Since two consecutive primes ending in the same last digit has a probability less than mere chance, then four consecutive primes ending in the digits $1,3,7,9$ in any order will be greater than chance. Examples of four primes ending in the four different digits are $11,13,17,19$ and $23,29,31,37.$ Probability is $\frac{4!}{4^4}=\frac{3}{32}$. There can be two consecutive sets of four each ending with the same four digits: $47,53,59,61$ and $53,59,61,67$ with probability $\frac{3}{128}$. There can be three consecutive sets as in $1061,1063,1069,1087$ and $1063,1069,1087,1091 $ and $1069,1087,1091,1093$ with probability $\frac{3}{512}$.
Does anyone want to run through the first few thousand primes to see how the results are better than mere chance four one set, two sets, three sets... having this last digit property? What's the largest string of consecutive primes having the most sets of four moving from left to right? Has anyone wondered if there are more groups of four like 11,13,17,19 with difference of first to last is eight than there are with a difference of greater than eight, as in 3331,3343,3347,3359? What is the most common difference of such groups of four?
