MATH EXPLANATIONS ONLY PLEASE
Question: What is more likely to BLOCK or PARRY?
if you don't care about <SHIELD STYLES>, ENGAGE, ARROWS, GUARD.
Consider the following OPTIONS:
1. to keep plugging pts into JUST one ITEM <SHIELD or PARRY>
30 Shield 6 Parry
2. to spread out pts between two items <PARRY & SHIELD>
23 Shield 20 Parry
--------------------------------------
I am Assuming the following is true, before making decision:
if LVL / 2 = BLOCK rate % round down?
if pts are increasing expansive as normal.
if your considering option 2. you have to take TWO ROLLS
if your considering option 1. you only have to ONE ROLL but lose in amount aloted pts but you have a greater chance on that one ROLL.
I assume someone who is good at math can figure this out
I haven't had stats in along time.
Assuming this is the problem? If someone can give a better reason one way or the other.
Option 1.
Roll ONE 3/100 chance of parry; Roll Two: 15/100 block
Option 2.
Roll ONE: 10/100 chance of parry; Roll TWO: 11/100 block
-------------------------
People in that past have just lumped these two SEPERATE ROLLS together can you do that, I don't think you can because they
are seperate rolls are they not?
-------------------------
If you assume you can get RvR pts % SC does that effect decision? <Master block and Parry> 3% to both? and +11 ~(5%) to one or both?
Gilez-50 Cleric Nimue
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Anonymous
I believe I may have solved my own problem since I was thinking about this question this weekend.
I believe this is the answer, if I am wrong send the correct
way to do this:
------------------
% Parry * % Shield = A
(Max % from above) + A = likely hood a block or parry is made.
------------------
Examples:
lets do the quarter test - lets assume I have a quarters and want to know if Heads lands up anytime out of two rolls.
%50 * %50 = .25
.5 + .25 = .75 chance of landing heads once
---------
So as to my own question now:
1. 30 Shield, 6 parry OR 2. 23 Shield 20 Parry
--------------
.15 * .03 = .0045
.15 + .0045 = .1545 basically .15
--------
.12 * .10 = .012
.12+.012 = .132 basically .13
--------
No matter what if you max SHIELD or max PARRY your going to beat splitting up points almost 90% of the time only time it even
comes close is within a few units from each other.
------------------------
CONCLUSION: Everyone who says put some into on then some into other to get more ablity to block/parry probably wrong.
------------------------
So 50 Shield ~= 42 Shield + 42 Parry
25%~=.21+.0441 or .25=.25
------------------------
But also note how much MASTER of BLOCKING or MASTER of Parry and SC +11 (shield or parry) could change this outcome.
------------------------
The old question EVADE Stealth question: I wondered so here's what I found out.
EVADE VII = .35%
EVADE VII > SHIELD 50 + PARRY 50
.35 > .31
EVADE VII < SHIELD 50 + PARRY 50 + EVADE I + MASTER BLOCKING + MASTER OF PARRY + SPELL CRAFTED +11 Parry +11 Shield
.35 < (~.40)
EVADE VII + DODGER III > SHIELD 50 + PARRY 50 + EVADE I + MASTER BLOCKING I + MASTER OF PARRY I + SPELL CRAFTED +11 Parry +11 Shield
.44 > (~.40)
----------------------------
SO YES!! EVADE VII IS BETTER Then FIGHTERS ablities and look what you had to give up verses them.
REALM pts (16) VERSES (2548) Skill pts + 2 SLOTS in SC +22 + 8 Realm pts and still 4% difference!!
By far EVADE wins hands down no wonder they kill my fighters.
<Gilez>
I believe this is the answer, if I am wrong send the correct
way to do this:
------------------
% Parry * % Shield = A
(Max % from above) + A = likely hood a block or parry is made.
------------------
Examples:
lets do the quarter test - lets assume I have a quarters and want to know if Heads lands up anytime out of two rolls.
%50 * %50 = .25
.5 + .25 = .75 chance of landing heads once
---------
So as to my own question now:
1. 30 Shield, 6 parry OR 2. 23 Shield 20 Parry
--------------
.15 * .03 = .0045
.15 + .0045 = .1545 basically .15
--------
.12 * .10 = .012
.12+.012 = .132 basically .13
--------
No matter what if you max SHIELD or max PARRY your going to beat splitting up points almost 90% of the time only time it even
comes close is within a few units from each other.
------------------------
CONCLUSION: Everyone who says put some into on then some into other to get more ablity to block/parry probably wrong.
------------------------
So 50 Shield ~= 42 Shield + 42 Parry
25%~=.21+.0441 or .25=.25
------------------------
But also note how much MASTER of BLOCKING or MASTER of Parry and SC +11 (shield or parry) could change this outcome.
------------------------
The old question EVADE Stealth question: I wondered so here's what I found out.
EVADE VII = .35%
EVADE VII > SHIELD 50 + PARRY 50
.35 > .31
EVADE VII < SHIELD 50 + PARRY 50 + EVADE I + MASTER BLOCKING + MASTER OF PARRY + SPELL CRAFTED +11 Parry +11 Shield
.35 < (~.40)
EVADE VII + DODGER III > SHIELD 50 + PARRY 50 + EVADE I + MASTER BLOCKING I + MASTER OF PARRY I + SPELL CRAFTED +11 Parry +11 Shield
.44 > (~.40)
----------------------------
SO YES!! EVADE VII IS BETTER Then FIGHTERS ablities and look what you had to give up verses them.
REALM pts (16) VERSES (2548) Skill pts + 2 SLOTS in SC +22 + 8 Realm pts and still 4% difference!!
By far EVADE wins hands down no wonder they kill my fighters.
