How many ways can cards be selected from a 52 card deck?

There are two ways to think of this question, "Student" attempted to answer the first one I'll go over. "Student" is looking at the question as "If you draw a card from a deck of 52 cards, what is the probability that the next card you draw will have a matching value (not suit)." Since the deck is now only 51 cards, and there are only 3 more matching values since 1 has been removed, the answer of 3/51 is correct. Now, a slightly different question would be "If you draw two cards from a deck of 52 cards, what is the probability that the cards will be a pair (matching values)." This is a little more complex. There are 52 * 51 combinations of cards in a deck, you can imagine that for every one of the 52 cards, there are 51 potential matches. This is 2,652. There are 4 suits of 13 cards, and for every card, there are 3 other cards that can form a pair with it. If you draw an ace, there are 3 other aces to draw. Lets consider this A1, A2, A3, and A4. Drawing A1 and A4 is the same as A4 and A1, so this is a combination problem rather than a permutation problem, so the question (for just aces) is, how many ways are there of drawing an ace? The answer is 6: A1,A2 A1,A3 A1,A4 A2,A3 A2,A4 A3,A4 So there are 6 possibilities per value, 13 values total; 6 * 13 is 78, or written out, there are 78 ways to draw two cards out the deck that will be pairs. We know from above that there are 2,652 possible ways to draw two cards out of the deck, so the probability that a randomly drawn set of two cards is a pair is 78/2,652, or 0.0294, or 2.94%.

I'm taking a review for a test and one of the problems is giving me a bit of trouble. I've gotten parts (a) and (b), but I can't figure out (c) and (d). The problem is as follows:

A player is dealt 6 cards from a standard 52-card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering questions a through d.

(a) How many ways can 6 cards be selected from a 52-card deck?

This one is just 52C6, so there are 20,358,520 ways that 6 cards can be selected from a 52-card deck.

(b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck?

This one is done by taking 4C3 = 4 (For each rank, there are four different suits, taken three at a time leaves four combinations) multiplied by the number of ranks there are, 13. 4 x 13 = 52 ways for a three of a kind.

And now here's where I'm having problems:

(c) The remaining 3 cards must be different from the 3 chosen and different from each other. After selecting the three of a kind, there are 12 different ranks of cards remaining in the deck that can be chosen. Of the 12 ranks remaining, the player chooses 3 of them and then selects one of the 4 cards in each of the three chosen ranks. How many ways can the player select the remaining 3 cards?

I haven't been able to figure this one out, even having the answer and trying to work backwards. The answer sheet for the review says the answer here is 14,080, but it doesn't explain how they got to that answer.

(d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and three cards that are not alike?

I'm assuming this one should be obvious given all of the other answers, but my brain is frazzled and I'm getting confused by it anyway. I know the General Multiplication Rule is P(E and F) = P(E) * P(F|E), but I'm not sure how to put the other answers into that equation.

Thanks for any clarification and assistance!

They seem harmless enough, 52 thin slices of laminated cardboard with colorful designs printed on their sides. Yet, as another illustration of the mantra that complexity begins from the most simple systems, the number of variations that these 52 cards can produce is virtually endless. The richness of most playing card games owes itself to this fact.

Permute this!

The number of possible permutations of 52 cards is 52!. I think the exclamation mark was chosen as the symbol for the factorial operator to highlight the fact that this function produces surprisingly large numbers in a very short time. If you have an old school pocket calculator, the kind that maxes out at 99,999,999, an attempt to calculate the factorial of any number greater than 11 results only in the none too helpful value of "Error". So if 12! will break a typical calculator, how large is 52!?

52! is the number of different ways you can arrange a single deck of cards. You can visualize this by constructing a randomly generated shuffle of the deck. Start with all the cards in one pile. Randomly select one of the 52 cards to be in position 1. Next, randomly select one of the remaining 51 cards for position 2, then one of the remaining 50 for position 3, and so on. Hence, the total number of ways you could arrange the cards is 52 * 51 * 50 * ... * 3 * 2 * 1, or 52!. Here's what that looks like:

80658175170943878571660636856403766975289505440883277824000000000000

This number is beyond astronomically large. I say beyond astronomically large because most numbers that we already consider to be astronomically large are mere infinitesmal fractions of this number. So, just how large is it? Let's try to wrap our puny human brains around the magnitude of this number with a fun little theoretical exercise. Start a timer that will count down the number of seconds from 52! to 0. We're going to see how much fun we can have before the timer counts down all the way.

