Calculating the probability of a crit with Perfect Strike

[Lastoth]

I've got a DPR sheet going and I am trying to show the effect of perfect strike on DPR (for the rounds it's used) but I am having some issues coming up with the right expression. Can anyone show me a good equation for this?

[Some call me Tim]

Let's assume your rolls are A and B.

max(A,B) is your attack roll.
min(A,B) is your confirmation roll.

Assume 20 is a crit threat and damage is 2 times.

IF max(A,B)=20 AND min(A,B)>=AC THEN DMG*2
ELSE IF max(A,B)>=AC THEN DMG
ELSE DMG=0

Adjust the pseudocode for whatever language you're using.

[slacks]

Someone should double check, but I think these work:

For AC 2-20:
[Perfect Strike Hit %] = 1-((([Target AC]-[Attack Bonus])-1)/20)^2
[Perfect Strike Critical %] = ((21-[Critical Threat Range])/20)^2

[Belafon]

Tim's example assumes you are using a random number generator. My own unwieldy beast of a spreadsheet was created to express the probability of things without a RNG, so if you are doing that and have already worked out normal crit values you can try this:

P(critical threat for 2-die perfect strike) = 1 - (P(not a threat on one die)*P(not a threat on one die))
P(crit conf for 2-die perfect strike) = P(crit conf on one die)

Before going on to the three-die case take a moment to think this through and realize that the chance of confirming a critical doesn't change. It doesn't matter that it's the lower of your two rolls. It's still just one die, and that value is independent of what was rolled on the other die. It can be a brain-burner, so take your time.

P(critical threat for 3-die perfect strike) = 1 - P(not a threat on one die)^3
P(crit conf for 3-die perfect strike) = 1 - P(not a crit conf on one die)^2

Hope this helps!

[Paul Zagieboylo]

slacks wrote:

Someone should double check, but I think these work:

For AC 2-20:
[Perfect Strike Hit %] = 1-((([Target AC]-[Attack Bonus])-1)/20)^2
[Perfect Strike Critical %] = ((21-[Critical Threat Range])/20)^2

The hit chance is right, but your crit chance requires a critical threat on both dice, which isn't right. You should just need a critical threat on one die and a hit on the other, and they can be in either order.

The way to figure this out is to realize that the probability of rolling a given number N or higher on a d20 is always (21-N)/20. So try to express probabilities in this form, and you'll get less confused.

I think the proper formula is as follows:

P(crit) = P(threat on 1st die)*P(hit on 2nd die) + P(hit but no threat on 1st die)*P(threat on 2nd die)
= ((21-Threat)/20)*((21-(AC-Atk))/20) + ((Threat-(AC-Atk))/20)*((21-Threat)/20)

Note that this assumes any critical threat is high enough to hit, and that the attacker still hits naturally on a 20 (and isn't relying on the "20 always hits" rule). If this is not the case (for example you have a keen sai but need a 20 (or worse, 21!) to hit), then the formula gets more complicated. In this case:

P(hit and threat on 1st die) = (21-MAX(Threat, MIN(20, AC-Atk))/20
P(hit on 2nd die) = (21-MIN(20, AC-Atk))/20

P(hit but no threat on 1st die) = MAX(0, (Threat-MIN(20, AC-Atk))/20)
P(hit and threat on 2nd die) = (21-MAX(Threat, MIN(20, AC-Atk))/20 (same as 1st die)

Then just take those and plug them into the expression above. I recommend making separate cells in your spreadsheet for each of these, and then putting them together into the final probability in a separate step.

Unfortunately Belafon is incorrect in his claim of independence. Although the values of the two dice are (clearly) independent, the "higher die" is not independent of the "lower die". You can see this more easily with d3s than d20s, because you can easily draw out the whole table yourself (it only has 9 entries instead of 400). Given that the "higher die" is 3, the probability of the "lower die" being at least 2 is only 3/5, not 2/3. I think Belafon's method is equivalent to double-counting the cases in which you roll a critical threat on both dice.

The three-die case gets kind of complicated. If you need it I can try to work it out.

[slacks]

@Paul
You are right I mixed some things up.

[Belafon]

Paul Zagieboylo wrote:

Unfortunately Belafon is incorrect in his claim of independence. Although the values of the two dice are (clearly) independent, the "higher die" is not independent of the "lower die". You can see this more easily with d3s than d20s, because you can easily draw out the whole table yourself (it only has 9 entries instead of 400). Given thatthe "higher die" is 3, the probability of the "lower die" being at least 2 is only 3/5, not 2/3. I think Belafon's method is equivalent to double-counting the cases in which you roll a critical threat on both dice.

I knew this was going to get me in trouble :) One of the coolest things about probability is how counterintuitive it is sometimes. And how much depends on how you state things. So, here we go:

My statement was that the chance of confirming a critical was independent. I could have been more exact and said that the chance of confirming a threatened critical is independent. There are three possible questions: 1) "What are my chances of threatening a critical?" 2) "If I threaten a critical, what are my chances of confirming it?" 3) "What are my chances of rolling a confirmed critical?" My spreadsheet math uses different cells for P(threaten critical) (first question) and P(confirm a threatened critical) (second question). Multiply the two together and you get the chance for rolling a confirmed critical (third question).

