XHKO-ing stuff

[Crystal_]

I thought this was interesting and I was willing to do it, which means that I did it. Here I explain how to calculate the chances of getting X(2,3,4...)HKOs.

First off, a 2HKO:

Imagine you have two dices with the numbers 1 to 6, and start thinking what could happen if you want the result of summing the two dices ; Possibilities of:
Getting a 2 = one (1,1)
Getting a 3 = two (1,2 ; 2,1)
Getting a 4 = three (1,3 ; 3,1 ; 2,2)
Getting a 5 = four (1,4 ; 4,1 ; 2,3 ; 3,2)
Getting a 6 = five (1,5 ; 5,1 ; 2,4 ; 4,2 ; 3,3)
Getting a 7 = six (1,6 ; 6,1 ; 2,5 ; 5,2 ; 3,4 ; 4,3)
Getting a 8 = five (2,6 ; 6,2 ; 3,5 ; 5,3 ; 4,4)
Getting a 9 = four (3,6 ; 6,3 ; 4,5 ; 5,4)
Getting a 10 = three (6,4 ; 4,6 ; 5,5)
Getting a 11 = two (6,5 ; 5,6)
Getting a 12 = one (6,6)

Now, think you have, instead of the dices, two random damages that can do between 240 and 265 to a Pokemon with 500 HP, or -if you prefer- that you have two big dices with the numbers 240 to 265.

First off, we have to determine the max and the min result:
Min = 240 + 240 = 480
Max = 265 + 265 = 530

Now, he have to see which results are capable of 2HKOing and which are not:
480 to 499 (20 possibilities) => NO
500 to 530 (31 possibilities) => YES
Total = 51 => 1, 2, 3,..., 24, 25, 26, 25, 24..., 1.

And now, lets apply the "dices rule".
Possibilities of getting a:
480 = one
481 = two
...
498 = nineteen
499 = twenty
The result is: 103460


Possibilities of getting a:
500 = 21
501 = 22
...
504 = 25
505 = 26
506 = 25
...
529 = 2
530 = 1
The result is: 237920

Now, turning the results in percentages:
Getting from 480 to 499 (Not a 2HKO) = 30'3%
Getting from 500 to 530 (2HKO) = 69'7%


Critical Hit factor:
Logically, if something is able score a 2HKO, if one of the moves does a CH you will always score the 2HKO (barring a miss).

For example our Pokemon has a 20% CH ratio, but it has two times to score a CH:

.........................................20% CH
100% CH / non CH -------------------------- = 20% CH
.................................100% CH / non CH


If 1st hit does NOT CH (80%):

.......................................20% CH
80% CH / non CH -------------------------- = 16% CH
...............................100% CH / non CH

Total = 20% + 16% = 36% chances of CH-ing.

So, the final division between 2 and 3HKOs must be a 64% of the total (100-36) being that 36%:

Chances of 2HKO-ing: 36 + (0'64*30'3%) = 55'39%


Accuracy factor:
The chances of missing should also be taken into account. A 255'd will not change much the result though. Remember that the 255 also affects to the CH formula, but in my example, the pokemon is supposed to have a 20% chances af CH-ing after doing the *255/256.

255/256*2(moves) = 0'9922

Therefore, the formula starts with a 99'22% of the total.
55'39%*0'9922= 54'96%
This is the final result that shows the possibilities of getting a 2HKO (or OHKO which will be: 0'2% multiplied * the chances of OHKO with the CH * 255/256).

3,4,5... HKOs are calculated in a different way but not much different.
Lets see, if you use three dices:
Chances of getting 3 = 1 (1,1,1)
Chances of getting 4 = 3 (1,1,2 ; 1,2,1 ; 2,1,1)
Chances of getting 5 = 6 (1,1,3 ; 1,3,1 ; 3,1,1 ; 2,2,1 ; 1,2,2 ; 2,1,2)

We can deduct that the sequence will be:
1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 36 , 28 , 21 , 15 , 10 , 6 , 3 , 1
So, you just have to change the *2s in the 2HKO formula to *3s, the *3s to *6s, the *4s to *10s... to get the 3HKO formula.

