The basic idea
When you deal a card and set it aside, the deck that remains is different from the deck you started with. Drawing without replacement means exactly that: each card removed stays out, so every later draw is made from a smaller, altered set. Drawing with replacement means returning the card (and reshuffling) so the deck is restored each time.
Almost all real card games use the first method. Cards in a hand do not go back into the stock. This is why card probability feels different from dice or coins, where each trial starts from the same position.
Two aces in a row
Take a standard 52-card deck with four aces. What is the chance the first two cards are both aces?
- The first card is an ace with probability 4/52.
- If it was, then 3 aces remain among 51 cards, so the second is an ace with probability 3/51.
- Multiply: (4/52) x (3/51) = 12/2652 = 1/221, about 0.45%.
Now compare with replacement. If the first ace goes back and the deck is reshuffled, the second draw is again 4/52, and the chance of two aces in a row is (4/52) x (4/52) = 16/2704 = 1/169, about 0.59%.
Removing the first ace makes a second ace less likely, because the aces are now thinner in the deck. The difference here is modest, but it grows when the deck is small or the group you are looking for is a large share of it.
Two hearts
For two hearts, the chain is 13/52 for the first card and then 12/51, because one heart is gone and one card in total is gone. The product is 156/2652 = 1/17, about 5.9%. With replacement it would be 13/52 x 13/52 = 1/16, or 6.25%. Again, drawing without replacement lowers the chance of a repeat of the same kind.
The reverse also holds. If you want the second card to be different from the first, removal helps. The chance that two cards drawn are of different suits is higher than it would be with replacement, since each suit has become slightly depleted relative to the others.
Dependent events
What is happening mathematically is that the draws are dependent events. The probability of the second card depends on the first. Written in the language of conditional probability, the chance of an ace on draw two given an ace on draw one is 3/51, but given a non-ace on draw one it is 4/51. Two different situations, two different numbers.
The multiplication used above, 4/52 x 3/51, is the chain rule applied step by step. Each fraction is the probability of that card given the ones before it. Probability trees draw the same thing as branches, with a fraction on each branch, and the answer for a path is the product along it.
A useful symmetry
There is a result that surprises many people. Before any card is seen, the second card in a shuffled deck is an ace with probability exactly 4/52, the same as the first. This looks like a contradiction with the numbers above, but it is not. The chance of ace-then-ace is 12/2652 and the chance of non-ace-then-ace is (48/52) x (4/51) = 192/2652. Adding gives 204/2652 = 1/13 = 4/52.
The lesson is that a deck’s order carries no preference for particular positions. Every position is equally likely to hold any given card. Probabilities change only when you gain information, such as seeing the first card. This is how a shuffled deck differs from a rigged one, and it is why you can draw from the bottom or top with the same odds.
At least one: counting what is left
Without replacement also affects “at least one” questions. What is the chance that a five-card hand contains at least one ace? The hands without an ace come from the 48 other cards, and the chance of that is the product (48/52) x (47/51) x (46/50) x (45/49) x (44/48). Working this out, it is about 65.9%. So the chance of at least one ace is about 34.1%. This is a typical use of complement counting.
The same number can be found by counting hands: there are 1,712,304 five-card hands from 48 cards and 2,598,960 five-card hands in all, and 1,712,304 / 2,598,960 is about 0.659. Both routes agree, which is a good habit to check, since a worked example with a wrong number is worse than none. A direct count in terms of binomial coefficients gives the same ratio.
What about large decks?
When the deck is very large compared with the number of cards drawn, removal hardly changes the proportions, and drawing with and without replacement give nearly the same answer. This is why simpler formulas for repeated trials sometimes work as an approximation in big populations. A 52-card deck is not large in that sense. Dealing 5 or 13 cards changes the make-up noticeably, and in games with a shoe of several decks, the effect is smaller but still present.
Why it matters in games
Removal of cards is the reason card tracking has any meaning at all. In blackjack, the remaining deck shifts as cards are shown, which changes the probabilities of future hands, and this is the mathematical basis behind the topic of card counting. In bridge, knowing which cards have been played changes the likely layout of the rest. In poker, your own cards and the cards on the table are removed from what an opponent could hold, which is part of working out hand combos.
None of this means that the previous result of a freshly shuffled deck tells you something about the next one. After a reshuffle the deck is restored, and the past is irrelevant. Where games have a built-in house edge, understanding removal does not remove a negative expectation, or turn a game with that property into a profitable one, and any gambling can lead to losses.