Ok, looking at some of the math here makes me sad. May I put the following in?
Given 2 dice being rolled, and they are used for hit/confirm....
You have 20x20 as the combinations of die rolls. 1/1, 1/2, 1/3, ... 20/18, 20/19, 20/20.
That's 400 combinations.
Given a 20 to crit and a 15 to confirm, you have 11 combinations that crit and confirm. 15/20, 16/20, 17/20, 18/20, 19/20, 20/20, 20/19, 20/18, 20/17, 20/16, 20/15.
Using the two dice rolled, your chances of having a confirmed crit are 11/400, or 0.0275.
Your chance of doing it using two different dice are :
Chance of a 20 on dice 1 : 1/20th, or 0.05
Chance of a 15 or higher on dice 2 : 6/20ths, or 0.3
Chance of confirming a crit on two dice rolled seperately : 0.05 * 0.3 = 0.015.
You are about 0.0125 more likely to get a confirmed crit using two dice in the manner specified by the ability than you are rolling separately. It's a bit counterintuitive, but that's how percentages and probability work. The reason is, with rolling two dice separately, you don't roll the second dice unless your first die rolls a 20. That means you are taking 5 possible ways of getting the crit out of the equation.
What mdt was trying to say is that you didn't actually say what the probabilities of getting a crit was in each case. You only give the number of possible outcomes that gives a crit, while forgetting to include the number of possible outcomes in total (which is needed for calculating the probability).
In Plan A, there are 400 possible outcomes. With the 11 possible successes, the probability is 11/400 (or 0.0275).
In Plan B, there are 8000 possible outcomes. With the 120 possible successes, the probability is 120/8000 (or 0.015).
Edit: There are actually more successes in Plan B. There are 39 possible ways to threaten, rather than the posited 20. (20/1,1/20,20/2,2/20 and so on up to 20/20). So, there are 39*6=234 possible successes, giving a probability of 234/8000 (or 0.02925).
So, Plan B does give more criticals, but not as many more as Rubia claims.
And I don't think you're right about the number of outcomes. Regardless of the number of dice rolled, the sample space is still a pair of numbers from 1 to 20.
I edited my post, since I noticed that the number of successes in Plan B was wrong originally.
Isn't the sample space still of size 400?
It can't be. There are 400 possible outcomes to the initial roll to see if you threaten in the first place.
Then, if you do, there are 20 possible outcomes to the roll to confirm.
Isn't the sample space still of size 400? We do extra things to determine two numbers, but it's still just two numbers. In other words, we're collapsing away a large number of irrelevant outcomes due to the max function in the first two dice.And I don't think you're right about the number of outcomes. Regardless of the number of dice rolled, the sample space is still a pair of numbers from 1 to 20.
I edited my post, since I noticed that the number of successes in Plan B was wrong originally.
Nope.
This is part of what I was trying to point out. If you take two dice separately, you do not have the same sample space. The reason being, that Die 1 only has 20 sample spaces, and only one of those sample spaces is a crit. Therefore, you have sample space 1 (since we are selecting for a crit), and sample space 20 on the second die. In other words, you only have a sample space of 20 combinations taking the dice one at a time, because you are self selecting only the 20 on the first die (since you do not roll the second die for 1 to 19, therefore they fall out of this equation). So, the sample size of rolling a crit using two dice separately is the sample size of a crit on die 1 (which is 1 in 20) and the sample size of 15 or higher in sample set two (which is 6 in 20) multiplied by each other (which is where they intersect).As to Plan A vs Plan B, I didn't even make any comments about your plan B. I was talking specifically about the original 2 dice vs 1 die + 1 confirm debate.
However, if you want me to comment on Plan B...
It's similar to 1 die + 1 confirm.
Your sample set for a crit is any roll where we have a 20 on either dice. That happens 39 times out of 400. That is, 39/400 = 0.0975.