<Gilez>
Anonymous
I think you're looking at your probabilities backwards...
for your quarter example, think of the probability of tossing at least one head as being the opposite of the probability of throwing all tails.
the probability of 2 tails is .5 * .5 = .25, so the probability of at least one head is 1 - .25 or .75.
Now your shield/parry... you are wanting to know, what are the odds I won't get hit, i.e. either block or parry. Well, this is basically, what's the probability I won't fail both, or 1 - (prob fail both rolls).
So, the chances of failing both is the product of the chance to fail each.
Case 1: 30 shield, 6 parry, or .15 success on shield, .03 success on parry. This is .85 fail shield, .97 fail parry, so .85 * .97 = .8245. Leaving .1755 chance to succeed on one or other. Round to 17% success.
case 2: 23 shield, 20 parry, or .12 shield, .10 parry, or failing .88 shield, .9 parry. .88 * .9 = .792 chance to fail both, or .208 chance to succeed on one. round to .20 chance of blocking an attack.
Hence, supporting the idea, medium on both is better than just high on one.
for your quarter example, think of the probability of tossing at least one head as being the opposite of the probability of throwing all tails.
the probability of 2 tails is .5 * .5 = .25, so the probability of at least one head is 1 - .25 or .75.
Now your shield/parry... you are wanting to know, what are the odds I won't get hit, i.e. either block or parry. Well, this is basically, what's the probability I won't fail both, or 1 - (prob fail both rolls).
So, the chances of failing both is the product of the chance to fail each.
Case 1: 30 shield, 6 parry, or .15 success on shield, .03 success on parry. This is .85 fail shield, .97 fail parry, so .85 * .97 = .8245. Leaving .1755 chance to succeed on one or other. Round to 17% success.
case 2: 23 shield, 20 parry, or .12 shield, .10 parry, or failing .88 shield, .9 parry. .88 * .9 = .792 chance to fail both, or .208 chance to succeed on one. round to .20 chance of blocking an attack.
Hence, supporting the idea, medium on both is better than just high on one.
Anonymous
this Gilez again
I guess I don't see how your comming up with these equations because your not doing 2 rolls like you need to in order to computer probablity.
==============
Your saying the quarter example is wrong and subsituing it with this type of explaination but no actual stat X something = Y equation.
--------------- YOUR EXAMPLE:
the probability of 2 tails is .5 * .5 = .25, so the probability of at least one head is 1 - .25 or .75.
I don't get the 1? what's that for?
--------------- where as MINE is this
% Parry * % Shield = A
(Max % from above) + A = likely hood a block or parry is made.
===================
What your not taken into account is that there are two rolls which are seperate from each other.
If I use your example, I can make your example make no sense.
Lets say I have 100% chance of block on ROUND one (ROLL 1)
I am using the simplest example I can think of...
Yours example ends up doing this
1 * 0 = 0
1 - 1 = 0 basically says you have 0 chance of bocking.
====================
Mine gives the correct result
1 * 0 = 0
(1) + 0 = 1 or 100% chance
====================
I guess I don't see how your comming up with these equations because your not doing 2 rolls like you need to in order to computer probablity.
==============
Your saying the quarter example is wrong and subsituing it with this type of explaination but no actual stat X something = Y equation.
--------------- YOUR EXAMPLE:
the probability of 2 tails is .5 * .5 = .25, so the probability of at least one head is 1 - .25 or .75.
I don't get the 1? what's that for?
--------------- where as MINE is this
% Parry * % Shield = A
(Max % from above) + A = likely hood a block or parry is made.
===================
What your not taken into account is that there are two rolls which are seperate from each other.
If I use your example, I can make your example make no sense.
Lets say I have 100% chance of block on ROUND one (ROLL 1)
I am using the simplest example I can think of...
Yours example ends up doing this
1 * 0 = 0
1 - 1 = 0 basically says you have 0 chance of bocking.
====================
Mine gives the correct result
1 * 0 = 0
(1) + 0 = 1 or 100% chance
====================
Anonymous
This is gilez forgot something, yeah I don't care if I am wrong but I want an equation from you that will work in all situations.
Gramma buys 99 out of 100 tickets for one drawing
Gramma buys 1 out of 100 tickets out of drawing two.
Then your saying splitting them up is better?
Which if she wouldn't of split up her points she would of had 100% chance of winning drawing ONE?
Gramma buys 99 out of 100 tickets for one drawing
Gramma buys 1 out of 100 tickets out of drawing two.
Then your saying splitting them up is better?
Which if she wouldn't of split up her points she would of had 100% chance of winning drawing ONE?
Anonymous
I might be taking your equation incorrectly, but however I still don't understand how your answer is explaining the fact that ANY either ROLL can stop the second ROLL in other words if your multiplying R1 * R2 but never taking into account that R1 or R2 can stop the other one from occuring. So IMO your taking an equation as tho I AM always THROWING TWO DICE down when in fact I am throwing ONE DICE seeing if I won then throwing the second one as a totally different ROLL.
Anonymous
R1 * R2
1 - (prob fail both rolls).
Where as my formula clearly takes into the account the
MAX ROLL that can occur stops the influence of 2nd (R1 or R2),
this may not be correct, but I know multiplying without taking
into account seperate rolls is not right either.
% Parry * % Shield = A
(Max % from above) + A = likely hood a block or parry is made.
Gilez
1 - (prob fail both rolls).
Where as my formula clearly takes into the account the
MAX ROLL that can occur stops the influence of 2nd (R1 or R2),
this may not be correct, but I know multiplying without taking
into account seperate rolls is not right either.
% Parry * % Shield = A
(Max % from above) + A = likely hood a block or parry is made.
Gilez
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