Shall we play a game?

Start by picking your favorite spot on the equator. You're going to walk around the world along the equator, but take a very leisurely pace of one step every billion years. The equatorial circumference of the Earth is 40,075,017 meters. Make sure to pack a deck of playing cards, so you can get in a few trillion hands of solitaire between steps. After you complete your round the world trip, remove one drop of water from the Pacific Ocean. Now do the same thing again: walk around the world at one billion years per step, removing one drop of water from the Pacific Ocean each time you circle the globe. The Pacific Ocean contains 707.6 million cubic kilometers of water. Continue until the ocean is empty. When it is, take one sheet of paper and place it flat on the ground. Now, fill the ocean back up and start the entire process all over again, adding a sheet of paper to the stack each time you’ve emptied the ocean.

Do this until the stack of paper reaches from the Earth to the Sun. Take a glance at the timer, you will see that the three left-most digits haven’t even changed. You still have 8.063e67 more seconds to go. 1 Astronomical Unit, the distance from the Earth to the Sun, is defined as 149,597,870.691 kilometers. So, take the stack of papers down and do it all over again. One thousand times more. Unfortunately, that still won’t do it. There are still more than 5.385e67 seconds remaining. You’re just about a third of the way done.

And you thought Sunday afternoons were boring

To pass the remaining time, start shuffling your deck of cards. Every billion years deal yourself a 5-card poker hand. Each time you get a royal flush, buy yourself a lottery ticket. A royal flush occurs in one out of every 649,740 hands. If that ticket wins the jackpot, throw a grain of sand into the Grand Canyon. Keep going and when you’ve filled up the canyon with sand, remove one ounce of rock from Mt. Everest. Now empty the canyon and start all over again. When you’ve levelled Mt. Everest, look at the timer, you still have 5.364e67 seconds remaining. Mt. Everest weighs about 357 trillion pounds. You barely made a dent. If you were to repeat this 255 times, you would still be looking at 3.024e64 seconds. The timer would finally reach zero sometime during your 256th attempt. Exercise for the reader: at what point exactly would the timer reach zero?

Back here on the ranch

Of course, in reality none of this could ever happen. Sorry to break it to you. The truth is, the Pacific Ocean will boil off as the Sun becomes a red giant before you could even take your fifth step in your first trek around the world. Somewhat more of an obstacle, however, is the fact that all the stars in the universe will eventually burn out leaving space a dark, ever-expanding void inhabited by a few scattered elementary particles drifting a tiny fraction of a degree above absolute zero. The exact details are still a bit fuzzy, but according to some reckonings of The Reckoning, all this could happen before you would've had a chance to reduce the vast Pacific by the amount of a few backyard swimming pools.

First, to clarify the problem, we are drawing two cards and hoping for a pair. The discussion at the link posted in the question is about drawing five cards and hoping for a pair.

There are two counting strategies floating around this page.

One strategy is to first choose any card from the available fifty-two cards. (Perhaps you draw A$\spadesuit$.) Regardless of your draw, there remain in the deck exactly three cards that will complete your first card to a pair. (For example, A$\clubsuit$, A$\heartsuit$, and A$\diamondsuit$.) This strategy counts every pair twice, however. (We wouldn't consider A$\spadesuit$ A$\heartsuit$ as a different hand than A$\heartsuit$ A$\spadesuit$.) We can correct this error by dividing by two, which gives us a final count of $\frac{52 \cdot 3}{2} = 78$.

We can avoid the overcounting and subsequent correction with a different strategy that describes the properties of a pair rather than draws the cards. Every pair is uniquely identified by its rank and two distinct suits of that rank. We first choose one of the thirteen ranks. (Perhaps we choose Ace.) Next, we need two of the four suits of that particular rank (such as $\spadesuit$ and $\heartsuit$) in such a way that order does not matter. This can be done in $\binom{4}{2}$ ways (or $C(4,2)$, if that notation is more familiar). This gives us a total count of $13 \cdot \binom{4}{2} = 78$.