Warning - maths ahead
Single hit probability
Start with defining 3 events:
\$M\$: A miss with a single attack
\$H\$: A hit with a single attack
\$C\$: A critical hit with a single attack
It is obvious that \$M\$ and \$H\$ are complimentary events. That is, if you miss you don't hit and vice versa, so: \$P(M) = 1 - P(H)\$.
It is also obvious that \$C\$ is a sub-event of \$H\$. That is, you can't get a critical hit unless you hit, so: \$P(H\cap C) = P(C)\$.
5e does not have any such thing as a critical miss so I am not going to deal with this.
We also define the target number \$t\$, as the number needed to roll or above to get a hit. \$t\$ must be between 2 and 20 inclusive, since a 1 always misses and a 20 always hits.
Normal attacks
$$\begin{align}
P(H) = h & = {21 - t \over 20} \\
P(M) = m & = {t - 1 \over 20} \\
P(C) & = {1 \over 20} \\
P(C|H) = c & = {P(C\cap H) \over P(H)} \\
& = {P(C) \over P(H)} \\
& = {1 \over 21-t} \\
\end{align}$$
Advantaged attacks
$$\begin{align}
P(H) = h & = 1 - \left({t - 1 \over 20}\right)^2 \\
& = {400 - (t-1)^2 \over 400} \\
P(M) = m & = \left({t - 1 \over 20}\right)^2 \\
P(C) &= 1 - \left({19 \over 20}\right)^2 \\
& = {39 \over 400} \\
P(C|H) = c & = {39 \over {400 - (t-1)^2}}{} \\
\end{align}$$
Disadvantaged attacks
$$\begin{align}
P(H) = h & = {(21 - t)^2 \over 400} \\
P(M) = m & = {400 - (21 - t)^2 \over 400} \\
P(C) &= \left({1 \over 20}\right)^2 \\
& = {1 \over 400} \\
P(C|H) = c & = {1 \over (21 - t)^2} \\
\end{align}$$
Now, you can pick your target number and get numerical solutions to these but I will continue to treat them algebraically. The important point to note is that the chance of critical hit given a hit was achieved is dependent on the target number so the quote that dismisses it as inconsequential in not valid. For example, if \$t\$ = 20 then every hit is a critical hit irrespective of advantage/disadvantage.
Multiple attack probability
This is a straightforward application of the Binomial Distribution.
Specifically, for \$n\$ attacks, the probability of \$k\$ hits is:
$$P(X=k) = \binom{n}{k}h^km^{1-k}$$
And the probability that of these \$k\$, \$j\$ will be critical hits is:
$$P(Y=j|X=k)= \binom{n}{k}h^km^{1-k}\binom{k}{j}c^j(1-c)^{1-j}$$
The mean number of hits is simply \$nh\$ and the mean number of critical hits is \$nhc\$.
Proposed alternative
Normal attacks
There is no difference.
Advantaged attacks
This gives a straight chance of a critical hit of \${1 \over {10}} {({40\over 400})}\$ where the actual chance is \$39 \over 400\$.
I won't take this any further because you can actually do this strictly in accordance with the rules with no additional work:
- Roll all the dice,
- reroll all the misses,
- look at all the dice now on the table (originals and rerrolls), 20s are critical hits and all the other hits are just hits.
Disadvantaged attacks
This obviously reduces the chance of a critical hit, from \$ 1 \over 400\$ to 0.
You can actually make this in accordance with RAW with one tiny additional step:
- Roll all the dice,
- all the misses are misses,
- reroll all the 20s away from the other hits, if these miss they miss, if they hit they hit and if you get a 20 they are criticals,
- reroll all the other hits, if they hit they hit and if they miss they miss - these cannot be criticals.
