What's the expected number of cards drawn to get a three of a kind?

Question

The deck contains 52 cards as usual, and we draw cards without replacement. How would the result change if we draw with replacement and shuffle after every draw?

Answer 1

I computed this as a finite-state absorbing Markov chain, as discussed in the comments, and the answer turned out to be $13.55$. I found thing surprisingly low (I had guessed it would be around $17$ or $18$), but I confirmed it with a simulation of $100,000$ trials, and it was bang on.

Here's the code if anyone is interested:

import numpy as np
'''
Draw cards froma starndard 52-card deck without replacement.
What is the expecetd number of darws until 3 of the same rank have been drawn?
Absorbing markov chain
States are (p,s) where p is the number of different pairs that have
been drawn, and s the number of different singletons.
Assume we start in state (0,1) (so one card drawn already.)  We have 0<s+p<=13.

'''
states=sorted([(p,s) for p in range(14) for s in range(14) if 0<s+p<=13])
n=len(states)
state2index = { }
for idx, state in enumerate(states):
    state2index[state] = idx
Q = np.zeros((n,n))
for (p,s) in states:
    cards = 52-s-2*p   # remaining cards in deck
    start = state2index[(p,s)]
    if p+s < 13:
        # draw one of 4 cards in each unseen rank
        single = state2index[(p,s+1)]
        Q[start,single] = 4*(13-p-s)/cards   
    if s > 0:
        #draw one of 3 cards in rank of each singleton
        pair = state2index[(p+1,s-1)]
        Q[start,pair] = 3*s/cards 
I = np.eye(n)
N = np.linalg.inv(I-Q)
steps=  ([email protected](n))[0]
print(1+steps)

and here's a histogram of the simulation.

I don't see how anyone could possibly do this as an interview question, but I think you could win a lot of bar bets with this.