Statistics of dropping dice

[jayem]

For high numbers of dice rolls you can pretty much assume that the lowest number will be a 1.

Spoiler: (For 10D10 it's about 34% chance)

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There's 0.9^10 of the numbers all being greater than 1 (35%)
There's 0.8^10 of all the numbers being greater than 2 (10.7%)
There's 0.7^10 of all the numbers being greater than 2 (2.8%)
etc...
So the chances that the lowest number is:
1=1.0^10*(1-0.9^10) (65%)
2=0.9^10*(1-0.8^10) (31.3%)
3=0.8^10*(1-0.7^10) (10.4%)
etc... [though from here you can kind of guess the 2nd half is pretty close to 1]
And for the average 'lowest roll' you can just add them up.
(1*0.65+2*0.313+3*0.104+4*0.7^10+5*0.6^10+6*0.5^10 +7*0.4^10+8*0.3^10+9*0.2^10+0.1^10)

So I get the actual average as 1.1.

So at that point for totals you can basically take 1 off the 'average'.

This of course is calculated independently of the total die roll (5.5*10). So although you can take the difference for an average thing, when dealing with individual cases or distributions you have to be a bit more cautious.

However in general it will be almost certainly to be 1 at the lower end (exactly one where total is less than 20), and will be higher at the far end (ending up at exactly 10 when the total is 100). But for much of the range the lowest is going to be between 1&2.

____________________________-For two die
For each dice, the effect of rolling with advantage is the same as having a 36 sided die with 11 (6's), 9 (5's) 7 (4's), 5 (3's), 3(2's),1 (1)
You can draw the table and count them [as described above], and see how the pattern generalises.

This looks linear (so I suspect 3 will be cubic or quartic), will have a look for a few small numbers but am busy.