# Probability of 3 people being born on consecutive days of the week

I started studying probabilities and I'm stuck trying to find the probability of 3 randomly chosen people on the street being born on consecutive days of the week (for example: Mon, Tue, Wed). This is what I tried to do:

First I manually wrote all the possible ways of having consecutive triples of days and found out there are 7 cases, then there are 7^3 possible ways of the 3 people having their birthdays in a week so the probability would be 7/7^3 = 0.02 (possitive cases / total cases)

But i know that approach won't work on bigger problems (like 3 consecutive days on a year) so I tried using combinatorics to find the positive cases using the following procedure:

The 1st person has the possibility of being born in any day of the week (7 cases)
The 2nd one must be born the day after or the day before the 1st one (2 cases)
The 3rd also has 2 options depending of what the 2nd one chose (2 cases).
This way there are 7x2x2 = 28 possible arrangement but i then divide that by 3! = 6 because i don't care about the order so Mon,Tue,Wed is the same as Wed,Mon,Tue in this case.
But I obtain a different answer than before (28/6) / 7^3 = 0.09 and having 28/6 possible cases doesn't make any sense.

So I wanted to know what I'm missing on the combinatorics approach, idk if I am missing cases or what is wrong.
Also sorry if i wrote too much and i appreciate all of your help

Actually, both answers are wrong. In the first one, you should account for the ordering in the numerator because you did in the denominator (i.e. $$7^3$$ already has the ordering). So, the correct answer is $$7\times 3! / 7^3=6/49$$. By the way, this logic would work just fine in regular years (i.e. no leap year), and it is a combinatorics approach, too.