As you have to have a new account after just one defeat, you will have to take that account through 1000 battles which are all wins. If we assume 0.5 as win rate for inactivity, we get to 0.5^1000=1E-301. Which would mean you would have to seed way more games than there are atoms in the known universe.
You can’t do this by just seeding accounts. If it is not a system glitch but legitimate, I can only think of stuff like squading. Or an exploit of the newbie pool. Which would be impossible today. Or knowing when you are bound to get a very good game next. Which would require 50 premiums back when they gave 20 top tier placements in ground arcade for a new GE premium. Stuff like that.
… But you are calculating the wrong probaility here.
Your calculation is for doing it on the first try. Not for trying over and over again, running multiple accounts simultanously. That’s were the trick comes in.
Do it for 10.000 accounts and the chance.changes a lot. If you look afterwards you don’t see the failed attempts only the successful ones. And then afterwards people ask themselves “how is it possible that someone achieved that?” Because they cannot see the failed attempts on the leaderboard.
This is why it makes no sense to calculate chances for events that already happened.
That’s why i called it a statistical trick, i looks like the chance of the event is astronomically low, but only because you hide the failed attempts.
So your jupiter example only makes sense if the 100 wins were achieved on the first try. This was the point of the derren brown video… You should have watched it, it’s great.
No, you wouldn’t. That’s not how you calculate the needed attempts.
This is like saying you need 365 people to find someone that has the same birthday as you… And therefore you need 182.5 people.to have a 50% chance of finding a person with the same.birthday as you… Which it total BS.
The number of people you need to have a 50% chance in reality is 23.
This is called the birthday problem, fairly interesting.
So no if you have a 1:10 chance it does not mean you need 10 tries to achieve it, it doesn’t work that way.
Again the chances to do this in one try are terrible but for n->infinity everything with a chance >0 however small is guaranteed to happen.
You can’t calculate the needed attempts. You can however calculate how many attempts (accounts) it would take on average to get one success (an account with 1000 wins and NO loss).
This is not like the birthday problem(see my addition to birthday problem below), because you have absolutely no choice in outcome. It isn’t just having the same birthday, no matter which day precisely. They all have to have birthday in the first half of the year, not a single exception. And you have a thousand people.
Or look at it this way: When you start looking at the passports for birthdays, you can’t be sure that you won’t have a match success until you have seen all passports. With the game here, you can already terminate the account in your first match, if you lose. The others don’t matter then.
Where did I make that claim? I gave you the probability of this streak happening for an account, based on constant p and independence. Now you can compute the expectation value of n attempts (accounts) with success probability p (1E-301). If we just look for an expectation value of “E=1” (successful account), then the amount of seeded accounts n would be E=np. So with 1E301 trial account, we would statistically “expect” one to come through. Of course, this doesn’t mean that after 518 attempts, you can’t possibly get lucky or that you will have a guaranteed success after 1E301 trials. But it makes very obvious, that by this method, given the assumptions, the life time of the game and total matches in tank arcade played, the existance of three such accounts is virtually impossible. As a comparison, the universe holds about 1E80 atoms. If every atom were a universe itself, with another 1E80 atoms in it, we would have reached 1E160 atoms, which is still incredibly less than 1E301.
Just in case you confuse this with player loss streaks within a larger set of battles: We are not looking for something like the chance of a continuous loss streak of 20 games in 10000, which is surprisingly likely. We have to win EVERY match within the account. One loss means restart from scratch.
Edit:
What you describe is actually wrong and not the birthday problem. The birthday problem is about any of the individuals in a group of size n sharing the same birthday without further specification. That is why it is relatively likely to happen even in small groups. When you have 367 random people, at least two of them must have identical birthdays.
But if you look at your birthday, none of them may have theirs on your birthday. In that case, each person has a 1/365 chance to share yours. If we have independence, it is now a binomial problem. If you take 252 random people, you pretty well have a chance of 50% of at least one of them having your birthday. If you take 365 random people, you are up to ~63 % chance of at least one of them having your birthday, while the expectation value is E=1 now. (assuming 365 days a year)
From viewing a server replay of _GOTTKAISER_ it’s definitely a bit weird. The first vehicle they play doesn’t have their name, nor are any of the view controls available. When the first vehicle is destroyed the next vehicle shows their name, the view controls are available, and the gameplay style is quite different.
Can only think it was a Gaijin bot at first, replaced by a player (e.g. late join). Maybe the stats are mis-calculated between the first vehicle played and then not including the player one correctly… something along those lines.
This is based on viewing one server replay so just a quick theory.
If you’re interested in seeing what is going on I’d suggest viewing a few more server replays.
He played 970 newbie battles within the replay server history.
While his wins count, his score is obviously not used in the charts. That is why it looks as if he is inactive.
The remaining question is, how does he manage to stay in the newbie pool for 970 battles?
Obviously, you don’t understand the order of the numbers we are talking about, the probability that you can make a streak of 1000 wins out of 10000 attempts is still zero. You can calculate such probabilities yourself in the game of heads and tails.
