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Five cards are dealt from a standard 52-card deck. What is the probability of drawing four face cards and an Ace? First, we do it analytically.

Combinations

We are choosing four face cards from the total possible 12 (and without replacement). Since the order of the deal doesn’t matter, it is a combination problem of the following form.

Now, calculate the ways of choosing one ace card from a total of four.

To get the required probability, we have to divide the product of the two with all the possible combinations (of selecting five cards from the deck of 52).

Finally, the probability is calculated by dividing the combination of getting four face cards and one ace card with the possibility of getting five cards from the deck.

We can execute the whole calculation using the following R code:

choose(12,4)*choose(4,1)/choose(52,5)

Monte Carlo

Let’s start building the deck as we did last time.

suits <- c("Diamonds", "Spades", "Hearts", "Clubs")
face <- c("Jack", "Queen", "King")
numb <- c("Deuce", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten")

face_card <- expand.grid(Face = face, Suit = suits)
face_card <- paste(face_card$Face, face_card$Suit)

numb_card <- expand.grid(Numb = numb, Suit = suits)
numb_card <- paste(numb_card$Numb, numb_card$Suit)

Aces <- paste("Ace", suits) 

deck <- c(Aces, numb_card, face_card)

Now start drawing five cards at random and check your hands. Repeat that a million times, and count the number of times you got what you wanted. And divide it by the million.

B <- 1000000

results <- replicate(B, {

hand <- sample(1:52, 5, replace = FALSE)
deal <- deck[hand]

match_1 <- sum(face_card %in% deal)
match_2 <- sum(Aces %in% deal)

if(match_1 == 4 & match_2 ==1) {
  counter <- 1 
}else{
  counter <- 0
}
  
})

sum(results) / B 

Let’s play some card games. Today we will create a deck of cards using R programming.

There are 52 cards in a deck. And they fall into four suits – Diamonds, Spades, Hearts and Clubs. Each of these suits can have nine numbers (2 – 10), three faces (Jack, Queen and King) or an Ace. For example, a card can be an Ace of Clubs, another may be a four of Hearts etc.

suits <- c("Diamonds", "Spades", "Hearts", "Clubs")
numbers <- c("Ace", "Deuce", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Jack", "Queen", "King")

We will use the expand.grid function in R to create a data frame with all combinations of suits and numbers.

deck <- expand.grid(Number = numbers, Suit = suits)

If you print the output from the previous command (as_tibble(deck)), you get the following data frame with 52 rows. The top five rows are shown below.

It still doesn’t look like cards, as they are separated by the columns of the data frame. So we will paste them into each other and create one deck of 52 entries, as follows.

deck <- paste(deck$Number, deck$Suit)

We will do the first two estimations now. How many kings are in the deck, and what is the probability of drawing a king from the deck?

kings <- paste("King", suits) 
deck %in% kings
sum(deck %in% kings) # output is 4
mean(deck %in% kings) # output is 0.07692308


# paste("King", suits) creates all kings (output:  "King Diamonds" "King Spades"   "King Hearts"   "King Clubs" )
# The command, %in%, searches for 'kings' inside the vector 'deck' and returns TRUE or FALSE, depending on whether it matched or not. 
# sum adds up all - each TRUE gets 1, and FALSE gets 0.
# mean gives the average: sum/total count