Probability Puzzles

Problem Statement

If you draw 2 cards from a shuffled 52 card deck, what is the probability that you’ll have a pair?

Solution

One way to think about this problem is to consider pulling 2 cards as a sequential process. Once the first card is pulled, it’s known and fixed, and there’re 51 cards left in the deck. Out of them only 3 would make a pair with the first one, so the probability would be 3/51.

Alternatively, one may think in terms of combinations and use the definition of a probability.

We need to count the number of ways to pull 2 cards out of 52, so a total number of possible outcomes is: $$ C_{52}^2 = {52! \over 50!~2!} = {52 * 51 \over 2} = 1326 $$ Pair represents a positive outcome of our experiment, and for a particular card type (let’s say for an Ace) we can calculate the number of possible pairs as: $$ C_{4}^{2} = {4! \over 2!~2!} = 6$$ Since we have 13 different types of cards in the deck, we need to multiply the number above by 13 in order to get the total number of possible pairs.

The probability we’re looking for is: $$\pmb{P = {{C_{4}^{2} * 13} \over C_{52}^{2}} = {78 \over 1326} = {3 \over 51} = 0.059}$$ Not convinced? Try the simulation below!

Simulation

show / hide simulation code


import random
from collections import namedtuple

deck = []
card = namedtuple('card', ['color', 'number'])
for color in ['a','b','c','d']:
    for n in range(1,14):
        deck.append(card(color, n))

nrounds = 100000
positive_outcomes = 0

for i in range(nrounds):
    sample = random.sample(deck, 2)
    positive_outcomes += sample[0].number == sample[1].number

simulated_probability = round(positive_outcomes / nrounds, 4)