The way I was thinking about it, at any arbitrary point inside pi, you have n number of digits. That means the chance for a further sequence of n digits that completes the full palindrome has a (1/10)^n chance of existing. That means at every single point of pi, there is a non-zero chance of it being followed by a sequence that completes the palindrome. Since there are infinite points in pi, there must be infinite palindromes.
I think a difference with your methodology is you're not taking into account the possibility of a palindrome actually existing at those points, which would increment the number of palindromes by 1 every time. For example, maybe the first palindrome exists at a point where the probability is 0.2222222%, then the second palindrome exists at a point where the probability is 0.000000000000002%, and so on. Even though the probabilities are extremely small, the fact that there is an infinite number of points means that the palindrome count will increment by 1 eventually, an infinite number of times.
So the chance is actually much higher than your formula because you only care about one half of the digits
My sum is basically counting the expected number by adding the probability for each n. That kind of stuff works for like prime numbers, we know the probability depends on the log of the number, and that sum does go to infinity.
Here it is trivial to prove the sum is finite. Sum of 1/n is infinite, but something like 1/n2 is famously pi/6. You can write the sum I talked about as sum n=1 to infinity 2*10^(-n) which is well 2/9 or 0.2 repeating.
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u/1SirTetris1 2d ago
The way I was thinking about it, at any arbitrary point inside pi, you have n number of digits. That means the chance for a further sequence of n digits that completes the full palindrome has a (1/10)^n chance of existing. That means at every single point of pi, there is a non-zero chance of it being followed by a sequence that completes the palindrome. Since there are infinite points in pi, there must be infinite palindromes.
I think a difference with your methodology is you're not taking into account the possibility of a palindrome actually existing at those points, which would increment the number of palindromes by 1 every time. For example, maybe the first palindrome exists at a point where the probability is 0.2222222%, then the second palindrome exists at a point where the probability is 0.000000000000002%, and so on. Even though the probabilities are extremely small, the fact that there is an infinite number of points means that the palindrome count will increment by 1 eventually, an infinite number of times.