# Estimate Probabilities of Card Games

## A practical example of how you can calculate Card Probabilities with Monte Carlo Simulation and Numerically

We are going to show how we can estimate card probabilities by applying Monte Carlo Simulation and how we can solve them numerically in **Python**. The first thing that we need to do is to create a deck of 52 cards.

**How to Generate a Deck of Cards**

importitertools, random# make a deck of cards

deck=list(itertools.product(['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K'],['Spade','Heart','Diamond','Club']))deck

And we get:

`[('A', 'Spade'),`

('A', 'Heart'),

('A', 'Diamond'),

('A', 'Club'),

('2', 'Spade'),

('2', 'Heart'),

('2', 'Diamond'),

('2', 'Club'),

('3', 'Spade'),

('3', 'Heart'),

('3', 'Diamond'),

('3', 'Club'),

('4', 'Spade'),

('4', 'Heart'),

('4', 'Diamond'),

('4', 'Club'),

('5', 'Spade'),

('5', 'Heart'),

...

**How to Shuffle the Deck**

`# shuffle the cards`

random.shuffle(deck)

deck[0:10]

And we get:

`[('6', 'Club'),`

('8', 'Spade'),

('J', 'Heart'),

('10', 'Heart'),

('Q', 'Spade'),

('7', 'Diamond'),

('K', 'Diamond'),

('J', 'Club'),

('J', 'Diamond'),

('A', 'Club')]

**How to Sort the Deck**

For some probabilities where the order does not matter, a good trick is to sort the cards. The following commands can be helpful.

# sort the deck

sorted(deck)# sort by face

sorted(deck, key=lambdax: x[0])# sort by suit

sorted(deck, key=lambdax: x[1])

**How to Remove Cards from the Deck**

Depending on the Games and the problem that we need to solve, sometimes there is a need to remove from the Deck the cards which have already been served. The commands that we can use are the following:

# assume that card is a tuple like ('J', 'Diamond')

deck.remove(card)# or we can use the pop command where we remove by the index

deck.pop(0)# or for the last card

deck.pop()

**Part 1: Estimate Card Probabilities with Monte Carlo Simulation**

**Question 1**: **What is the probability that when two cards are drawn from a deck of cards without a replacement that both of them will be Ace?**

Let’s say that we were not familiar with formulas or that the problem was more complicated. We could find an approximate probability with a Monte Carlo Simulation (10M Simulations)

N=10000000

double_aces=0forhandsinrange(N):

# shuffle the cards

random.shuffle(deck)

aces=[d[0]fordindeck[0:2]].count('A')

ifaces==2:

double_aces+=1prob=double_aces/N

prob

And we get `0.0045214`

where the actual probability is **0.0045**

**Question 2: What is the probability of two Aces in 5 card hand without replacement**.

FYI: In case you want to solve it explicitly by applying mathematical formulas:

fromscipy.statsimporthypergeom# k is the number of required aces

# M is the total number of the cards in a deck

# n how many aces are in the deck

# N how many aces cards we will gethypergeom.pmf(k=2,M=52, n=4, N=5)

And we get

. Let’s try to solve it by applying simulation:**0.03992981808107859**

N=10000000

double_aces=0forhandsinrange(N):

# shuffle the cards

random.shuffle(deck)

aces=[d[0]fordindeck[0:5]].count('A')

ifaces==2:

# print(deck[0:2])

double_aces+=1prob=double_aces/N

prob

And we get `0.0398805.`

Again, quite close to the actual probability

**Question 3: What is the probability of being dealt a flush (5 cards of all the same suit) from the first 5 cards in a deck?**

If you would like to see the mathematical solution of this question you can visit PredictiveHacks. Let’s solve it again by applying a Monte Carlo Simulation.

N=10000000flushes=0forhandsinrange(N):

# shuffle the cards

random.shuffle(deck)

flush=[d[1]fordindeck[0:5]]

iflen(set(flush))==1:

flushes+=1

prob=flushes/N

prob

And we get `0.0019823`

which is quite close to the actual probability which is **0.00198**.

