Distribution of matches between two shuffled decks

Take two desks of cards and shuffle them. They can be standard 52-card decks, though the number of cards in the decks doesn't matter as long as they're the same and the decks are fairly large.

Now count the number of times the two desks match, i.e. how many times the same card is in the same position in both desks. The number of matches is random, and its distribution is approximately Poisson with mean 1. Let's do a simulation and see how close the results come to the predicted outcome.

Here's the Python code:

import numpy as npfrom scipy.stats import poissonimport matplotlib.pyplot as pltdef count_zeros(x): return len(x[x==0])num_reps = 10000deck_size = 52matches = np.zeros(deck_size+1, dtype=int)# Simulationfor _ in range(num_reps): # Shuffle two decks a = np.random.permutation(deck_size) b = np.random.permutation(deck_size) # Count how often they match num_matches = count_zeros(a-b) matches[ num_matches ] += 1# Cut off outputs too small to seecutoff = 8# Matches predicted by a Poisson distribution with mean 1.predicted = [num_reps*poisson(1).pmf(i) for i in range(cutoff)]# Plot resultsx = np.arange(cutoff)w = 0.3 # bar widthplt.bar(x, matches[0:cutoff], w)plt.bar(x+w, predicted, w)plt.legend(["actual", "predicted"])plt.xlabel("matches")plt.ylabel("frequency")plt.savefig("shuffled_desk_matches.svg")plt.show()

And here's the output based on 10,000 simulations:

About 1/3 of the time, you get no matches, another 1/3 of the time you get one match, and the rest of the time you get more. More precisely, according to the Poisson model zero matches and one match are both have probability 1/e.