## Code

`binomial(4, 1)`

`4`

Calculating the probability of drawing a full house

Julia

Probability

Published

March 29, 2024

I recently saw someone mention that they received an interview question for a DS position in which they were asked to calculate the probability of drawing a full house when drawing 5 cards from a standard 52-card deck.

So let’s solve that in Julia.

The function we want is `binomial(n::Integer, k::Integer)`

, which returns the binomial coefficient – the number of ways to choose `k`

out of `n`

items.

Let’s look at some examples. First, if we try 4C1 (4 choose 1), we expect to just get 4 – there are 4 different ways to choose 1 item from a group of 4 items.

Now imagine we choose 2 different items from a group of 4. We expect to get 6 (assuming we don’t care about order, i.e. that 1,2 is the same as 2,1):

- 1, 2
- 1, 3
- 1, 4
- 2, 3
- 2, 4
- 3, 4

So let’s solve the actual problem now. A full house is 5 cards comprising 3-of-a-kind and a pair. There are 52 cards in a deck – 4 suits comprising 13 unique values (2, 3, …, Ace) each.

The approach here is to calculate the number of ways to get a full house and divide that by the number of ways to draw 5 cards from a deck. We can start with the number of ways to draw 5 cards from a deck (the denominator) first, since it’s the most straightforward:

Then let’s calculate the number of ways we can get three of a kind. There are 13 different card values and 4 different suits. We need to choose 1 value with 3 different suits:

`binomial(13, 1)`

gives us the number of ways to choose 1 value from 13 options (which is just 13)`binomial(4, 3)`

gives us the number of ways to choose 3 different suits from 4 possible options

And then since this is probability, we multiply everything together:

Then we do the same thing for drawing a pair. There are now 12 different card values (we can’t get a pair of the value that we already drew three-of-a-kind for), and we need to choose 1 value with 2 different suits:

And from here, we can estimate the probability of a full house by multiplying and dividing:

So there’s a 0.144% chance of drawing a full house from a typical 52-card deck.

We could also take a simulation-approach to solving this. First, let’s create a deck of cards.

```
52-element Vector{Int64}:
1
2
3
4
5
6
7
8
9
10
11
12
13
⋮
2
3
4
5
6
7
8
9
10
11
12
13
```

Then we’ll create a few functions to help us with the simulation:

`make_hands()`

will draw`n`

5-card hands from the deck;`is_full_house()`

will check whether any given hand is a full house;`count_full_house()`

takes a vector of hands and counts the number of them that have a full house

```
function make_hands(deck::AbstractVector{<:Integer}, n::Int64)
v = Vector{Vector{Int64}}(undef, n)
for i in 1:n
v[i] = sample(deck, 5; replace=false)
end
return v
end
function is_full_house(hand::AbstractVector{<:Integer})
return extrema(values(countmap(hand))) == (2, 3)
end
function count_full_house(hands::Vector{Vector{Int64}})
s = 0
for i in eachindex(hands)
if is_full_house(hands[i])
s += 1
end
end
return s
end
```

`count_full_house (generic function with 1 method)`

Then from here we just run our simulation.

And we see that we get roughly the same answer as we did previously.

BibTeX citation:

```
@online{ekholm2024,
author = {Ekholm, Eric},
title = {Probability of {Drawing} a {Full} {House}},
date = {2024-03-29},
url = {https://www.ericekholm.com/posts/full-house},
langid = {en}
}
```

For attribution, please cite this work as:

Ekholm, Eric. 2024. “Probability of Drawing a Full House.”
March 29, 2024. https://www.ericekholm.com/posts/full-house.