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File: afd9861a092b660⋯.jpg (139.47 KB, 1240x775, 8:5, Maths.jpg)

e65613 No.13799272

so /v/, lets say that there is… a video game

in that video game, there is a certain weapon:

Fire Razor

Greataxe

69-69 Physical Damage+1-10 Fire Damage

3.6 Speed

Special: on strike, this weapon attempts to hit twice; this weapon's bigger attack from the 2 hits always connects, disregarding critical hits, and the smaller attack from the 2 hits always misses, also disregarding critical hits

Special: for both swings, if the fire damage dealt is below 5 after either or both calculations, then both instances of fire damage are boosted to 5, though one of them still misses anyway

so this weapon takes 69 physical+1-10 fire, rolls it twice, then gives you the higher number, but THEN if either attack's fire damage is below 5, then it or they is or are increased to 5 fire damage

so I ask you /v/: how do you calculate the average fire damage dealt by Fire Razor against an enemy who has 0% fire resistance and cannot be critically stricken, and what IS the average fire damage dealt against that enemy?

252c1c No.13799319

>>13799272

>how do you calculate the average fire damage dealt by Fire Razor

You don't

You get a weapon which isn't a piece of shit.


7e5381 No.13799344

Boy, these dubs threads are getting really elaborate


8c9948 No.13799359

>>13799344

Checked!


69d907 No.13799477

File: 5655ed4b0404e24⋯.jpg (90.97 KB, 650x422, 325:211, 1819bdaf399f1f889a3888ab53….jpg)

Working our way up:

5 is a result of:

Rolling 4 or less twice in a row (40% of 40%) = 16%

Rolling 5 the first time, then rolling 5 or less (50% of 10%) = 5%

Rolling 4 or less then rolling a 5 (10% of 40%) = 4%

Total = 25%

6 is a result of:

Rolling 5 or less, then rolling a 6 (10% of 50%) = 5%

Rolling a 6 the first time, then rolling 6 or less (60% of 10%) Total = 6%

Total = 11%

7 is a result of:

Rolling 6 or less, then rolling a 7 (10% of 60%) = 6%

Rolling a 7 the first time, then rolling 7 or less (70% of 10) = (7%)

Total = 13%

The pattern of each number above 5 being 2% more likely than the previous continues up to 10 so in the end:

5 = 25%

6 = 11%

7 = 13%

8 = 15%

9 = 17%

10 = 19%

69 will be a part of any average already, thus we can take the average of the result of the additional roll and add 69 to get the average damage.

((5*25)+(6*11)+(7*13)+(8*15)+(9*17)+(10*19))/100 = 7.45

Thus the average damage is 76.45.


69d907 No.13799571

Give me more problems OP you faggot.


89b0b4 No.13799618

>>13799477

Impressive. I am a retard. Where are you getting those percentages from? Please elaborate.


cf9838 No.13799626

>>13799477

>what's the average fire damage

>fire damage is 1-10 per attack

>the fire damage is 76.45

that's a lot of effort to get the completely wrong answer by not paying attention to the question. but you got nice dubs, faggot


5ece15 No.13799630

File: da7b9df1c29f318⋯.jpg (52.9 KB, 534x800, 267:400, man-in-jacket-talking-on-p….jpg)

>>13799477

How fascinating. I follow the logic easily since you formatted your reply so clearly but how do you arrive at the first set of percentages in each case? (40% of 40%), (50% of 10%), etc. And is their a specific name so I can easily google further info? Please respond, I want to get smart and drive women wild with my maths.


4f580b No.13799633

File: de38796116ba69b⋯.jpg (27.53 KB, 500x500, 1:1, de38796116ba69ba4f7ae4a8ef….jpg)


e90365 No.13799634

File: 64584b6a204ba13⋯.jpg (159.88 KB, 1280x720, 16:9, Gonsteam.jpg)

Uhhhh…


69d907 No.13799645

>>13799618

>>13799630

I explained it in the post, for example the chance of rolling 4 or less (assuming a fair dice) twice in a row is 16%, because 40% of 40% is 16%. It's just basic probability.


0d8d77 No.13799649

>>13799272

>how do you calculate the average fire damage dealt by Fire Razor against an enemy who has 0% fire resistance and cannot be critically stricken, and what IS the average fire damage dealt against that enemy?

You can do it by using a fucking table. Can't be arsed to MSPaint this, but it should be easy enough to follow. (5*25+6*11+7*13+8*15+9*17+10*19)/100 = 7.45


4f580b No.13799656

>>13799626

Try rereading his post you retard.

>((5*25)+(6*11)+(7*13)+(8*15)+(9*17)+(10*19))/100 = 7.45


1bbaa2 No.13799684

I see someone is playing Grim Dawn


0b9c42 No.13799715

>>13799633

Impressive.


0d8d77 No.13799717

File: 857a3a14d5cede7⋯.png (Spoiler Image, 18.74 KB, 128x128, 1:1, 1440106916340.png)

Alright, here's a problem for you. Every now and then a shit thread pops up that just needs dubsposting. Each Bateman post has a 0.11% (1/9) probability to get dubs. Let's assume that each and every post in the thread is not a rampant shitposter and will only repost dubs with a 33% (1/3) probability a second time, and with 10% (1/10) probability a third time. After a third time, he stops posting. When a thread has three doubles posts, a poster may check one of those, adding to his post count independently of earlier dubsposting, so an anon gets a free post per three dubs posted.

Assume that a thread is considered properly shitted up when it has 11 (eleven) dubs in it and that time doesn't matter for this problem - every anon lurks the thread after exhausting his postcount. How much anons do you need to dubspost a thread with 80% accuracy at the minimum? What's the probability for 5 anons to successfully shit up the thread?


0d8d77 No.13799722

>>13799717

>each and every post

poster


4f580b No.13799802

>>13799717

>0.11%

That's one tenth and one hundreth of 1%. Did you even graduate highs chool? Middle school? Learn to fucking type probabilities. You mean 11%

>Each Bateman post has a 0.11% 11% (1/9) probability to get dubs.

00, 11, 22, 33, 44, 55, 66, 77, 88, 99. 10 possible dubs from the numbers 00-99. that's 10/100, a 1/10 (10%) chance. How the fuck did you get 0.11% 11%?




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