The simplest answer to the question of how they did it, according to the principle of Occam’s razor, is that this guy found a system vulnerability. It’s amazing that you chose the least likely option as your answer to my question.
Because i mean even trying to get a deterministic value would be stupid. An expected value is the only one making sense. That’s like saying you cannot calculate how many dice throws you nee to get a six. Sure you cannot calculate a deterministic value. But it is really stupid to expect a deterministic value since that is an oxymoron. But of course you can calculate the number of dice rolls you need to expect a six, its 3.8.
So of course it isn’t impossible. And of course we can do this in this case.
Why would you even expect a deterministic value? That is impossible when it comes to probaility distributions.
Lol E=np? No that is simply BS. That’s not how you do it. Otherwise it would be 6 dice rolls to expect a six but in reality it is 3.8. don’t make up formulas, use the proper math…
That’s really not how it works.
What you calculated is the number of tries n where you get an average of one six. That is not the same thing as the expected value of n to get one six.
The formula you need is
1-(1-p)^n=0.5
And then you need to find the n where this is the case.
Look you could have at least looked it up, not make up a formula.
This is Dunning and Kruger really. And I have no interest in showing who is who. I pointed out your obvious mistakes with the birthday problem, I am not going to lecture you on the meaning of expectation value of binomial distributions. Anyone interested can look it up. You don’t pay me. ;-)
1-(1-p)^n :
And keep editing, I actually did all that above, when I corrected your birthday problem to n=252 and n= 365. But you don’t even realise it. Have a nice day.
" A binomial distribution’s expected value, or mean, is calculated by multiplying the number of trials (n) by the probability of successes (p), or n × p. For example, the expected value of the number of heads in 100 trials of heads or tails is 50, or (100 × 0.5)."
Well sort of you claim to know stochastics but fail miserably.
You Don’t know what you are talking about. But you say it with confidence.
The expected value is the first momentum of a distribution. It is the point where you have a 50% of the successful cases are lower and 50% are higher. The middle point so to speak.
n*p is only for uniform distributions like the results of a dice roll. But since we are not talking about the results of a dice roll but the number of attempts to get a desired result it isn’t a uniform distribution but the distribution:
1-(1-p)^n
With p being the chance to succeed on the first try.
Oh god you read the wiki article and didn’t understand it. Ok this is comedy gold.
I love it. Ok it says there are 253 pairs to consider when you have 23 people. It doesn’t mean that you need 253 people to have a 50% chance. You need 23, the 253 comes from the fact that each of the 23 people have 22 other people they can possibly have a common birthday with, since (23-1)=22, which creates 253 possible birthday pairs (23*22)/2=253 (gaussian sum formula). You took that for 253 people?
Geez you really do suck at this.
No the answer is still 23. And for people like you who obvioulsy have a problem with the math, there is a calculator:
Damn you are bad at this.
Edit: on a side note, i like how the wiki article for the Birthday problem calls it counter intuitive because the result is much lower than what people expect it to be.
Because it’s the same what happens with the 100 victories, you guys think intuitively that it needs much more tries than it really does and therefore cannot wrap your head around it.
Oh yeah i wrote that wrong.
It was meant to be the classical proplem.
I can see the error. Fine. That’s on me.
I still don’t get how you got to E=np. That’s still very wrong.
Now E=np is more baffeling since you seem to have used the correct formula for the birthday probaility.
As you can see this has been done in low tier, there are squads that chase AB winrates that get easily 90% since few people play tge objectives [WinFX is i recall, not sure if they are still around]. In low tier it is even easier for them. Then the expected value drops a lot. There is even a thread in the old forum where the squad explains it, and it is just used as a flex. With a winrate of 95% which is very doable for them with a new account against bots, it is just 116 tries, with mostly bots as your enemies the winrate might be even higher. The inactivity comes from high altitude bombing.
I don’t want to get into demagoguery, but I have to say that the main point I was trying to make is that, according to the principle of Occam’s razor, out of two theories that explain the same thing, the simplest one should be chosen. In this particular case, the theory that this player found a way to cheat the system is more realistic, and Dodo_Dud proved it to be true, this player/players found a way to constantly get into battles with bots. You defended theories about 10,000 accounts and the incredible coincidence of winning 1,000 battles in a row. The probability of this is close to 0, which is why I used the analogy with Russell’s teapot.
I don’t know how you do it Dodo. This thread was so painful to read. It’s like an infinite number of monkeys on an infinite number of typewriters were asked to write the most annoying thread about probability imaginable…
(Hint: All you have to do is figure out the number of accounts that would be needed to do the "launch an account and play it til it wins then discard it after the first loss to keep your one perfect account for 1000 games " trick, and compare that to the base size of their account table (currently 10^9 from the last ban list, with 160 million total accounts ever created so far in 10 years) to see the problem. If every single one of those accounts ever created in War Thunder was played that way, only ever played in ground AB, then the losing half of all accounts ever discarded on first loss, the longest streak you’d have with the one surviving account at the end would still only be 2^27, or 27 wins.)
In any case, yeah, they’re exploiting the newb pool somehow, as was established up thread. I would focus on that.