Now, your chances of a 15 or higher on the third die are 6/20ths, or, as we saw earlier, 0.3.
The chances of a crit confirmation using Plan B is 0.0975 * 0.3 = 0.02925. Basically, you gain 0.002 by adding the third die. Statistically, that's not a large increase over the specified method used by the ability.
Ok, looking at some of the math here makes me sad. May I put the following in?
Deleted stuff that wasn't relevant...
I recommend that you read my post more carefully. The question wasn't whether Perfect Strike is better than rolling two dice separately. That's crystal clear. In fact, that's written clearly in the description of Plan A in my previous post.
The difference is between Plan A and Plan B. I entreat you to read my post carefully before further obfuscating the issue.
Rubia
Actually, it wasn't very clear from a statistical standpoint, which is why I cleared it up. As to your plan B, as I said in the post above, I never said I was addressing your plan B.
Yeah, I spaced the fact that the probability of each item in the sample space isn't the same. Sigh. I can't edit my original post, so here's the modification:...
Sorry for the errors, guys.
LOL,
No worries. Statistics are a pain to get down. I have a MST in mathematics, and I still have to sit and think about it for a bit. And you can twist them to just about any argument you want, one of the reasons they call it Lies, Damn Lies, and Statistics. :)
I don't argue with the stats, but I felt like trying to render a less mathematical version:
Assume that every time you made an attack roll, you rolled to hit and then to confirm (and just ignored the confirmation if you didn't threaten). This really isn't any different from what happens now (the only difference is in throwing the second die and ignoring it, instead of not throwing it). What Perfect Strike lets you do is swap the two, so that what would have been a wasted confirmation of a threat that didn't happen becomes a threat, and what would have been a miss becomes a failed confirmation. You sacrificed confirmation of a crit, in order to hit in the first place. If you were instead to roll a third die to confirm, you wouldn't be making that trade - you'd just be rolling two dice to hit in the first place. Which is a perfectly viable mechanic, but a very different one from what Perfect Strike does, and thus the text specifying what it does do is necessary.
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Your supposition is precluding possibilities (a 20) as a given. But it's not a given.stuff...it has nothing to do with the gobbledegook of crit-confirmation because, once you've chosen either dice to be the attack die, the odds that the remaining one confirmed a threat are the same as any randomly thrown nth dice, and it does not matter if Schroadinger's box is open (the die is already rolled and on the table) or closed (a 3rd die yet unrolled in the instance the "leftover" die may not be used).
And specifically, this is not true---the probability of confirming is not necessarily the same. The counter-example is constructed as follows:
Suppose that you can threaten a critical on a 19 or 20. Any perfect strike roll that results in at least one roll of 20 does not contradict your premise. However, suppose that your highest result was 19. The roll was (19,X), for some X. However, you know that X is at most 19, and therefore, its value cannot include a 20. However, a third die could include a 20. Thus, the chance of rolling the nth die is different from that of the non-chosen die from perfect strike.
You can, of course, extend this argument arbitrarily for any critical threat range. Thus, your premise fails.
Rubia
The problem here is you're getting tunnel vision into equations, and not appreciating what is obvious: The player chooses one die of two for his attack role. Provided it's a critical-threat, the odds that the remaining die confirms the threat are exactly the same as any random die. That the threat-confirmation die was "rolled in advance" as a mechanic of the feat doesn't change the odds that it rolled what it did.
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Assume that every time you made an attack roll, you rolled to hit and then to confirm (and just ignored the confirmation if you didn't threaten). This really isn't any different from what happens now (the only difference is in throwing the second die and ignoring it, instead of not throwing it). What Perfect Strike lets you do is swap the two, so that what would have been a wasted confirmation of a threat that didn't happen becomes a threat, and what would have been a miss becomes a failed confirmation. You sacrificed confirmation of a crit, in order to hit in the first place. If you were instead to roll a third die to confirm, you wouldn't be making that trade - you'd just be rolling two dice to hit in the first place. Which is a perfectly viable mechanic, but a very different one from what Perfect Strike does, and thus the text specifying what it does do is necessary.
