Probability of a run of $n$ cards of the same color? A magician friend hit me with this one: given a shuffled deck, what is the probability that it contains at least one run of at least $n$ cards of the same color — for example, four reds or four blacks in a row?
 A: It's not easy to give an exact result, but if you want a simple upper bound, you can compute the expected number of runs of length $n$ easily.  Any given block of $n$ cards is a run with probability
$$
\frac{{{52-n}\choose{26}}}{{52}\choose{26}}=\frac{(52-n)!26!}{(26-n)!52!},
$$
so the expected number of runs of length $n$ is just
$$
E_n = (53-n)\frac{(52-n)!26!}{(26-n)!52!}=\frac{53\cdot {{26}\choose{n}}}{{53}\choose{n}}.
$$
(Here we are allowing non-maximal runs in the count... so a run of length $8$, say, contains three runs of length $6$.)  The expected number of runs of length $n$ is greater than or equal to the probability of there being at least one run of length $n$.  The first nontrivial result comes for $n=6$: the probability of at least one run of length $6$ satisfies
$$
P_6 \le E_6 = \frac{53\cdot{{26}\choose{6}}}{{53}\choose{6}}=\frac{26\cdot 25\cdot 24\cdot 23 \cdot 22\cdot 21}{52 \cdot 51\cdot 50 \cdot 49\cdot 48}=0.5315...
$$
Since the exact result is $P_6=0.46424\ldots$, this isn't too loose; and it will yield closer approximations for longer run lengths.
