Might there be an $n^{\text{th}}$ digit of $\pi$ where the sequence becomes palindromic?

Question

Assuming $n>1$, would it be reasonable to think there is an $n^{\text{th}}$ digit of $\pi$ where stopping there would yield a palindromic number $(3.14159...951413)$?

Would it be more likely that this occurs an infinite amount of times or never?

Answer 1

If $\pi$ is a normal number (probable, but not proved) then any finite sequence appears infinitely often, so any particular palindrome appears infinitely often somewhere. But that does not say one ever occurs at the start.

That is certainly possible, but I think very unlikely. It might be possible to prove that for normal numbers it happens with probability $0$. It's easy to show it does not happen in $\pi$ up to, say, $n=10000$ (I haven't checked). Then you can see that if it does not happen at $2n$ but does happen at $2n+1$ it must almost happen at $2n$, and that says that the second half of the expansion so far must almost mirror the first half. For normal numbers that happens with probability about $1/10^n$.

Answer 2

It is a bit strange to speak of the ‘probability’ of a single event that has already occurred. The decimal expansion of $\pi$ either does, or does not begin with a palindromic sequence, so the probability is either $0$ or $1$, we just don't know which. Probability is an expression of uncertainty about future events, usually a sequence of similar, repeatable events such as die rolls.

If we imagine generating a random real number by rolling a 10-sided die for every digit, we can compute the probability of getting a number that starts with a palindrome. It is exactly $$\frac19\approx 11.1\%.$$ Most such numbers— $90\%$ in fact­—simply begin with the same two digits: they look like $3.3\ldots$ for example. Of the $10\%$ that don't look like this, a further $90\%$ look like $3.553\ldots$.

If we imagine that we have rolled the $10$-sided die $100$ times so far, it is possible that the next $100$ rolls will exactly mirror the first $100$, and we will get a $200$-digit palindrome. The probability of this happening is $10^{-100}$. We might also get even luckier and get an even longer palindrome; the chance of this happening is one-ninth as large, only $\frac1910^{-100}$.

These vanishingly small probabilities are hard to understand because they have no resemblance to anything we encounter in everyday life. The chance of winning a big U.S. lottery, worth tens or hundreds of millions of dollars, is on the order of $\frac1{1000000000} = 10^{-9}$. If you bought one lottery ticket for every drawing, your chances of winning the jackpot every week for a million years would be $10^{-16}$ or thereabouts. I don't know how to express how much more enormously likely this would be than the $10^{-100}$ probability of rolling a particular $100$-digit sequence.

Our current understanding of the digits of $\pi$ is essentially zero. Even if we know the first $n$ digits, current mathematics gives us absolutely no information that would let us guess the next digit without actually calculating what it is.

But we do know a great many of the initial digits of $\pi$, about 61 trillion digits. (That is $61\cdot 10^{12}$.) We know that $\pi$ does not begin with a palindrome for at least this long. If we imagine an experiment in which we were to roll the $10$-sided die $61$ trillion times, and it just happened to produce the first $61$ trillion digits of $\pi$, we would know that there had been no palindrome so far, and we can ask what would be the probability that continued rolling would eventually produce a palindrome. This probability is around $10^{-62000000000000}$. I couldn't describe $10^{-100}$ earlier. I can't even describe how much harder it is to understand $10^{-62000000000000}$; I don't even know where to begin.

You asked:

Would it be more likely that this occurs an infinite amount of times or never?

Given the current state of mathematical knowledge, “never” is so much more likely than “even once” that you and I are never going to understand, even slightly, how much so.

However, it is conceivable that, once we compute the next $62000000000000$ digits of $\pi$ we will find that they exactly mirror the first $62000000000000$. In that case I think we would have to consider several possible explanations, including that some vastly superior intelligence was messing with us.