Within the context of the player rolling two dice simultaneously, the text is meaningless.
The problem here is you're getting tunnel vision into equations, and not appreciating what is obvious: The player chooses one die of two for his attack role. Provided it's a critical-threat, the odds that the remaining die confirms the threat are exactly the same as any random die. That the threat-confirmation die was "rolled in advance" as a mechanic of the feat doesn't change the odds that it rolled what it did.
The bolded section is wrong. The odds that the remaining die are not the same as any random die, because you already picked the higher of the two as the attack roll. Example: if you rolled an 18 and a 15, then the attack roll was an 18, and the confirmation roll isn't chosen from the [1,20] space, it's chosen from the [1,18] space. There's less of a chance of it confirming than a randomly rolled die made after you've threatened.
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The only point of contention is whether or not crits confirm differently when, after two dice are rolled and one die yields a thread, you are required to either A: accept the other as the confirmer, or B: use an external third die.
-- If you are arguing anything else in this thread, then you have misunderstood the discussion.
* * * * *
A (Perfect Strike as written):
....[roll 2d20, choose one for attack, the other to confirm any threat]
vs.
B: (dual-roll without the line of text regarding confirmation):
...[roll 2d20, choose one for attack, ignore the other, roll a third d20 to confirm any threat]
* * * * *
Test #1: You need a 20 to hit (and therefore two 20s to crit). Two 20s are rolled.
Result: Player of A goes "Yay!"; and player of B goes "Dude, I've been punked! I have to throw away a perfectly good 20 to reroll a third die!"
-- In the case of two 20s, it would seem starkly obvious that B gets shafted by having to ignore his second 20; but if one "concludes" there, they're overlooking the fact that A is required to keep BAD rolls. For instance:
Test #2: You need a 20 to hit. A 20 and a 1 are rolled.
Result: A is required to keep the 1 as confirming die. B rolls a third die and has a chance to confirm the crit.
-- This same is true of any instance of 20 + (not 20). A is always required to keep, and B always rolls a third die.
In any instance of [20 + (anything)] when the 20 is kept as the attack die, the odds that A's 2nd die or B's reroll die are a 20 are exactly the same: 5%.
-- This same is true of any instance of 20 + (not 20). A is always required to keep, and B always rolls a third die.In any instance of [20 + (anything)] when the 20 is kept as the attack die, the odds that A's 2nd die or B's reroll die are a 20 are exactly the same: 5%.
You are ignoring the fact that if you roll two dice for the first die roll, then you have a greater chance of rolling a 20 for that first roll.
Chances of rolling a 20 on one die = 1/20
Chance of rolling a 20 on either of two dice = 39/400
1/20 = 0.05
39/400 = 0.0975
In other words, you are almost twice as likely to get a 20 using 2 dice, roll extra to confirm, over 1 dice, roll one to confirm. That is what people have been trying to tell you over and over again, but you don't seem to get it.
You are looking only at the confirmation die, and not what went into getting the 20 in the first place. That is why you do not understand why there is a difference between roll 2 use one as confirmation and roll 2, roll a third as confirmation.
I find it funny that an argument about a single phrase being wasted space is now 74 posts long.
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I'm sorry mike schneider but I believe you are missing the importance of rolling the 2 dice at once. The importance and difference is....
For hitting/missing: You can take the highest one of 2 and have a greater chance of hitting.
For Criticals: If either of the dice are a possible critical you can take it and confirm with the other. Because you have 2 dice rolling your chance of actually scoring a possible crit is increase.
... I hope this short explanation helps.
Edit: After re-reading your original post I will say that you are correct in saying there is no statistical difference between rolling 2 taking the highest (using the other to confirm) and rolling 2 taking the highest (then ignoring the lower and roll a 3rd die to confirm). It's possible that you understand the significance of rolling 2 initially. Sorry If my statement was too rude.