I use different cells because there are feats, traits, and weapon properties that can affect your chance to confirm a crit but not to hit in the first place. Some people may use one big honking formula in a single cell but I prefer smaller bites that give more granularity to my numbers. I also find they are easier to modify for special cases like the 3-dice case (see above for numbers) (and are easier to troubleshoot).

[Paul Zagieboylo]

@Belafon It still isn't true, in the case of Perfect Strike. Rolling a critical threat and confirming that threat are not independent in the presence of Perfect Strike, even though they are in all other cases. This is what my example using d3s was meant to illustrate, but let's try again, using real numbers (gasp!).

Zahariel is Perfect Striking with his keen sai, so he rolls two dice. If either of them come up 19 or 20, he threatens a critical, and automatically looks at the other one to confirm it. (If he rolled a third die to confirm the threat, rather than looking at the other one, then they would be independent as you suggest, but he doesn't.) Now, he only needs to roll 11 to confirm a critical hit. Given that one of the dice came up 19 or 20, what are the chances that the other one is at least 11?

Surprisingly, it's less than 1/2. Draw out the entire matrix of results for these two dice (20x20 = 400 entries, I'll wait). Now, in 76 cases, one of the dice is 19 or 20, so these outcomes are the ones that threaten a critical. Out of these cases, only 36 of them have the other die 11 or higher, so his chance of confirming a random critical is 36/76 = 9/19, just less than 1/2, which is his chance to confirm a normal critical hit by rolling an independent die.

"What the hell?" I hear you ask. "Surely," you assert, "using Perfect Strike should make me more likely to get critical hits, not less." And it does. The probability of any given threat being confirmed is very slightly less, but the probability of getting a threat at all is almost doubled (or tripled for a high-level weapon master monk)! Zahariel's normal chance to roll a critical threat is 1/10, just like everyone else with a 19-20 weapon; his chance to confirm these threats is exactly 1/2, so he scores a critical hit on 1/20 of his normal attacks. With Perfect Strike, he rolls a threat on 19/100 of his attacks, which is much higher; he confirms 9/19 of these, so the chance to critically hit with a Perfect Strike is 9/100, much better than 5/100 for normal attacks.

[Belafon]

Paul Zagieboylo wrote:

@Belafon It still isn't true, in the case of Perfect Strike. Rolling a critical threat and confirming that threat are not independent in the presence of Perfect Strike, even though they are in all other cases. This is what my example using d3s was meant to illustrate, but let's try again, using real numbers (gasp!).

I think you are still misunderstanding the statement "Your chances of confirming a threatened critical are no different if you are using perfect strike than just rolling one die after a normal threat." The presupposition is that a critical is threatened.

Let's use your D3 example. We'll assume we threaten on a 3 and confirm on a 2. Normally, if I roll a 3, I re-roll and if it's a 2 or 3 I confirm. So my chance of confirming a threatened critical is 2/3. Now let's go to Perfect Strike. We roll two dice, we'll call them "red" and "blue." If I roll a 3 on the red die, the blue die is 1, 2, or 3. If I roll a 3 on the blue die, the red die is 1, 2, or 3. If one die threatens, the chance of a confirming roll on the other is 2/3, the same as a "normal" confirmation roll.

You can try this at home with a D4 (since they exist, unlike our hypothetical D3s). Assume you threaten on a 4 and confirm on a 3. Over a very large number of attempts (we're talking hundreds if not thousands here), you will find that you confirm 1/2 of your threatened criticals. Remember to only count IF a 4 appears on at least one of the dice. Better yet, do a bunch of Monte Carlo runs.

The problem with your matrix is that it is a summary of all possible states before a critical is threatened. Once we have the information that a critical is threatened on at least one die, the problem collapses to a question of what is on the other die. The probability of rolling two 20s on 2D20 is 1/400. But GIVEN that you rolled at least one 20, the chances that the other die is also a 20 is 1/20, not 1/400.

[Belafon]

The simplest example I can think of:

Suppose I am going to flip two fair coins. If I ask you what the probability of them both coming up heads is, you will say (correctly) that it's 25%. I flip them and one lands on the table heads-up in front of you while the other falls on the floor. What is the probability that the second coin is heads?

[slacks]

The numbers that confirm a critical are a subset of the numbers that create a critical threat. So Belafon is right that if you roll one die and it could create a critical threat then the chance of confirming the critical is the same. If you roll under AC on the second die then you don't critical, if you roll over AC but under the first die then you confirm with the second die, if you roll over AC and over the first die then you confirm with the first die.

[Paul Zagieboylo]

Belafon wrote:

The problem with your matrix is that it is a summary of all possible states before a critical is threatened. Once we have the information that a critical is threatened on at least one die, the problem collapses to a question of what is on the other die. The probability of rolling two 20s on 2D20 is 1/400. But GIVEN that you rolled at least one 20, the chances that the other die is also a 20 is 1/20, not 1/400.

