Last year I wrote a blog post pointing out that the D&D rulebooks are WRONG when they say that "Advantage" is equivalent to +5. It's actually equivalent to just +3.25.
A couple days ago, the excellent youtube channel "Stand-Up Maths" did a video on exactly this topic, and came to the same conclusion I had (though he proceeded to push the math much further because, y'know, he's a mathematician, and I'm not).
The average roll of 2d20k1, aka "d20 with Advantage", is just 13.825.
https://www.youtube.com/watch?v=X_DdGRjtwAo
This is most important for DMs when planning adventures, but can also be really critical for PCs who have some reliable way of achieving advantage on a common die roll. If you use the recommended short-hand equivalency that the rulebook suggests and think of advantage as if it were +5, you will over-estimate the power of the PCs and generally make things much harder for the players than you intended.
The video also helpfully provides formulas for similar rolls of various dice sizes.
Average roll of n-sided die "with advantage" is ((1/6n) x (n+1) x (4n-1))
So average of 2d6k1 is 4.472 (1/36 x 7 x 23)
So average of 2d20k1 is 13.825 (1/120 x 21 x 79)
The average of any 2k1 roll is roughly 2/3 the number of sides. The larger the die, the closer they get to that 2/3rd limit.
Average of 3dnk1 is ((1/4n) x (n+1) x (3n-1))
So 3d6k1 is 4.9583 ((1/24) x (7) x (17))
So 3d20k1 is 15.4875 ((1/80) x (21) x (59))
The average of any 3k1 roll is roughly 3/4 the number of sides.
Average of any 4k1 is roughly 4/5 the number of sides.
Average of xdnk1 is ((x/(x+1))n +1/2) where x is number of dice rolled, and n is number of sides on each die. So 5dNk1 is 5/6 N, 6dNk1 is 6/7 N, etc. It's a really cool formula that makes it super easy to figure out the average roll on some seemingly complicated dice pools.
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