Read all the thread. I think I understand where Mike is trying to lead us;
(I think) He means that if you roll the two dices and take the highest (1st), let's say in this case a critical threat; the chances to confirm using a 3rd dice or the 2nd one are the same. Which is true. We're talking about a d20. BUT! It's only common logic that it has to be said in advance : I reroll my threat with a 3rd, or I take the 2nd dice.
If you have the option to reroll or not, of course it increases odds to confirm.
I think the phrasing of the feat was designed this way to prevent cheating, by choosing to reroll a 3rd or not. Simple as that. The phrasing is not superfluous.
With all the arguments on these boards, I can understand the designers to write more precise rules to avoid cheese.
(I think) He means that if you roll the two dices and take the highest (1st), let's say in this case a critical threat; the chances to confirm using a 3rd dice or the 2nd one are the same. Which is true.
The maths shown in several posts above say that is actually not true. There is a difference, although it is small.
(I think) He means that if you roll the two dices and take the highest (1st), let's say in this case a critical threat; the chances to confirm using a 3rd dice or the 2nd one are the same. Which is true.The maths shown in several posts above say that is actually not true. There is a difference, although it is small.
It's because Mike is framing his statement in the wrong way. He's creating a mathematical disconnect. Here's an example...
Three adventurers stay at an inn. The innkeeper charges them 3 silver for their room. The adventurer's shrug and each coughs up a silver piece. Later, the innkeeper realizes he made a mistake on the charges, and he hands 5 copper to his serving wench. "Here, those three adventurer's in room 1? Give them these, I overcharged them."
The wench, walking up the stairs, is trying to figure out how to split 5 coppers up amongst three adventurers. Figuring they don't know the exact refund, she gives them 3 coppers, one each, and keeps two coppers for herself.
Now, each Adventurer paid 9 coppers for his room. Three times 9 is 27. Plus the 2 coppers in the wench's pocket makes 29 coppers.
Where's the missing copper piece?
This is basically what Mike is doing with the way he is working, he's framing the issue incorrectly.
Aaaaaaaaaah missed this particular one... I lay down the weapons, and since I can't edit my post... Here's why I am proven wrong :
Suppose that you can threaten a critical on a 19 or 20. Any perfect strike roll that results in at least one roll of 20 does not contradict your premise. However, suppose that your highest result was 19. The roll was (19,X), for some X. However, you know that X is at most 19, and therefore, its value cannot include a 20. However, a third die could include a 20. Thus, the chance of rolling the nth die is different from that of the non-chosen die from perfect strike.
Like Are said... there IS a difference, though it's one hell of a small one :P
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To come in with another perspective on explaining it. The reason why rolling a third die is mechanically different (and increases the chances of confirming a critical) is because by rolling a third die you are putting possibilities back into the "pot" as it were. With Plan A, there are only 400 possible combinations of numbers: {1,1},{1,2},{1,3} etc. However in Plan B there are 8000 combinations: {1,1,1},{1,1,2},{1,1,3}, etc. The question really becomes: Are there possibilities in one Plan that the other plan doesn't have? The answer is: Yes. In fact, there are 14 out of 8000 additional possibilities in Plan B than there are in Plan A.
Here's a few examples, by equalizing the possibilities:
Plan A:
Roll 20 and 20, then roll any number on a d20. I threat and confirm.
Roll 20 and 19, then roll any number on a d20. I threat and confirm.
Roll 20 and 14, then roll any number on a d20. I threat, but only hit.
Plan B:
Roll 20 and 20, then roll any number 15 and above. I threat and confirm.
Roll 20 and 19, then roll any number 15 and above. I threat and confirm.
Roll 20 and 14, then roll any number 15 and above. I threat and confirm.
If we take those possibilities above for a confirmed critical and add them together here is what we get.
