# Probability of two people having a phone number that ends in 99?

I’m a history major, so I’m hoping a mathematically minded person much smarter than me can help me out.

I believe a single person manages two Twitter accounts that have interacted in the past. The password reset number for both accounts end in 99. One twitter account has 11,400 followers and the other has 3,600.

Im trying to prove the point that the odds of the two following each other, and interacting, and both having account numbers that end in the same two digits, is unlikely.

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The chance that a phone number (on the assumption they are equally distributed) ends specifically in $$99$$ is $$\frac 1{100}$$ and the chance that two independent phone numbers both end in $$99$$ is $$\frac 1{10\ 000}$$. However, the chance that two randomly chosen phone numbers end in the same two digits is $$\frac 1{100}$$ because the first can end in anything, the second just has to match it. With thousands of followers out there, this is not unlikely at all.

• The million dollar question, then, is the odds that they have the same last numbers AND they interact. Need some more info to tackle that one... – MikeY Dec 20 '18 at 20:53
• If you interact with $100$ people, it is rather likely (about $63\%$) that at least one has the same last numbers as you do. – Ross Millikan Dec 20 '18 at 21:07
• "on the assumption they are equally distributed" - It should be observed that this is not a valid assumption for e.g. the North American Numbering Plan and probably for most other numbering plans worldwide. There are numerous complicated and arbitrary restrictions on number validity, and the rules have changed over time, so that numbers may be distributed differently based on their age as well as the current rules. – Kevin Dec 21 '18 at 0:03

Following on from Ross Millikan's answer and comment, here's how to think about the probability of there being a match somewhere. I'll take the $$3600$$-follower account as the example.

What is the likelihood that none of the 3600 will have a number whose last two digits match those of the account owner?

Each person has a $$\frac{99}{100}$$ chance of having the wrong digits. For there to be no match, every single person must manage this.

So for $$3600$$ people, the chance of no match is

$$\left(\frac{99}{100}\right)^{3600}≈1.94×10^{-16}$$

which is about $$1$$ in $$5,167,000,000,000,000$$.

So, say, roughly a millionth as likely as picking a specific person at random from the world's population. In other words, a match among 3600 followers is virtually certain.

If you add the requirement that the matching digits must be 99, the chance of a match remains the same but there's only a $$\frac{1}{100}$$ chance that the account owner's number ends in $$99$$. So the probability goes down from "almost certain" to "almost exactly $$\frac{1}{100}$$".

This answer assumes that one Twitter account following another constitutes both accounts interacting—if it doesn't, then you need an estimate of how many followers are interacted with (and a way to define what counts as interaction).

Edit: In fact, we need to be very careful about defining what event we're finding the probability of. For example, does the match and mutual following have to happen between

• two specific accounts, chosen randomly from the whole of Twitter?
• ditto, but chosen from the accounts you've encountered (so you can notice them)?
• a specific account and a randomly chosen follower?
• a specific account and any follower (as calculated here)?

Defining what coincidence we're actually looking at is often the trickiest bit.

In addition to the answers provided by Ross Millikan and timtfj, both of which point out that the odds of a match are in fact far greater than you think, I would like to point out that I think there is also a good bit of psychology at work here.

For example, you noted that the last two digits were $$99$$. Well, that's a pretty striking combination of digits; more striking than something like $$36$$. So, in a way, your brain went like

"Hey, interesting, $$99$$ ... and wow, there is $$99$$ again! My God!"

But would your brain have lit up that same way had both accounts ended with $$36$$? Probably not, or at least less so. In other words, your perception on the 'unlikeliness' of this event was partially driven by the nature of these very numbers; you probably would not have perceived both accounts to both end with $$36$$ to be as equally unlikely ... even though mathematically it is.

Although, maybe I should take that latter part back: I don't know how those passwords are generated, but if that was something chosen by the users themselves, then maybe both accounts ending in $$99$$ was in fact more likely to happen than both accounts ending in $$36$$, since people are more likely to choose passwords that have some kind of pattern to it they can remember. Indeed, if this is the case, then the probability of no one else matching $$99$$ at the end is even smaller than the calculations by timtjf actually show!

Finally, how long were these passwords? If there were lots of digits (or other characters) there, note that there would have been even more possible matches with digits of other characters. So the fact that it was the last two digits, rather than any pair of digits, probably did a good bit of psychological work as well: had the second and third digit of each password have been $$99$$, you probably would not have been that surprised, and you might in fact not even noticed it.

In the end, then, I am saying that something psychologically jumped out at you, but we tend to give that far more importance than it should. Millions of events happen every day ... and most of them don't jump out at us. But out of those millions, some of them are bound to be a little 'weird', 'unusual', and do jump out at us. Does that mean anything? Probably not.