If you have a prime number (other than two and five) then the last digit of is 1, 3, 7, or 9. If for some reason we believe that prime numbers are randomly distributed (after all, the last digits 1,3,7, and 9 appear equally likely among primes), then we expect if a prime numbers ends in 1, then the next prime number also ends in 1 with probability of 1/4. A recent study has shown otherwise, statistically, and has proved it mathematically, assuming the Hardy–Littlewood *k*-tuple conjecture is true. In fact the statistical data shows the probability is less than 1/4. Here is an article explaining the main content of this study.

And if you want to check it for yourself, here is a little code that finds the empirical probability of two consecutive prime numbers ending in the same last digit:

n = 100000
Primes = primes_first_n(n)
q = 10
Qrimes = []
for p in Primes:
Qrimes += [p % q]
count = [ 0 for i in range(q) ]
for i in range(n-1):
a = Qrimes[i]
if a == Qrimes[i+1]:
count[a] += 1
#print(Primes[i],Primes[i+1])
(sum(count)/(n)).n(32)

0.153300000

And here is a modification that shows how the probability changes as grows:

n = 1000000
Primes = primes_first_n(n)
q = 10
Qrimes = []
for p in Primes:
Qrimes += [p % q]
Probs = []
count = [ 0 for i in range(q) ]
for i in range(n-1):
a = Qrimes[i]
if a == Qrimes[i+1]:
count[a] += 1
Probs += [[i,(sum(count)/(i))]]
P = point(Probs)
P

As grows (horizontal axis) the empirical probability of having two consecutive prime numbers with the same last digit (vertical axis) changes but remains well below 1/4.

The fun part is by noting the last digit of a number is just the remainder mod 10, then you can ask about what happens to the remainders mod some other number . In the paper they show the probability is still less than what one would expect mod any number , which sounds paradoxical.

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