quote Pants1000
quote yarbinger
quote Pants1000
quote yarbinger
quote pancox2
Your AP Cap at number 1. The order for the others doesn't matter. You might want to put the VS Cap 1st to give it 75% chance of proccing but I personally wouldn't.
I've been looking to figure out the middle card bonus. What I'm currently hypothesizing is that the bonus to the center card is (1 - base chance) / 2. So for Average (50%), the middle card receives a 25% bonus (75% total). For Rel. High (67%), the middle card receives 16.67% bonus (83.33% total). For High (75%), the middle card receives 12.5% bonus (87.5%).
Is this how you came up with the 75% chance for VS to proc in position 1? And has anyone else confirmed this?
I had the same thought. My thinking was that the game checks the first card twice, and the math works out to the same numbers you have.
Sorry, I'm not sure what you mean by "checks the first card twice".
So like, the game checks it first at 50% and then checks a second time for 50% of that, bringing us to 75%?
And for High, it checks first at 75% and then again at 75% of that, giving us .75 + .75^2 = 1.3125?
Would you mind calculating high for me real quick? That should help me understand, haha.
I don't know this is how it works, but if it checked the first card twice, the math would be like this.
This also assumes the 50/67/75 numbers, and I don't know if they're actually right.
For average if it checked twice, there would be a 50/50 chance for success, then another 50/50 chance. .5+.5*.5=.75
For high one the first attempt there's a 75% chance of success, then if it failed (25% chance) it would have another 75% chance. .75+.25*.75=.9375
Again this is purely conjecture, but it would be easy to program, and it seems to fit my experience.
Okay, that makes a lot of sense. Thanks a lot for the response! The first time I messed with something like that outside of school was when I was trying to calculate the probability of losing in Roulette 7 times in a row, lol. It's the same as 1 - (1 - x)^n. In this case, 1 - (1 - .75)^2 = 1 - .25^2 = .9375. That does seem pretty accurate. Also, it would seem, if checking twice is the case, that the checks are consecutive and (I don't believe I've ever had a center card proc after any other proc) are the first procs to be checked.
Thanks again!
|Also Quoting yarbinger, from another post, but same page I believe, this thread|
Yusuke's numbers, after factoring in how it relates to a 3 proc, do not support this:
High = 90% (+1% or 91%)
Rel High = 79% (+2% or 81%)
Avg = 50% (+5% or 55%)
My calculations provide the following using my hypothesis as stated above:
High = 89.65% (92.29% after middle card bonus)
Rel High = 79.01% (83.95% after middle card bonus)
Avg = 50% (59.38% after middle card bonus)
It appears Yusuke did something similar, only using the 3 proc chance as his base instead of middle card proc chance and divided his result by 5 due to each card. For instance, for High: (1 - 0.9) / 2 / 5 = 0.01 aka 1%. Rel High: (1 - .8) / 2 / 5 = 0.02 aka 2%. Avg: (1 - 0.5) / 2 / 5 = 0.05 aka 5%.
If anyone can correct me, especially Yusuke, please do so. I want to understand this completely.
If anyone is wondering how I calculated this, I wrote the following program using MATLAB (3/4 for High, 2/3 for Rel High, and 1/2 for Avg), but it should be easily modified to fit whatever language you want to use. The program works for different card proc chances, so feel free to plug in any % for P1-P5. The second P1 redefines P1 as the modified chance due to being in position 1, according to my earlier hypothesis.
P1 = 3/4;
P2 = 3/4;
P3 = 3/4;
P4 = 3/4;
P5 = 3/4;
P1 = (1 - P1) / 2 + P1;
nC0 = (1-P1)*(1-P2)*(1-P3)*(1-P4)*(1-P5)
nC1 = (P1)*(1-P2)*(1-P3)*(1-P4)*(1-P5) + ....
(1-P1)*(P2)*(1-P3)*(1-P4)*(1-P5) + ....
(1-P1)*(1-P2)*(P3)*(1-P4)*(1-P5) + ....
(1-P1)*(1-P2)*(1-P3)*(P4)*(1-P5) + ....
(1-P1)*(1-P2)*(1-P3)*(1-P4)*(P5)
nC2 = (P1)*(P2)*(1-P3)*(1-P4)*(1-P5) + ....
(P1)*(1-P2)*(P3)*(1-P4)*(1-P5) + ....
