In a deck of $36$ cards ($9$ cards per color, $4$ colors) what is the probability to have $2$ jacks (or more) that follow each other?
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$\begingroup$There are $36\choose 4$ ways to position the jacks. There are $33\choose 4$ ways to position the jacks without adjacencies (imagine to first lay aside three neutral cards and later insert them after the first three jacks). Therefore the answer is $$1-\frac{33\choose 4}{36\choose 4}=1-\frac{33\cdot 32\cdot 31\cdot 30}{36\cdot 35\cdot 34\cdot 33}=\frac{109}{357}. $$
$\endgroup$ $\begingroup$Lets count the possibilities that all jacks are separated. Put the 32 non-jacks into a row. Now there are $\binom{33}{4} = 40920$ possibilities to choose the positions of the jacks.
The total number of possibilities is $\binom{36}{4} = 58905$.
So the result is $$1 - \frac{40920}{58905} = \frac{109}{357} \approx 30.5\%. $$
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