TITLE: What is the probability of shuffling a deck and not getting a repeating card for the whole deck?
QUESTION [5 upvotes]: What are the odds of shuffling a deck of 52 cards and going through each card one by one without getting any repeats (like 2 kings back to back).
I'm bad at this kind of deduction, but at a high level it seems roughly like 1 in 13 odds over 52 iterations. 
Can anyone provide a more accurate approach to solving this?

REPLY [3 votes]: In short: the odds of having no two consecutive cards with the same rank in a shuffled deck is about 1 out of 22, or 4.5%.
One way to approximate a solution to this problem is to assume that each set of 2 adjacent cards is independent of the other 50 cards:
$$ P\Big(\bigcap_{i=2}^{52} Rank(card_i) \neq Rank(card_{i-1}) \Big) \approx \prod_{i=2}^{52}P\Big(Rank(card_i) \neq Rank(card_{i-1})\Big) = \Big(\frac{48}{51} \Big)^{51} = 0.0454176  $$
This number seems to approximately agree with some of the Monte Carlo simulations done below.  Intuitively, I would tend to believe that the total dependence between two cards and the other 50 would be negligibly small, or possibly even zero.  I am not sure how to justify this independence assumption, however.
Monte Carlo simulation:
check_deck <-function(deck){
  for(i in 2:52){
    if(deck[i] == deck[i-1]){
      return(0)
    }
  }
  return(1)
}

N = 10000000   # takes a few minutes
count = 0 
for(i in 1:N){
  d = sample(rep(1:13,4))  
  count = count + check_deck(d)
}
p = count/N

Results: 
$\qquad \hat{p}  = 0.0454795$
95% Confidence interval:
$$\hat{p} \pm 1.96 \sqrt{\frac{1}{N}*\hat{p}(1-\hat{p})} = \big[0.04535,\ 0.04561\big]$$
So we can be 95% sure the true value lies somewhere in the range $[0.04535,\ 0.04561]$. In other words, we can be fairly sure that we can expect about one out of every 22 randomly shuffled decks to have no adjacent cards with the same rank.