Probability of a pair of results in a series of attempts?

Short and simple, how would I calculate the chance of getting both a 20 and a [16-20] in 3 rolls of a 20-sided die?

Would it be the chance of getting a single 20 in 3 rolls * the chance of getting the range in 2 rolls?

First let’s clarify the question. Are you asking for the probability that, if you roll three twenty-sided dice (or roll one of them three times in a row), at least one of them comes up 20, and at least one of the other two is in the 16-20 range?

The probability that at least one comes up 20 would be 1 - (19/20)^3 = 0.142625, because (19/20)^3 is the probability that this does not happen: that all three are some number other than 20.

Of the other two dice, the probability that at least one is in the 16-20 range would be 1 - (3/4)^2 = 0.4375.

The probability that both these things happen, then, would be 0.142625 * 0.4375 = 0.0624.

At least, that’s what looks right to me. Anyone want to check my logic?

I think Thulow’s method isn’t quite right because of the possibilities of multiple 20’s is not adequately taken into account.
Your first pass allows for there to be two twenties, but then it is unclear what you mean by “the other two dice”

The easiest way is probably to work through all the combinations.

Let T be a twenty, S be a 16-19, X be a 1-15.

The winning possibilities are

TTT, TTS, TSS, and TSX, as well as all possible re-orderings of these (for example SXT , STS etc.)

So all that remains is to work out the probability of an instance of each win and the number of reordering of each win.

TTT= (1 order)(probability=1/201/201/20) = 1/8000
TTS= (3 orders) (probability=1/201/20
4/20)=12/8000
TSS= (3 orders)* (probability =1/204/204/20)=48/8000
TSX= (6 orders)* (probability =1/204/2015/20)=360/8000

Grand total probability =421/8000 = 0.0526

Thudlow Boink, you correctly interpreted what I was asking. Just wanted to respond to your question ASAP.

And Buck Godot’s answer makes sense. Barring any surprises I guess that means thread solved. Thanks for the quick responses guys!

No wait, that (my previous answer) can’t be right. The probability has to be some fraction of 8000 (=20^3), which that answer is not. I think that method may have counted some outcomes twice.

The brute force approach of running through all 20^3 possibilities yields 466/8000 = 0.05825, if I’ve figured it right.

You missed
TTX = 3 1/201/20*15/20 = 45/8000

And indeed, 466 = 421 + 45

Thanks, septimus. I like Buck Godot’s approach, especially now that you’ve corrected it so that it yields the same answer as what I got.

For the record, I got my second answer by writing a short computer program (with three nested loops) to check through all the possibilities. But, in retrospect, here’s what it amounts to, as I explained it to myself when I was trying to figure out whether it was Buck or I who had made a mistake somewhere.

First, look at all rolls in which die #1 comes up 20. There are 20*20 such, and 7/16 (= 1 - (15/20)(15/20)) of them have a 15-20 on at least one of the other two dice. This accounts for 175 possibilities.

Then look at all rolls in which D#2 is 20 but D#1 is not. There are 19*20 such, and 1 - (15/19)(15/20) of them have a 15-20 on at least one of the other two dice, keeping in mind that we’ve eliminated the possibility that D#1 could have a 20. This accounts for 155 possibilities.

Then look at all rolls in which D#3 is 20 but D#1 and D#2 are not. There are 19*19 suck, and 1 - (15/19)(15/19) of them have a 15-20 (really, a 15-19) on at least one of the other two. This accounts for 136v possibilities.

And 175 + 155 + 136 = 466.

But I do think Buck Godot’s approach is more elegant.

Well, answers can be fast, cheap or correct, pick two out of three.

I guess I picked fast and cheap. :stuck_out_tongue:

My first thought on reading this is that someone is trying to calculate the odds of getting a critical hit. I think a Greatsword has a critical threat range of 18-20 if that helps.

That was my thought too. The (correct) answers above assume that the order you get the results doesn’t matter, but that’s not the case for critical hit calculations. As I recall, a weapon threatens a critical when you roll in a certain range, and after that you roll again; a twenty gives you a critical. I don’t see where the third roll comes from.

Yes and no…

I was playing around with some game design ideas of my own. The general gist of it is a hack & slash style, quick and dirty amalgamation of mechanics from D&D 3E, Final Fantasy Tactics, and WoW. Still on the fence whether it would be for an app, or a boxed tabletop game ala Heroes’ Quest. I’d been thinking about it ever since I started learning about D&D* for Neverwinter Nights and Penny Arcade’s D&D 4E podcasts. What spurred me to start putting pen to paper was reading about this new game, Super Dungeon Explore, by Soda Pop Miniatures.

Anyway, to make a long story short (too late!) I was tinkering with trying to make a character with abilities to strike a bunch of times with standard attack rolls, or once with a to-hit bonus and expanded crit range. The idea was that the multiple attacks should run a better chance of doing some damage, and have the possibility of doing the most, but the single would generally do more whether comparing crits or regular hits.

The example figures where off the top of my head and actually a little backwards as a result. But not to worry, the method of finding the probability for specific combinations on multiple rolls was the important part.

Thanks again guys.

*I never knew anyone growing up who played Dungeons & Dragons, so I never got a chance to play it myself.

In 3rd edition, at least, the first roll determines whether it’s a hit, based on your attack bonuses and your opponent’s armor class, with a roll of 20 always being a hit regardless of AC or bonuses. Then, if it’s a hit and if the number rolled is in the weapon’s critical threat range (which varies from weapon to weapon, and can be anywhere from 20 for a nonmagical axe to 15-20 for a keen magical scimitar), you roll another attack roll to confirm the crit. If the second attack roll is enough (with all bonuses and considering the armor class) that it would have been a hit, then it’s a critical hit and does extra damage.