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For example in a infinitely repeating series such as $\frac{110}{111}=0.\overline{990}$, what would be the probability of selecting a 0 in the series generated by the infinitely repeating decimals?

I thought that the answer seemed obvious seeing that in each repeating segment, there are two 9s and one 0, so the probability of selecting a 0 would be $\frac{1}{3}$. However, couldn't a bijection be created between each of the 9s and 0s? And so, the probability would be $\frac{1}{2}$?

This is super counterintuitive, and if this isn't true, what's the difference between this and the proof for the number of natural numbers and natural even numbers being equal?

dsjoint
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  • As there are infinitely many digits in the expression, you need to specify what you mean by the probability (you can't say your selection is "uniformly random" on the infinite list, for example). Usual thing would be to let $p_n$ be the probability if you choose uniformly out of the first $n$, and then try to let $n$ go to $\infty$. In this case, you get $\frac 13$ as you predicted. – lulu Dec 03 '15 at 12:06
  • yes, that is what was meant. sorry for the lack of rigor. – dsjoint Dec 03 '15 at 12:18
  • Are you able to work out $p_n$? It is almost exactly $\frac 13$. Specifically: working up to $n^{th}$ slot. Let $n=3k+i$, $i\in{0,1,2}$. Then there are exactly $k$ $0's$, so $p_n=\frac {k}{3k+i}$. In particular, amongst the first $n$ digits there certainly isn't a bijection between the $9's$ and the $0's$. – lulu Dec 03 '15 at 12:30

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The short answer is just that infinities are weird sometimes (or a lot of times) and have a lot of counter-intuitive properties.

Let's say we assign labels to the decimals like $d_1, d_2, d_3, \ldots$ As $n$ tends to infinity, then the probability of you selecting a $d_i$ associated with a $0$ out of all $d_i$ such that $1 \leq i \leq n$ will tend to $1/3$. I think that this is the best we can do.

Let's try ignoring the limit and just try directly selecting a label instead out of all of the infinite number of labels. This is the same as trying to select a random natural number. Unfortunately, there is no uniform random distribution on the natural numbers, and so we cannot do this. As a thought experiment, what would a random natural number look like? For example, how large should it be?

Yes, we can create a bijection between the $9$s and $0$s, and this is similar to the bijection between the natural numbers and even natural numbers. However, their cardinalities being the same (hopefully) doesn't contradict with our intuition on the limit we took earlier. The trouble comes when we try to think of the infinite sequence all at once, but as noted earlier, we can't actually take a random natural number directly, nor can we take a random digit from the infinite decimal expansion of $\frac{110}{111}$.

mathochist
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  • how can we show that it tends to 1/3 algebraically? – dsjoint Dec 03 '15 at 12:16
  • I am not sure what it would mean to do that. For each fixed $n$ in the limit, we can solve that case by counting the $9$s and $0$s and comparing their ratios. In the infinite sequence case, we can't even uniformly randomly choose a decimal to begin with, so it doesn't make sense to solve for the probability of a choice which doesn't exist. – mathochist Dec 03 '15 at 12:33
  • @blank Suppose we randomly select from among the first $n$ digits. Call probability that the selected digit is a zero $P_n$. Can you prove that we have:$$\dfrac13-\dfrac1n<P_n\le\dfrac13$$Equivalently, can you prove that:$$0\le\dfrac13-P_n<\dfrac1n$$These each imply that $P_n$ tends to $\dfrac13$, by the squeeze theorem. – Akiva Weinberger Dec 03 '15 at 15:59
  • @AkivaWeinberger, how would you prove something like that? sorry. my level of mathematics is currently only high school level. – dsjoint Dec 04 '15 at 09:54
  • Perhaps it's easier to do casework. Case 1: $n$ is a multiple of three (of the form $3k$). Case 2: $n$ is one more than a multiple of three (of the form $3k+1$). Case 3: $n$ is two more than a multiple of three (of the form $3k+2$). What's $P_n$ for each case? It would help to try a few examples of each. – Akiva Weinberger Dec 04 '15 at 13:14