what is the chance of a deck of cards not having adjacent suits or numbers?

Question

For example, if we have an Ace of spades, the next card cannot be an Ace nor spades.

Edit: Assuming that we pick one particular shuffle amongst all possible ones

Answer 1

To get a complete enumeration and thus an exact value for the probability, we can classify ranks according to which suits are left in them. There are $2^4$ different suit patterns for the $13$ ranks, onto which they can be distributed in $\binom{16+13-1}{16-1}=\binom{28}{15}=37442160$ different ways. We also have to remember which card was last played, for which there are at most $52$ possibilities, so we have at most $52\cdot37442160=1946992320\approx2\cdot10^9$ different states to process. We can do this stepping through the $52$ cards one at a time and tallying the number of admissible ways there are to reach the possible states.

Here's Java code that does this. The result is that

$$ 1609436968954808435644946271743718475717055824496860279603200 $$

different deals out of the total of

$$ 52!=80658175170943878571660636856403766975289505440883277824000000000000 $$

are admissible, which yields a probability

\begin{eqnarray*} &\frac{1609436968954808435644946271743718475717055824496860279603200 }{80658175170943878571660636856403766975289505440883277824000000000000}\\\\&=\frac{5258385993865270320943168907713343815039252427}{263528070245000096386506036857149662453104640000000000}\\\\&\approx1.9954\cdot10^{-8} \end{eqnarray*}

in good agreement with Ross' estimate and quasi's simulation.

Note that I didn't make use of the permutational symmetry among the suits. That would require a bit more programming effort, but would further substantially reduce the number of states that need to be processed.

Answer 2

Roughly speaking each card after the first has a $\frac {15}{51}$ chance of causing a failure, or a $\frac {36}{51}$ chance of success. This ignores correlations between the cards, but should not be far wrong. To succeed $51$ times in a row, the chances would be $$\left ( \frac {36}{51}\right)^{51} \approx 1.9\cdot 10^{-8}$$