TL;DR

A GM shouldn't roll all-or-none saves. If reduced rolling is necessary, instead they should figure the expected number of saves, then add a d4 and subtract a d4. Below are pictured the results of this method for various numbers of enemies and probabilities of saving.

Read on to see how these are derived.

All-or-none is a bad idea. But the GM need not roll individually, either.

All-or-none is a different game.

We'll stipulate that in the long run (say, a thousand spells with saves) there's no difference. But we're not looking at a caster with a thousand spell slots looking at an army of goblins. We're looking at a caster with a half-dozen spell slots they'll burn before some different set of rules comes into play. (Death saves, for instance.)

The GM's all-or-none scheme makes combat more "swingy." Extreme outcomes--the caster sniping an encounter or the caster being completely ineffectual are hugely more-likely than they were before. If your GM wants to play a different game than the one in the rulebooks, that's fine. But this sort of change should be done in consultation with players, out of an agreement that you'd all like more randomness and goblin dice.

The probability that any individual enemy saves is unchanged...

Suppose your enemy needs to roll a 9 or higher to save. There's a 60% probability of this happening after the GM says "goblin 1..." and rolls a d20. There's also a 60% probability of a successful save after the GM says "goblin 2..." and rolls a d20, and there's a 60% probability of a successful save after the GM says "I'm rolling for all goblins in the area..." and rolls a d20.

...but the distribution of the number of goblins saving is radically different.

Let's assume there are 5 goblins in the area and the same required roll as above. Below is pictured the probability of a number of goblins saving if each is rolled individually or if the whole group is rolled-for at once. (Click for larger image.)

Things have changed. A lot.

All-or-not does horrible violence to mathematics.

Seriously: the GM would come closer to the "right" distribution if they just rolled a d5 to determine how many got hit. A uniform distribution would be closer to the truth than stuffing all of the probability into the two extreme cases.

Let's dig into the probabilities a bit.

First of all, the probablity that \$n\$ of \$N\$ enemies in the area make their save, when any individual enemy's probability of doing so is \$p\$, is given by $$P(n \text{ of } N, \text{ given } p) = \binom{n}{N} \times p^n \times (1-p)^{N-n}$$

(Read more about the binomial distribution.)

With that in hand, let's look at how the two inputs--\$N\$ and \$p\$--really affect things.

First, let's stick with a \$p\$ of 0.6, and play with \$N\$. Here are tabulated the probabilities of some \$n\$ of \$N\$ enemies saving, for various \$N\$:

\begin{array}{c| c c c c c c c c} & N \\ n\text{ successes} & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline 0 & 0.00 & 0.00 & 0.00 & 0.01 & 0.03 & 0.06 & 0.16 & 0.4 \\ 1 & 0.01 & 0.02 & 0.04 & 0.08 & 0.15 & 0.29 & 0.48 & 0.6 \\ 2 & 0.04 & 0.08 & 0.14 & 0.23 & 0.35 & 0.43 & 0.36 & \\ 3 & 0.12 & 0.19 & 0.28 & 0.35 & 0.35 & 0.22 & & \\ 4 & 0.23 & 0.29 & 0.31 & 0.26 & 0.13 & & & \\ 5 & 0.28 & 0.26 & 0.19 & 0.08 & & & & \\ 6 & 0.21 & 0.13 & 0.05 & & & & & \\ 7 & 0.09 & 0.03 & & & & & & \\ 8 & 0.02 & & & & & & & \\ \end{array}

Things to notice:

• The distribution always has its peak value at the expected number of saves (\$p\cdot N\$, rounded to the nearest 1). The expected number of saves is also the most likely. (That's not always the case in probability, but for these scenarios it is.)
• The smaller \$N\$ is, the sharper that peak is. Alternatively, the larger \$N\$ is, the flatter the distribution is.

Second, let's look at what happens with varying \$p\$ values. For this we'll stick to your original \$N=5\$ and tabulate:

\begin{array}{c| ccccccc} & p \\ n\text{ successes} & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 & 0.7 & 0.8 \\ \hline 0 & 0.33 & 0.17 & 0.08 & 0.03 & 0.01 & 0.00 & 0.00 \\ 1 & 0.41 & 0.36 & 0.26 & 0.16 & 0.08 & 0.03 & 0.01 \\ 2 & 0.20 & 0.31 & 0.35 & 0.31 & 0.23 & 0.13 & 0.05 \\ 3 & 0.05 & 0.13 & 0.23 & 0.31 & 0.35 & 0.31 & 0.20 \\ 4 & 0.01 & 0.03 & 0.08 & 0.16 & 0.26 & 0.36 & 0.41 \\ 5 & 0.00 & 0.00 & 0.01 & 0.03 & 0.08 & 0.17 & 0.33 \\ \end{array}

Things to notice:

• Again, the distribution always has its peak value at the expected number of saves (\$p\cdot N\$, rounded to the nearest 1). The expected number of saves is also the most likely.
• The peak shifts up and down with \$p\$, and the "shoulders" around it develop asymmetries as the peak nears the ends of the distribution.

But what to do!?

You want to convince your GM to roll all the saves. But they don't want to. Here's a better way: hand the GM a better alternative, better for them and better for you. It needs to:

• work quickly, preferably with a single roll,
• peak at the expectation value,
• be more peaked for low \$N\$, flatter for large \$N\$,
• it'd be a nice add-on to maintain the asymmetry we see as \$p\$ gets extreme.

\$N=1-4\$

Roll the saves. The GM should have 4d20 at hand at all times--multiattack with (dis)advantage, anyone?

\$5\leq N\leq20\$

The number saving is going to be simulated by

$$(\text{expected value}) + \text{d}4 - \text{d}4$$

Below are pictured some comparisons of this method to the actual distributions, were the saves rolled (click for larger images):

You can see that the outcomes generally lie within a few percent of desired. One bit of trickery: the roll at N=5 occasionally generates a negative result (because the expected value is so low). Here's how to handle those results: a result of -1 is counted as the expected value, a result of -2 is counted as the next-higher value. (That a result of -3 is nonsensical is left as an exercise to the reader. That the underlying mathematics are symmetric high-low and the same scheme works when we see results larger than N is left as an exercise to the reader.)

Larger \$N\$

For N larger than 20 bump the die size to a d6. Below is pictured a result using this scheme.

For N larger than, say, 40, this sort of scheme really starts to break down--the distribution generated by dX-dX doesn't handle the flatness of fifty saves well.

I'd point out at this point--needing fifty opponents to save--you're looking at a powerful large-area spell like circle of death: you're 13th level at least. If you haven't worked this out with your GM by then, ping me.