Math and EXP Gach

I like knowing random numbers no one else gives a pomegranate about, so this has been bugging me for a while.

EXP Gach tickets give set EXP amounts:

Code:

100k 50k 30k 20k 10k 5k 3k 2k 1k 500 300 200 100

Each of which is equally likely(I assume). That means if you could graph the possible outcomes of every possible reward for, say, 30 tickets, there would be a few peaks: Highest and lowest Net Exp gain (both the least probable. 1/13^30 I think) and the most likely outcome(s).
Now, having not finished Highschool math, I have no idea how I can calculate all the outcomes and their probabilities without 13^30 different calculations. I could put the net Exp Gain on a Standard Normal Curve and run with numbers there, but how would I get the Standard Deviation for that curve? (I assume the mean with just be the Expected Value) Take the standard Deviation of the 13 different Exp Values? o_0

I'm confugled. Could someone help?

Variance is defined as the average of the squared distance of each term from the mean (for distribution X, and mean mu, sum((x-mu)^2)/size(X) for all x in X)

If you add 2 together, you get that variance(X+Y) = variance(X)+variance(Y)+covariance(X,Y), the standard deviation is the square root of this. Since they're independent (hopefully) you get covariance of zero, meaning the standard deviation over 30 tickets would be sqrt(30*variance(1 ticket)), or sqrt(30)*stdev(1 ticket)


I don't believe that assuming they're equally likely is correct though, it's not true for the easter exp rewards, or the red envelopes at chinese new year.

However if they are, mean is 17084 exp. stdev is 28000 exp. In 30 tickets, that's a mean of 512000 exp, stdev of 153000 exp.

Variance is defined as the average of the squared distance of each term from the mean (for distribution X, and mean mu, sum((x-mu)^2)/size(X) for all x in X)

If you add 2 together, you get that variance(X+Y) = variance(X)+variance(Y)+covariance(X,Y), the standard deviation is the square root of this. Since they're independent (hopefully) you get covariance of zero, meaning the standard deviation over 30 tickets would be sqrt(30*variance(1 ticket)), or sqrt(30)*stdev(1 ticket)


I don't believe that assuming they're equally likely is correct though, it's not true for the easter exp rewards, or the red envelopes at chinese new year.

However if they are, mean is 17084 exp. stdev is 28000 exp. In 30 tickets, that's a mean of 512000 exp, stdev of 153000 exp.

Oh, I get it... kinda.. ish? Haven't started on variance(x+y) and covariance yet, so you lost me there. xD I'll talk about this with mah teacher so I can get a better idea next week. Thanks.

Think of the possible exps as shapes in a bag.

13 different shapes, 30 shapes per bag(30 ticket exp gach run)

this would give us:

(13+30-1)!/[30!(13-1)!] = 11058116888 possible bags each with different shapes.

But coming back to the question, the exps are numbers. Imagine the exp as mass, and you are weighing each bag. You should actually have less number of bags with different mass because of some of the masses being multiples of each other. For example(in hundreds)

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5

is the same as

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2

Maybe this will help you out a bit lol...

Think of the possible exps as shapes in a bag.

13 different shapes, 30 shapes per bag(30 ticket exp gach run)

this would give us:

(13+30-1)!/[30!(13-1)!] = 11058116888 possible bags each with different shapes.

But coming back to the question, the exps are numbers. Imagine the exp as mass, and you are weighing each bag. You should actually have less number of bags with different mass because of some of the masses being multiples of each other. For example(in hundreds)

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5

is the same as

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2

Maybe this will help you out a bit lol...

I get what you're saying, but that doesn't help with finding the probabilities. D:

Oh, I get it... kinda.. ish? Haven't started on variance(x+y) and covariance yet, so you lost me there. xD I'll talk about this with mah teacher so I can get a better idea next week. Thanks.

If you don't learn it in high school, you'll learn it in an intro stats course in university.

If you don't learn it in high school, you'll learn it in an intro stats course in university.

I'm still in high school, halfway through a Stats Course. :f6:
Edit: Lol, I read fast, thought you said "if you didn't learn it", ignore dis post.

Unfortunately, all of the EXP points aren't given at an equal rate, so this kind of destroys the entire point of your calculations.

Unfortunately, all of the EXP points aren't given at an equal rate, so this kind of destroys the entire point of your calculations.

Well, if we can find the actual rates, it's not hard to incorporate them.
I don't suppose you have those for us, do you?

Well if you're going to go with your original idea, the only way I think to get a real answer is to do all the calculations. If you have any programming experience/background use it.

Hm.

Incidentally I'm attempting to write a script to calculate the number of hits needed to kill a given monster with a given damage range, which is kind of the same thing except with a range of integers instead of specific discrete values.

I'm sort of cheating, though, so my progress wouldn't be particularly useful to you.

Finding number of hits to kill a monster is a pain and I never did it in any way other than by experimental results >_> (write a script that imitates attacks, run it for a while, see how many hits it takes to kill)

Yes, that it is.

I'm trying to define the damage distribution as a piecewise function, which works, but I'm only on 4HKO and I haven't found any really useful patterns yet.

Yes, that it is.

I'm trying to define the damage distribution as a piecewise function, which works, but I'm only on 4HKO and I haven't found any really useful patterns yet.

I tried doing that a long time ago but I didn't really get anywhere =(

There's also the brute-force method, but I'd frown upon that.

I tried doing that a long time ago but I didn't really get anywhere =(

There's also the brute-force method, but I'd frown upon that.

What brute-force? As in, making a huge array and counting up all the possibilities?...

Well if you're going to go with your original idea, the only way I think to get a real answer is to do all the calculations. If you have any programming experience/background use it.

HTML doesn't count... does it? Besides, it's more fun to cheat with Stereo's method.

What brute-force? As in, making a huge array and counting up all the possibilities?...

Yes lmao xD Simple design, horribly inefficient. Who knows, throw in an OutOfMemoryException in there...

13 loopz ftw? I say I'll run for like 5 mins.