def a bunny irl

You WILL love this person. It doesn't matter who you loved before; any feelings you had, any promises you made, they will become inconsequential as soon as you lock eyes with the stranger Fate has picked for you.

There's no way to stop it.

There's no way to say "no."

You will meet someone and with a single glance, both of you will become someone new, someone who's now bound to this stranger whether you like it or not, want it or not.

You WILL love this person. It doesn't matter who you loved before; any feelings you had, any promises you made, they will become inconsequential as soon as you lock eyes with the stranger Fate has picked for you.

There's no way to stop it.

There's no way to say "no."

You will meet someone and with a single glance, both of you will become someone new, someone who's now bound to this stranger whether you like it or not, want it or not.

You WILL love this person. It doesn't matter who you loved before; any feelings you had, any promises you made, they will become inconsequential as soon as you lock eyes with the stranger Fate has picked for you.

There's no way to stop it.

There's no way to say "no."

You will meet someone and with a single glance, both of you will become someone new, someone who's now bound to this stranger whether you like it or not, want it or not.

Little fish eats his foods 

"After they overheard that ICE was at the courthouse to arrest someone, they improperly accessed court databases to determine who was not born in the United States," a DOJ detention filing says. "They then snuck every suspected illegal alien who was at the courthouse out a back door, where ICE, who was waiting in the parking lot for their target to leave the building, could not see them."

What a completely inconsequential color palette

Fortunately, death saving throws are a lot simpler than the general case of "X of outcome 1 before Y of outcome 2" because (probability of outcome 1 + probability of outcome 2) equals 1; the presence of a "neither" case can throw a monkey into the wrench.

Since there's no "neither" outcome, the number of rolls you can possibly make is bounded; "three successes before three failures", for example, will never exceed five rolls, because after the fifth roll you're guaranteed to have at least three of one outcome or the other. That makes it possible to enumerate every possible sequence of outcomes:

SSS SSFS SFSS FSSS SSFFS SFSFS FSSFS SFFSS FSFSS FFSSS FFSSF FSFSF SFFSF FSSFF SFSFF SSFFF FFSF FSFF SFFF FFF

Let's assume that our S (success) result is rolling a 4 on 1d4, and our F (failure) result is rolling anything else. We can then work out the odds of each case, like so:

SSS = 1/64 = 16/1024 SSFS = 3/256 = 12/1024 SFSS = 3/256 = 12/1024 FSSS = 3/256 = 12/1024 SSFFS = 9/1024 SFSFS = 9/1024 FSSFS = 9/1024 SFFSS = 9/1024 FSFSS = 9/1024 FFSSS = 9/1024 FFSSF = 27/1024 FSFSF = 27/1024 SFFSF = 27/1024 FSSFF = 27/1024 SFSFF = 27/1024 SSFFF = 27/1024 FFSF = 27/256 = 108/1024 FSFF = 27/256 = 108/1024 SFFF = 27/256 = 108/1024 FFF = 27/64 = 432/1024

Adding up our target outcomes, we get 106/1024 odds of three successes before three failures, and 918/1024 odds of three failures before three successes. (If these don't add up to 1, you know you missed a case somewhere.)

Once we've got the symmetrical case (i.e., X successes before X failures), we can deal with asymmetrical cases simply by changing up which rows in the probability tree we've already constructed count as which outcome – we needn't start from scratch. For example, the odds of two successes before three failures moves all the 27/1024 rows in the above table into the "overall success" bucket, and we arrive at 268/1024 odds. Similarly, "three successes before two failures" would flip the 9/1024 rows over to the "overall failure" bucket.

In this case, n is the number of save throws, k is the number of necessary throws for success, and p is the probability of a successful save throw.

The only complication is n varies depending on when you get enough of one outcome. If your first three saves succeed, you're not rolling anymore. But we can work around it by just considering what would have happened if we kept rolling after reaching our goal.

In other words, "probability you get X of outcome 1 before getting Y of outcome 2" is equivalent to "probability you get at least X of outcome 1 over X+Y-1 attempts".

In your example, there is a 16/1024 chance of rolling three successes in a row. If we rolled two additional times, the probabilities would have broken down like this: SSSSS = 1/1024 SSSSF = 3/1024 SSSFS = 3/1024 SSSFF = 9/1024

And they add up to 16/1024.

So the general formula would be something like