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What is the probability of two two-digit integers, with reversed digits (i.e., 13 and 31, or 45 and 54) appearing in a set of two integers?

Assume the first integer in the set of two can be any integer between 00 and 99 (inclusive). The second integer can be any other integer between 00 and 99 (inclusive); it can't be the first integer repeated again.

I enjoy math a lot, but I know little enough about the fine tunings of it to get anything out of this problem that seems remotely likely. All of my guesses show what I assume to be far too high a likelihood.

This thought experiment arrives from my place of work, where I discovered two serial numbers that were identical, except that the last two digits were reversed. Because these were items of the same make and model, from the same delivery, the serial number could not be repeated and they were manufactured near-enough apart that the first seven digits were guaranteed to be identical, so I began to wonder what the chances were of those last two digits being the same, yet reversed.

Jety Lefr
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    Your first integer has a $\frac 9{10}$ chance of having distinct digits. The second integer has a $\frac 1{99}$ chance of being the reverse of the first. What chances have you come up with? – abiessu Sep 07 '22 at 12:23
  • For what its worth, coincidences happen all the time. Just remember that there are many opportunities for coincidences to occur. "I discovered two serial numbers that..." out of how many serial numbers that you looked at? If you looked at a bit over a dozen, then it is more likely than not that this will occur at least once rather than not occur at all. It is not just that you compare the first and second serial numbers, but also compare the first and third, and first and fourth and... and second and third, and second and fourth, etc... – JMoravitz Sep 07 '22 at 12:35
  • Similarly, even if you were talking only about two serial numbers at a time... if this didn't occur today, it might also have occurred yesterday, or the day before that, or the day before that. Its almost more unusual that a person never experiences what they would consider a coincidence of some sort over their life. – JMoravitz Sep 07 '22 at 12:39
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    @AndrewChin See the comment of abiessu. The first number can not (for example) be 11, because the second number must be distinct from the first number. – user2661923 Sep 07 '22 at 13:07

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Taking my comment a bit further, first note that in $\frac {90}{100}$ of two-digit integers $u$ where $01$ is considered one of these, there are two distinct digits in the number $u$. Then of these which have distinct digits, there is exactly $\frac 1{99}$ of the remaining $99$ two-digit integers which is the digital reverse of $u$. This gives a $\frac 1{110}$ probability that such a pair will occur when two such two-digit integers are chosen at random.

What is interesting is that we get the same probability of having two repeated-digit integers chosen if we have the same without-replacement restriction: $u$ has probability $\frac {10}{100}$ and $v$ has probability $\frac 9{99}$, giving the probability of choosing such a pair as $\frac 1{110}$.

So here's a question: how many distinct describable probability questions can be made about choosing pairs of two-digit integers which each have this same likelihood of occurring?

abiessu
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