No it is not (the correct value of the conditional probability is 1/39). This is where you go wrong. The probability of "rolling at least one 20" is not the same as the probability of "rolling a 20 on the red die", nor is it the same as "rolling a 20 on the blue die", nor even is it the sum of these. It's less than the sum by exactly the chance that you got 20 on both dice, and this is where you're overcounting.

You have to start with the full matrix to get the right answer. Go back to the definitions. The conditional probability of an event P(B|A) is always P(B and A) / P(A). This is the only correct way to calculate a conditional probability, and it works whether or not A and B are actually independent events. Here we have A: "a critical was threatened" and B: "a threatened critical was confirmed." In the d3 example, P(A) is 5/9, which is the chance of rolling at least one 3 on 2 d3s. P(B and A) is the probability of scoring an actual critical hit. The only outcomes this includes are (2,3), (3,3), and (3,2), so the probability P(B and A) is 3/9. Therefore, the probability P(B|A) is just P(B and A)/P(A), which is (3/9)/(5/9) = 3/5, not 2/3.

Belafon wrote:

Let's use your D3 example. We'll assume we threaten on a 3 and confirm on a 2. Normally, if I roll a 3, I re-roll and if it's a 2 or 3 I confirm. So my chance of confirming a threatened critical is 2/3. Now let's go to Perfect Strike. We roll two dice, we'll call them "red" and "blue." If I roll a 3 on the red die, the blue die is 1, 2, or 3. If I roll a 3 on the blue die, the red die is 1, 2, or 3. If one die threatens, the chance of a confirming roll on the other is 2/3, the same as a "normal" confirmation roll.

The fallacy in this argument is an incorrect application of the excluded middle. You took the statement "a critical was threatened" and rewrote it as "either the red die threatened, or the blue die threatened", and then implicitly treated these as disjoint possiblities by addressing them separately. The problem is, they aren't; you could be in both possibilities at once, and you end up counting twice any outcome that is in both.

The correct rewriting is "either only the red die threatened, or only the blue die threatened, or both did". These three possibilities are now disjoint, and you can treat them separately.

Belafon" wrote:

Suppose I am going to flip two fair coins. If I ask you what the probability of them both coming up heads is, you will say (correctly) that it's 25%. I flip them and one lands on the table heads-up in front of you while the other falls on the floor. What is the probability that the second coin is heads?

This example is fallacious: it is not equivalent to the Perfect Strike example, because the coins are being treated differently. Here you are only considering the case in which the red coin (die) threatens a critical. In this case the events are independent, as expected.

For Perfect Strike, they aren't. The Perfect Strike example with coins is "Suppose I flip two fair coins, but they both fall on the floor. After some searching around, you find one of them (but you don't know which), and it came up heads. What is the probability that the other also came up heads?" The answer is 1/3, not 1/2. This last example is exactly identical to the Boy or Girl paradox, most famously posed by Martin Gardner. Go there and read up.

But don't believe me on faith. I would yell at my students if they did that (yes, I spent several years teaching mathematics and computer science at the college level for Carnegie Mellon). Go write that Monte Carlo simulation. Post the code and the results.

[Shisumo]

I think someone worked out the probability calculus for the three-die version of Perfect Strike on the DPR Olymipcs thread - I don't remember the result, save that it involved factoring X^3 + Y^3 + Z^3 and only taking some of the terms. Searching the site should pull it up, though.

EDIT: Actually, here it is. Though how you turn that into code for a spreadsheet I've no idea...

[Belafon]

Paul Zagieboylo wrote:

Belafon wrote:

Suppose I am going to flip two fair coins. If I ask you what the probability of them both coming up heads is, you will say (correctly) that it's 25%. I flip them and one lands on the table heads-up in front of you while the other falls on the floor. What is the probability that the second coin is heads?

This example is fallacious: it is not equivalent to the Perfect Strike example, because the coins are being treated differently. Here you are only considering the case in which the red coin (die) threatens a critical. In this case the events are independent, as expected.

For Perfect Strike, they aren't. The Perfect Strike example with coins is "Suppose I flip two fair coins, but they both fall on the floor. After some searching around, you find one of them (but you don't know which), and it came up heads. What is the probability that the other also came up heads?" The answer is 1/3, not 1/2. This last example is exactly identical to the Boy or Girl paradox, most famously posed by Martin Gardner. Go there and read up.

Going to spoiler text here because we aren't helping the original poster:

Spoiler:

I did oversimplify the coin flip, at the expense of exactitude. Your way was indeed correct. My first draft ended up in the Monty Hall situation of "I show you one coin, which is heads" (in which situation the odds of the other being heads is 1/3) and I didn't get specific enough in my final example. However your own example, the one I was intending to illustrate, ends up in the Boy or Girl ambiguity interpretation of "From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2."

That very interpretation is the one I am trying to illustrate. Probability problems just depend so much on interpretation. The math is completely repeatable but differences in setup can result in more "correct" answers than almost any other hard science. I don't want to get too much more into this, so I think we're going to have to agree to disagree.