If you use four dices to get the 4HKO formula you have:
1 , 4 , 10 , 19 , 32 , ... , 4 , 1

Etc.

[t3h Icy]

Your math is slightly wrong, but you have the concepts right. All attacks have 39 possibilities and depending on the damage range, can have the same number multiple times, and if it's large enough (often), the damage can skip over many numbers (for example, Tauros's Hyper Beam will never do exactly 523 damage to Chansey).

Look into this if you're interested in learning more.

[Crystal_]

So, do you mean that if a Pokemon can do between 500 and 550 (I know it is not exacly done with the random factor of the damage formula), there will be 12 numbers that can not do? Like 3-7-11-...-43-47, following a logical neutral serie? Are you sure?

+---------+
|Example 1|
+---------+

A Gengar uses Thunderbolt against Gyarados. For this example, we'll assume that
they both have perfect Stats, so Gengar's Special will be 358 and Gyarados's
will be 298. They are also both Level 100. Thunderbolt's Base power is 95.

Damage (Attacker is L100) = ((0.84 * A * B / D) + 2) * M * R / 255
Damage (Attacker is L100) = ((0.84 * 358 * 95 / 298) + 2) * M * R / 255
Damage (Attacker is L100) = ~97 * M * R / 255 (don't round down to 97 until the
formula is completed)

Thunderbolt is Super Effective against Water and Flying, Thunderbolt's type does
not match Gengar's and this attack is not a Critical Hit. So M is 4.

Damage (Attacker is L100) = ~97 * 4 * R / 255
Damage (Attacker is L100) = ~391 * R / 255
Damage (Attacker is L100) = ~1.5 * R

Since R is random from 217 and 255, we'll check both to find the minimum and
maximum that Thunderbolt can do.

Minimum = ~1.5 * 217
Minimum = 333

Maximum = ~1.5 * 255
Maximum = 391

So, for example Gengar can do 334 (1'5*218 = 334'53) but can not do 335 because 1'5*219 = 336'07. Instead, it will do 336 with this random. [Im doing that as if the ~1'5 IS EXACTLY a 1'5].
Do you mean that? It is logical, isnt it [39=255-217]? I didnt know that though.

Anyway, if the 39 numbers used have that proporcion (like:

Used, used, not used, used, used, not used...
or
3 times, 3 times, 2 times, 3 times, 2 times, 3 times, 3 times, 2 times, 3 times, 2 times...
)
the formula will not change much, just a bit less chances of getting the best XHKO (since the decimals are rounded down i suppose [like the Gengar's 334'53 that becomes a 334]). But just something like ~0'05 %.

Pretty much like if we suprime the 2s and the 5s of the dices. The proporcion will be nearly the same (in this dice case) or nearly the same in my damage example.

[t3h Icy]

Pretty much. Just make sure if you're planning to do accurate counts to use all 39 possibilities and that you know the actual damage for each of the 39 values.

What you could do is use Excel (or another spreadsheet program), and create a 40x40 table, having the top-left corner be whatever (a title I guess), and then the 39 rows and columns each be the 39 values of the damages, then add them up in the cells.

Basically, the simple formula is 39^i with i being the number of attacks used. The result is the number of possible outcomes (and if you put 0, the outcome is 1; the HP is always the same since no attacks were used).

If you have some programming abilities, you can also make a quick program to do this for you and it'll output the percentages really quickly. The reason why I don't ususally make any of these things public myself is that RBY is dynamic and only a few match-ups involve no switching and only attacking moves (Alakazam vs Chansey involves healing, Articuno vs Jolteon may involve switching, etc).

[Crystal_]

Nah. Too much.