(P1)*(1-P2)*(1-P3)*(P4)*(1-P5) + ....
(P1)*(1-P2)*(1-P3)*(1-P4)*(P5) + ....
(1-P1)*(P2)*(P3)*(1-P4)*(1-P5) + ....
(1-P1)*(P2)*(1-P3)*(P4)*(1-P5) + ....
(1-P1)*(P2)*(1-P3)*(1-P4)*(P5) + ....
(1-P1)*(1-P2)*(P3)*(P4)*(1-P5) + ....
(1-P1)*(1-P2)*(P3)*(1-P4)*(P5) + ....
(1-P1)*(1-P2)*(1-P3)*(P4)*(P5)
Final_Probability = (nC0 + nC1 + nC2)
quote SantaLx
quote yarbinger
If you would like, I can demonstrate it.
Yes, please
In your first message you just give proc rates not the formula.
The probability of getting 3 procs with 5 cards that can proc can be determined by calculating and then adding the probabilities that a deck will get 0 procs, 1 proc, and 2 procs, and then subtracting that from 1. I will abbreviate "n choose k" as nCk. n represents the number of attempts (in this case we have 5 cards). k represents the number of successful attempts. For this example, I will use 5 cards with Avg proc rates (50%).
nCk = n! / (k! * (n - k)!) >> This represents the number of ways your outcome can occur
nCk * (chance of occuring)^number of times it occurs * (chance of not occuring)^number of times it does not occur
5C0 = 5! / (0! * 5!) = 5! / 5! = 1 >> There is only one combination for receiving 0 procs
1 * (1/2)^0 * (1/2)^5 = 1/32
5C1 = 5! / (1! * 4!) = 5! / 4! = 5 >> There are 5 combinations for receiving 1 proc
5 * (1/2)^1 * (1/2)^4 = 5 * 1/32 = 5/32
5C2 = 5! / (2! * 3!) = (5 * 4) / 2 = 10 >> There are 10 combinations for receiving 2 procs
10 * (1/2)^2 * (1/2)^3 = 10 * 1/32 = 10/32
This is all we need to know, since a 3 proc includes the probability of a 4 proc and a 5 proc (as 3 is max). From here, we add those chances together:
(1 + 5 + 10) / 32 = 16 / 32 = 1 / 2
Subtract that from 1:
1 - (1/2) = 1/2
Convert to a %:
1/2 = 0.5 >> 50%
And that's our probability of getting a 3 proc using 5 Avg cards. This method works for any % as long as each card has the same %. If you want to calculate the chances using cards with a different %, then eliminate nCk from the formula, and instead, calculate each combination individually. The script I use for this is the following:
P1 = 3/4;
P2 = 3/4;
P3 = 3/4;
P4 = 3/4;
P5 = 3/4;
P1 = (1 - P1) / 2 + P1;
nC0 = (1-P1)*(1-P2)*(1-P3)*(1-P4)*(1-P5)
nC1 = (P1)*(1-P2)*(1-P3)*(1-P4)*(1-P5) + ....
(1-P1)*(P2)*(1-P3)*(1-P4)*(1-P5) + ....
(1-P1)*(1-P2)*(P3)*(1-P4)*(1-P5) + ....
(1-P1)*(1-P2)*(1-P3)*(P4)*(1-P5) + ....
(1-P1)*(1-P2)*(1-P3)*(1-P4)*(P5)
nC2 = (P1)*(P2)*(1-P3)*(1-P4)*(1-P5) + ....
(P1)*(1-P2)*(P3)*(1-P4)*(1-P5) + ....
(P1)*(1-P2)*(1-P3)*(P4)*(1-P5) + ....
(P1)*(1-P2)*(1-P3)*(1-P4)*(P5) + ....
(1-P1)*(P2)*(P3)*(1-P4)*(1-P5) + ....
(1-P1)*(P2)*(1-P3)*(P4)*(1-P5) + ....
(1-P1)*(P2)*(1-P3)*(1-P4)*(P5) + ....
(1-P1)*(1-P2)*(P3)*(P4)*(1-P5) + ....
(1-P1)*(1-P2)*(P3)*(1-P4)*(P5) + ....
(1-P1)*(1-P2)*(1-P3)*(P4)*(P5)
Final_Probability = 1 - (nC0 + nC1 + nC2)
Hopefully, this clarifies things for you. If you have any more questions, feel free to ask.