$ subtracting one because there is one less suit to choose from.

Same logic applies for $C(2,1)$ and $C(1,1)$ and then $C(4,4)$ and $C(48,1)$ lets me choose any of the remaining $48$ cards from all suits because the last card doesn't matter.

Then answer ends up being something ridiculously big (not a fraction) and I don't know where I went wrong because I believe my logic is sound. What would be the correct answer and why?

Your numerator makes it seem like there is a distinction between the "first" chosen suit and the "second" chosen suit and so on... The denominator there is no way of knowing which was "first" or "second", they are all the same. If you want order to matter then you need order to matter for both numerator and denominator.

Your numerator counted the number of ordered five-card hands where very specifically all suits are present among the first four of the cards. If you wanted order to matter in the denominator like it does for the numerator then use $P(52,5)$ instead (i.e. $52\times 51\times 50\times 49\times 48$)... but then your numerator misses the mark in another way as it neglects to account for scenarios where the repeated suit occurs twice within the first four cards and the fifth card is a new suit such as $A\spadesuit A\heartsuit K\spadesuit A\diamondsuit A\clubsuit$

For a correct approach

• Via Inclusion-exclusion:

Count the number of five card hands. Remove the number of five card hands missing at least a particular suit and which suit that was (among possibly missing more). Add back the number of five card hands missing at least two particular suits, subtracting missing at least three particular suits, etc... You can end at that point since it is impossible to be missing four or more suits at a time here (but more complicated scenarios may have you continuing to alternate between adding and subtracting even further)

$$\dfrac{\binom{52}{5}-\binom{4}{1}\binom{39}{5}+\binom{4}{2}\binom{26}{5}-\binom{4}{3}\binom{13}{5}}{\binom{52}{5}}$$

• Directly:

Choose which suit is repeated. Choose the two cards for that suit simultaneously. For each suit isn't repeated (in alphabetical order) choose which card for that suit is used.

$$\dfrac{\binom{4}{1}\binom{13}{2}13^3}{\binom{52}{5}}$$

If you insist on phrasing this such that order matters... then make sure it matters for numerator and denominator...

Choose the positions of the two cards in the hand for the suit that is repeated. Choose the suit for those cards. Choose the rank of the first, and then the rank of the second. Then for the remaining unused positions, choose a remaining suit and a remaining rank to be used. Take the ratio with respect to the number of ordered hands possible.

$$\dfrac{\binom{5}{2}\cdot 4\cdot 13\cdot 12\cdot 3\cdot 13\cdot 2\cdot 13\cdot 1\cdot 13}{52\cdot 51\cdot 50\cdot 49\cdot 48}$$

First of all, $ \displaystyle {4 \choose 1}{4 \choose 2}{4 \choose 3}{4 \choose 4}$ will order the cards from $4$ suits. It does not matter which suit is chosen first or second or third or fourth. What matters is what cards are chosen in the hand.

Second, after you have chosen first $4$ cards, you multiply by $ \displaystyle {48 \choose 1}$ but that brings in duplicates again. If you chose $ \color {red} {\diamondsuit A} $ as one of the first four cards and $ \color {red} {\heartsuit A} $ as the fifth card as part of ${48 \choose 1}$, you will also have a combination where you chose $ \color {red} {\heartsuit A} $ as one of the first four cards and $ \color {red} {\diamondsuit A} $ as the fifth card, with other three cards remaining the same.

So to remove the duplicates, you can divide by $2$ and write it as,

$ \displaystyle {13 \choose 1}^4 {48 \choose 1} / \left(2 \cdot {52 \choose 5} \right)$

But I would suggest you approach it as follows -

Choose the suit you will have $2$ cards of and then choose $2$ cards from that suit.

So the desired probability is,

$ \displaystyle {4 \choose 1} {13 \choose 2}{13 \choose 1}^3 / {52 \choose 5}$