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a) Four of a kind (Contains four cards of equal face value)

So for this one, we want four cards that have the same face value, different suit. And the last card can be any remaining card.

There are 13 ranks, (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K). We have $\binom{13}{1}$ ways to pick 1 rank out of 13. For each of these ranks, we want to pick 4 cards that are all the same rank (and it is forcefully implied that these 4 cards will differ in suits). So $\binom{13}{1}\binom{4}{4}$, and for each of these ways, we have $\binom{48}{1}$ way to pick 1 card out of the remaining deck of 48 cards, because we picked 4 cards already. Thus, $$\frac{\binom{13}{1}\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$$ is the total number of ways.

b) Full House (Three cards of equal face value, and two others of equal face value). So i.e: 3, 3, 3, 2, 2 would be a full house where the three 3's and two 2's are distinguishable (different suit).

So there are 13 ranks again, $\binom{13}{1}$ ways to pick 1 rank out of 13 total. For each of these ways, we want to pick three cards of the same rank. So $\binom{13}{1}\binom{4}{3}$. Now we want two more cards that are of equal face value that differ from the other three cards picked earlier, so there are 12 ranks left, and $\binom{12}{1}$ ways to pick 1 rank out of 12 remaining. For each of these ways, we have $\binom{4}{2}$ ways to pick 2 cards out of the 4 suits belonging to the same rank, thus $\binom{12}{1}\binom{4}{2}$. The total number of ways is: $$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}$$

c) Three of a kind. (Three cards of equal face value, and two cards with face values that differ from each other and the other three).

13 ranks, we want to pick 3 cards of the same rank. $\binom{13}{1}\binom{4}{3}$. Now we want two cards that have different face values from the three picked, and each other. Now here is where I'm a bit confused, the ranks MUST be different, but the suits can be the same. So proceeding, $\binom{12}{1} \binom{4}{1}$ to pick 1 card of a different face value, but could be same suit, and then $\binom{11}{1}\binom{4}{1}$ to pick another card with a different face value, but could be a different suit. Multiply these altogether and divide by the denominator, and we have our total ways.

Is my work correct?

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  • $\begingroup$ Very nearly. Your numerator is the number of total ways to achieve these combinations. Dividing by $52 \choose 5$ gives you the probability of achieving those hands. $\endgroup$ – Robert Shore Feb 14 at 19:36
  • $\begingroup$ That's what I said..... $\endgroup$ – Stuy Feb 14 at 19:39
  • $\begingroup$ Then you'll want to clarify the last line of sections (a) and (b), which both refer to the quotient as "[t]he total number of ways" rather than the probability. $\endgroup$ – Robert Shore Feb 14 at 20:00
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You correctly calculated the probabilities of four of a kind and a full house, not the total number of ways. The number of ways these hands can be obtained is given by the numerators of your probabilities.

What is the probability of three of a kind?

As you know, there are $\binom{52}{5}$ possible five-card hands that can be drawn from a standard deck.

However, your count of the favorable cases is not quite right. There are $\binom{13}{1}$ ways to choose the rank from which three cards are drawn and $\binom{4}{3}$ ways to select cards of that rank, as you found. The remaining two cards must come from different ranks. There are $\binom{12}{2}$ ways to select two ranks from which one card is drawn and $\binom{4}{1}$ ways to choose one card from each of the two ranks. Hence, the number of favorable cases is $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$ Therefore, the probability of three of a kind is $$\frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}^2}{\dbinom{52}{5}}$$

What was your mistake?

You count each hand twice since the order in which the singletons are selected does not matter. For instance, if the hand is $\color{red}{5\heartsuit}, \color{red}{5\diamondsuit}, 5\spadesuit, 7\clubsuit, \color{red}{10\diamondsuit}$, you count it once when you select $7\clubsuit$ as the card you are drawing from the remaining $12$ ranks in the deck and $\color{red}{10\diamondsuit}$ as the card you are drawing from the remaining $11$ ranks in the deck and once when you select $\color{red}{10\diamondsuit}$ as the card you are drawing from the remaining $12$ ranks in the deck and $7\clubsuit$ as the card you are drawing from the remaining $11$ ranks in the deck. However, both selections result in the same hand. Therefore, you need to divide your answer by $2$. Notice that $$\frac{1}{2}\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{11}{1}\binom{4}{1} = \binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$

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  • $\begingroup$ I don't know how to identify my own mistake, like I don't know how order doesn't matter in this case? How am I overcounting? $\endgroup$ – Stuy Feb 16 at 19:56
  • $\begingroup$ What matters is which cards are selected. We need to choose the rank from which three cards are drawn, three cards of that rank, the two ranks from which a single card is drawn, and one card from each of those ranks. In your attempt, you distinguished between the first single card you draw and the second one you draw by first selecting a card from one of the remaining $12$ ranks and then selecting a card from one of the remaining $11$ ranks. However, you still get the same hand if you select the cards in the opposite order, which means you have counted every hand twice. $\endgroup$ – N. F. Taussig Feb 17 at 0:07

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