**Question 4: What is the probability of being dealt a royal flush from the first 5 cards in a deck?**

The actual probability of this case is around 0.00000154. Let’s see if the Monte Carlo works with such small numbers.

# royal flushN=10000000

royal_flushes=0forhandsinrange(N):

# shuffle the cards

random.shuffle(deck)

flush=[d[1]fordindeck[0:5]]

face=[d[0]fordindeck[0:5]]

iflen(set(flush))==1andsorted(['A','J', 'Q', 'K', '10'])==sorted(face):

royal_flushes+=1prob=royal_flushes/N

prob

And we get `1.5e-06`

which is a very good estimation.

# Part 2: Calculate Exact Card Probabilities Numerically

Above, we showed how we can calculate the Card Probabilities explicitly by applying mathematical formulas and how we can estimate them by applying Monte Carlo Simulation. Now, we will show how we can get the exact probability using Python. This is not always applicable but let’s try to solve the questions of Part 1.

The logic here is to generate all the possible combinations and then to calculate the ratio. Let’s see how we can get all the possible 5-hand cards of a 52-card deck.

# importing modulesimportitertoolsfromitertoolsimportcombinations# make a deck of cards

deck=list(itertools.product(['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K'], ['Spade','Heart','Diamond','Club']))# get the (52 5) combinations

all_possible_by_5_combinations=list(combinations(deck,5))len(all_possible_by_5_combinations)

And we get `2598960`

as expected.

**Question 1: What is the probability that when two cards are drawn from a deck of cards without a replacement that both of them will be Ace?**

# importing modulesimportitertoolsfromitertoolsimportcombinations

# make a deck of cardsdeck=list(itertools.product(['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K'], ['Spade','Heart','Diamond','Club']))# get the (52 2) combinationsall_possible_by_2_combinations=list(combinations(deck,2))Aces=0forcardinall_possible_by_2_combinations:

if[d[0]fordincard].count('A')==2:

Aces+=1prob=Aces/len(all_possible_by_2_combinations)

prob

And we get `0.004524886877828055.`

**Question 2: What is the probability of two Aces in 5 card hand without replacement**.

# get the (52 5) combinationsall_possible_by_5_combinations=list(combinations(deck,5))Aces=0forcardinall_possible_by_5_combinations:

if[d[0]fordincard].count('A')==2:

Aces+=1prob=Aces/len(all_possible_by_5_combinations)

prob

And we get `0.03992981808107859`

**Question 3: What is the probability of being dealt a flush (5 cards of all the same suit) from the first 5 cards in a deck?**

# get the (52 5) combinations

all_possible_by_5_combinations=list(combinations(deck,5))flushes=0forcardinall_possible_by_5_combinations:

flush=[d[1]fordincard]

iflen(set(flush))==1:

flushes+=1prob=flushes/len(all_possible_by_5_combinations)

prob

And we get `0.0019807923169267707`

**Question 4: What is the probability of being dealt a royal flush from the first 5 cards in a deck?**

# get the (52 5) combinationsall_possible_by_5_combinations=list(combinations(deck,5))royal_flushes=0forcardinall_possible_by_5_combinations:

flush=[d[1]fordincard]

face=[d[0]fordincard]

iflen(set(flush))==1andsorted(['A','J', 'Q', 'K', '10'])==sorted(face):

royal_flushes+=1prob=royal_flushes/len(all_possible_by_5_combinations)

prob

And we get `1.5390771693292702e-06`

# Discussion

Computing Power enables us to estimate and even calculate complicated probabilities without being necessary to be experts in Statistics and Probabilities. In this post, we provided some relatively simple examples. The same logic can be extended to more advanced and complicated problems in many fields, like simulating card games, lotteries, gambling games etc.

*Originally published at https://**predictivehacks**.com.*