Plan A:
1st die rolls a 20: 6*20=120 (2nd die needs a 15, 3rd can roll anything)
1st die rolls a 19: 1*20=20 (2nd die needs a 20, 3rd can roll anything)
1st die rolls an 18: 1*20=20
...
1st die rolls a 15: 1*20=20
1st die rolls a 14: 0*1=0 (Even with a 20 on second die, no chance of confirmation)
(6*20)+(5*20)=220
Plan B:
1st die rolls a 20: 20*6=120 (2nd die can roll anything, 3rd needs a 15)
1st die rolls a 19: 1*6=6 (2nd die needs a 20, 3rd needs a 15)
1st die rolls an 18: 1*6=6
...
1st die rolls a 1: 1*6=6
(20*6)+(19*6)=234
TLDR version: Rolling a third die adds more possibility of a confirmed critical overall. While it may cause some rolls from Plan A to fail to crit, it will add more possibilities with Plan B to succeed at critting than it takes out.
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Looking at the numbers, I realized that adding a third die roll increases the number of probabilities of a critical equal by the (# needed to hit enemy -1). This is easiest seen at the maximum numbers.
Plan A for 20 to hit:
Need two 20s, and then any number rolled on third die:
{20,20,20} or {20,20,1-19}
1+19=20/8000
Plan B for 20 to hit:
Need one 20 on either die, then a 20 on the third die:
{20,20,20} or {20,1-19,20} or {1-19,20,20}
1+19+19=39/8000
Then looking at needing a 2 to hit:
Plan A:
{20,20,1-20} or {2-19,20,1-20} or {20,2-19,1-20}
20+(18*20)+(18*20)=740/8000
Plan B:
{20,20,2-20} or {20,1-19,2-20} or {1-19,20,2-20}
19+(19*19)+(19*19)=741/8000
Hopefully this helps you see how the third die roll is mechanically different as well...
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Flip a coin (total endings) : 2 = F
Flip a coin and get heads (amount) : 1 : 50%
Flip a coin and get tails (amount) : 1 : 50%
Flip a coin and get heads or tails : 2 : 100%
Flip 2 coins and get 1 heads (01|10)&(00|01|10|11): 2 out of 4
Flip 2 coins and get 2 heads (11)&(00|01|10|11): 1 out of 4
Flip 3 coins and get 1 heads (001|010|011|100|101|110|111)&(000|001|010|011|100|101|110|111): 7 out of 8
Flip 3 coins and get 2 heads (011|101|110|111)&(000|001|010|011|100|101|110|111): 4 out of 8
if 1 head equates to a threat, and 2 heads equate to a confirm
then given that the first 2 are flipped together, threats happen 2/4 (50%), and confirmations happen 1/4 (25%)
or .5*.25 = .125
then given that the first 2 are flipped together and 1 tail may be ignored and re-flipped (turning it to 3 flips), threats happen 7/8 (87.5%), and confirmations happen 4/8 (50%)
or .875 * .50 = . 4375
however the actual course (for three flips is) is:
Flip 2 coins and get 1 heads (01|10)&(00|01|10|11): 2 out of 4
followed by flipping a third
Flip 3 coins and get 2 heads (011|101|110|111)&(000|001|010|011|100|101|110|111): 4 out of 8
of .5*.5 = .25
notice that all of the good outcomes for 3 (011|101|110|111) include the good outcomes for 2 (01|10) which alludes to the fact that in order to flip 3 and get 2 heads you must first flip 2 and get at least 1 head
now as .25 > .125 I believe I have proven that (through binary math) 2 flips for threat and another flip for confirm is statistically better than 2 flips and statistically worse than 2 and a re-roll (ie 3 for all cases)
and as a outro
tails = 0
heads = 1
3 coins = (0|1) + (0|1) + (0|1)
(000|001|010|011|100|101|110|111) & (000|001|010|011|100|101|110|111)
= 8/8
= 100%
and all of my cases are tautologically correct