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Found 8 results

  1. A contributor of reddit has done a 99% UCL calc on a personal data set and came up with the following: https://www.reddit.com/r/WorldofTanks/comments/6dg5tf/the_918_matchmaker_saga_part_two_new_data_and/ I have no idea of whether this could be discerned from the API, but this kind of ninja nerf would be such a WG kind of thing to do. I wonder if the OP is a member here, it'd be interesting to have him present to our math guys.
  2. So, while trying to redo some 5 crew skills, I became interested in how accurate the info an popular understanding of firefighting was. It's a small sample, but I'll add a few different ones later to validate results Methodology: Target: 215b - it's got a very exposed and soft fuel tank for a lot of HP. Shooter mt-25 - good pen vs low alpha Went through replays with a hitlog and recorded each fire damage, hp at time, tick for module damage, number of ticks. Threw out any where fire killed tank. Groupings: Test 1 & 2 Skills: FF 100% + BIA Test 3 & 4: BIA Results: - Fire damage is based on Hull HP - FF has no effect on reducing module damage, - The first tick of 1st fire is ~3.9% of max HP (+/- 0.3%) regardless of FF - The first tick of 2nd fire is~3.5% of max hp(+/- 0.3%) Regardless of FF - FF increases the inter-tick reduction of fire damage by ~50% (wow, close to what was written) (In test it reduced each tick by 10 +/- 1, non FF reduced each tick by 5 +/- 1) - fires burn until tick damage = ~0 - FF reduces ticks by great e - The first fire seems to cost ~22% +/- 3 maxHP with FF The first fire seems to cost ~40% +/- 5 maxHP without FF - Second fire seems to cost ~15% with FF - Not enough data for cost without FF Conclusions: Basic fire equation FireDam = (MaxHP*~0.039(0.95n-(.05*X))) X = FF skill If people wish to test this data, feel free to post here and I'll try and update as it goes. Edit: Special thanks to Etrenety, a random NA player who was kind enough to run through a number of test runs.
  3. Hey guys, so I need help. I am a student at the University of Southern California and the University of California at Los Angeles graduating this year with both a degree in Mathematics and Physics. I have already submitted my Thesis which I wrote on the mathematical exploration and principles of the Yang-Mills existence and mass gap dilemma. However, I would like other input from my fellow mathematicians. I should mention that I am trying to solve and provide a proof for it for my dissertation. Also, there is a million dollar prize for solving the problem which I will share with the people that help me evenly excluding myself as I will decline my share. Thanks again, Kosizch. (Currently in class, so I will be avaliable when I get home in about 4 hours). ANY AND ALL HELP IS APPRECIATED.
  4. In this topic I'll try to give an idea, how WG converts your damage into MoE rating, measured in percents. After each battle your "moving average damage" is recalculated as Dnew = (1 - 2/101) * Dold + 2/101 * (damage_battle + max(assist_track_battle , assist_radio_battle) ) When you buy a new tank, your initial moving damage is set to zero. After hundreds of battles it approaches your plain average damage+assist (if you play constantly good). Your current value of moving damage can be found in battle_results cache files, if you use Phalynx converter. Writing down moving damage and MoE rating after each battle, it is possible to plot a corresponding curve (red color): For every tank that I tested, I got a piecewise linear function. For percents 0,20,40,65,75,85,95,100 they define a corresponding damage+assist value (which is always divisible by 50, and every day in can slightly change), and use a linear interpolation for intermediate values. The question, that you might be interested in, is "Where the hell these 5% of players, that do 4k damage+assist on average?" Of course, there are such players, but their number is less than ~0.1%. How could WG obtain such a number, 5% for 4k damage? We can find a hint, if we have a look on a distribution of damage in single battles. I analysed about 60k replays from wotreplays site, and plotted a distribution of damage+assist for players, excluding replay uploaders. The result is presented by a black curve on the plot above. As you can see, the curves look quite similar, though there are significant difference, so we can't conclude, that WG treats just single battles to make their tables. My guess is that they treat player's day statistics, no matter how many battles was played. And then they compare it with your moving damage. That's why you don't see these legions of unicums: WG compares average damage for 100 battles with average damage for one day.
  5. Been seeing a rash of the usual arguments on the official forums: game is rigged, etc. The usual crap. What mystifies me, however, is the frequency with which people will make flatly false statements, like "I had nine out of eleven matches last night with a 40% or below chance to win. That can't happen randomly." I mean, yes it can. Is the concept of sample size really that counter-intuitive? Is it really counter-intuitive that people need to collect data before making such bold statements about a game being rigged? Is it really that difficult to understand that chance plays a larger role in a small sample size, but evens out as the sample size grows? Is it truly counterintuitive that streaks happen in series of random events? Do they really expect a small sample to quickly even out? Do they expect that if they flip a coin ten times, they'll get five heads and five tails...and if they do another ten flops, they'll get another five heads and five tails? I find it impossible to believe that this stuff is really so counter-intuitive. Deltavolt told one guy he needed to take a statistics course to see why what he was saying was wrong, and all I could think was [i[why can't he see this without a statistics course? It's that intuitive he shouldn't need it. Bizarre. Almost every single person that goes to college in America takes calculus, for some weird reason, and yet simple statistical concepts are beyond most people. Come on, math people. Does this seem normal to you? Is it all just people believing what they want to be true, or are basic statistics really this opaque to most people?
  6. I posted the below in the WoT Forums and I wanted to offer it up to the experts here to make sure my understanding and math were correct. It is a basic watered down version of the complexities of the aiming mechanics with some examples to help illustrate the wording. Thank you. The aiming algorithm is different than the pen/damage algorithm as it utilizes the gun value known as distribution...or effective accuracy. It is a value given of where your shell will land at 100m from the target. The higher the value, the wider the shot. The further the object, the chance for a miss increases on a linear scale. If I understand on a basic level this means that a gun with a .40 dispersion will land, around 95% of the time, within a .40m circle at a distance of 100 meters. Think about that. Two men of average height standing at opposite ends of a football field looks roughly to be about the size of half a dime to each other. This gives us a rough ratio of 0.375" : 72" or a scale factor of .00521. https://www.youtube.com/embed/HpiZEo_QYlc Now, move your target to 200 meters and if I understand this correctly, roughly 95% of your shells will land in a .80 meter diameter circle. At 300 meters, 95% of your shells should land in a 1.20 meter diameter circle. Remember 1 meter = 3.28084 feet. If I did my math correctly: The size of a TOG II facing you at 0 meters is 10.13m wide X 3.05m tall. At 100 meters it would appear to be approximately 2" wide x 5/8" tall. This gives you a rectangular area of 1.2504". So, a .40m distribution is a circle 1.31234 feet (or 15.74808 inches) in diameter. TOG II Maths: Width = 10.13m = 33.2' = 398.4" * .00521 = 2.076" Height = 3.05m = 10' = 120" * .00521 = 0.6252" I am surprised we hit anything at all! Thank goodness that sniper mode gives us a 8x or 16x zoom for magnification (couldn't find stated value).
  7. I freely admit that I am a math bumbass, and a bad player. Nor am I a numbers driven personality. I've got 6689 battles and a ~43% WR. I don't know how to calculate the following questions: how many games will it take to move a WR to 45 to 75% in 5% incriments. I guess anouther way to say this is, what are the numbers needed to move through the rateing colors starting at red make any assumptions neeed as to play improvement in order to get the numbers to work I've been told not to stress out over stats and just work on improving my game. I realize that I may not be able to attain superstar status, not everyone is capable of that. However, if the numbers I need to reach each new color on the scale are unrealistic I need to know that too. Perhaps then I won't feel like I've been beating my head against a brick wall. I WILL NOT crerate a new account and start over, so that is not what this is about. thanks from an ardent math phobic.
  8. Let's do an experiment. Let's say you flip a coin with a 50% chance of landing heads. You flip the coin X times. What is the longest streak of continuous heads (or tails) you can possibly hope to achieve? My answer is below: The longest streak of heads (or tails), out of a series of X throws, you can expect is approximately that of log2(X) More on Streaks Here's why. The chance of a streak of N consecutive heads is (0.5)N, this is because the odds of flipping four heads in a row, for example, would be (0.5 x 0.5 x 0.5 x 0.5), or 0.0625. Now look at a streak of N consecutive heads, it is bound to start SOMEWHERE. So, in a series of X flips, we have a certain number of starting points, if there are G starting points for a series, the average number of times a streak of N will occur is therefore G (0.5)N. For example, if there were 100 start points, one would expect about three sequences of length 5 or more, because 100 x 0.03125 is 3.125. Hence, we would expect six streaks of five, three streaks of five heads, and three more of five tails, on average. What if we had a streak of 10 in a row out of 100 starting points (i.e. 110 flips)? The average number of streaks of ten heads in a row out of 100 is that of 100 x (0.5)10, which is about 0.1 (0.977, to be precise), doubling that, the odds of having a streak of ten heads or tails, in a row would be double that, 0.2 Let's see where this gets us. If three players flipped a coin 109 times (109 coins means 100 start points), the odds of at least ONE of them getting ten heads or tails in a row is about 50%. So, how do we find what the longest expected streak is? In general, when X(0.5)N is greater than 1, we can expect at least one streak of the sort to occur, when X(0.5)N is less than 1, we can expect the streak NOT to occur. The point between these two can be approximated to log2X, to give us the longest expected streak. Note that X is the number of STARTING points, not the number of FLIPS. Let's use an example, if I flipped a coin 32 times, I would have 28 possible starting points for a streak of 5, 27 start points for a streak of 6, and 26 for a streak of 7. On average, I can expect 0.42 streaks of 7 to occur, meaning that it is unlikely, but I could expect about 2 streaks of 5 to occur. So what do we have here? Let's assume a player determines his games purely by luck. He plays 1000 games. His longest expected streak would be about 10 games, and he can expect to have about two such streaks. The Best Part: This is if the player has a 50% chance of winning, good players drag their odds of losing streaks down, bad players drag them up For the purposes of this experiment, we shall ignore draws Let's say we have five players playing 1009 games (for 1000 start points), let's calculate the number of times they can expect to have a ten game losing streak (we can calculate the losing streak by NOT doubling the number) A 40%er would expect to have SIX such losing streaks A 45%er would expect to have just 2.5 such losing streaks A 50%er, as said earlier, could expect to have just ONE such streak A 55%er is unlikely to have such a streak, he could expect just 0.34 of a streak, meaning he could expect to have such a streak after about TWO such cycles of 1000 games, and even then, the odds of a streak of 10 losses after that many games would just be seven in ten. A 60%er could expect just 0.1 of such a streak, he would expect to have just a 50% chance of having such a streak after SIX cycles. All this, of course, is theoretical, and doesn't account for factors like playing form, but it's interesting to know. A bad player on bad form could easily be expected to have losing streaks of more than that, and a good player is certainly not exempt from the 60% chance either.
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