论坛: 英语 rss-feed

这是英语的讨论版块。在这个论坛,你必须使用这种语言发帖,使用其他语言的帖子都将删除,恕不另行通知.论坛有不同的主题分类,发帖的时候请注意发到合适的主题分类里。

英语 >> 疑难解答

The Position Experience thread (18)


il Numpty >> 星期二 十一月 29 - 12:08, 已编辑 星期二 十一月 29 - 13:22

I've occasionally wondered how long it takes for a player to reach 1000 position experience. So thought I'd set myself a little challenge to work it out. 

* This all comes with the massive caveat that all the results are based on the formula for position xp earned per match, which I believe is correct but is not proven. 

Summary of results

I'll start with the results, since most people probably won't be bothered with the detail. 

If a player starts with zero position xp and always plays in the designated position; that is, as a keeper, defender, midfielder or forward; then this is how many matches are needed to reach the following amounts of position xp:

  • 100 xp - 11 matches
  • 200 xp - 23 matches
  • 300 xp - 36 matches
  • 400 xp - 51 matches
  • 500 xp - 69 matches
  • 600 xp - 92 matches 
  • 700 xp - 120 matches 
  • 800 xp - 161 matches
  • 900 xp - 230 matches
  • 999.5 xp - 757 matches (we never actually reach 1000)

(This is without using position training)

il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
il Numpty >> 星期二 十一月 29 - 12:09, 已编辑 星期天 三月 19 - 18:20

Now for the detail and the various formulae for position experience.

Position xp earned per match 

When a player plays in his designated position - a defender in defence etc. - his position experience increases by 1% of the difference between 1000 and his current amount. 

Maybe a simpler way to understand is that for every match played the player loses 1% of position xp from all 4 positions and then gets an extra 10 for the position played (and an extra 5 for any secondary position). 

For example, at zero xp he gets 10 for his first match, at 100 xp he gets 9  for a match and at 500 xp he gets 5 for a match and so on. 

  • Position xp earned per match = [1000 - Current] / 100
  •  = 10 - Current / 100 (same)

Note - I believe that this is correct but all position xp values are rounded to the nearest integer so it has not been proven definitively. 

Total position xp

If we start at zero then the total position xp earned is as follows:

Total (T) = 10 + 9.9 + 9.801 + 9.70299 + ... 

  • Total (T) = 10 * [1 + 0.99 + 0.99^2 + 0.99^3 + ... 0.99^(n-1)] 

- Where n is the number of matches played

This is called a geometric progression and can be simplified like so: 

  • Total Position xp (T) = 1000 * [1 - 0.99^n] 

(Ask me if you want to know how I did that!)

This gives us the formula for calculating how much position xp has been earned after n matches have been played (in the correct position). 

As n gets very large (tends to infinity) then Total position xp = 1000, which is what we should expect. 

How many matches?

If we want to know how many matches are needed to reach a given total, then we need to reformat the equation. Solving for n we get this:

  • n = log [1 - T/1000] / log 0.99  

This gives us how many matches are needed to reach a given total T if we start from zero. We'll obviously need to round each result up to the nearest integer and this gives us:

  • 100 xp - 11 matches
  • 200 xp - 23 matches
  • 300 xp - 36 matches
  • 400 xp - 51 matches
  • 500 xp - 69 matches
  • 600 xp - 92 matches 
  • 700 xp - 120 matches 
  • 800 xp - 161 matches
  • 900 xp - 230 matches

We can never reach 1000 xp - without position training - but we can get to 999.5. 

  • 999.5 xp - 757 matches 
il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
pl GOBOYGO >> 星期二 十一月 29 - 14:10

Holy s* bro, respect for the details. Half of it I don't even understand but I believe your figures ;-)

pl GOBOYGO
尚未出师
注册于2021-12-19
il Numpty >> 星期二 十一月 29 - 17:07, 已编辑 星期五 二月 10 - 10:16

Thanks.

il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
us Mike >> 星期四 十二月 1 - 02:01

Good work Numpty.  These fit fairly well.  I have this new guy who has only played 13 matches, but is at 187 attacking xp which is on the high end of what you would expect.

https://rockingsoccer.com/en/soccer/info/player-2635075

Some of my older teen players fit better.

us Mike
小有名气
注册于2015-05-29
us Colonials FC
il Numpty >> 星期四 十二月 1 - 03:18, 已编辑 星期四 十二月 1 - 11:50

Cheers @Mike.

Bear in mind the formula works starting from zero. If he's on 187 after 13 matches then he probably started on around 70-75.  After a bit of playing with my formula it suggests that starting on 74 would get to 187.4 after 13 matches.

The same issue will apply to your older teens. They will all have started on some level of position xp. But the effect reduces with more matches so will be less noticeable. 

Just to add that I've not researched starting amounts of position xp. But as far as I know the initial position xp of youngsters depends on their age and the distribution across the 4 positions depends on your youth centre settings. Plus, of course, there's a random element.

il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
us Xandy >> 星期四 十二月 1 - 16:02, 已编辑 星期四 十二月 1 - 16:03

Thanks Numpty

And they wondered in Math Class when would you ever use this in life lol

us Xandy
小有名气
注册于2021-05-11
us Plano happyfeet
il Numpty >> 星期四 十二月 1 - 16:12, 已编辑 星期四 二月 23 - 12:07

General formula 

Perhaps it might be useful to have a more general formula that doesn't rely on the starting amount being zero. 

If C is the current or starting amount of position xp then we get this: 

  • T = C * 0.99^n + 1000 - 1000 * 0.99^n 
  • T = 1000 - 0.99^n * [1000 - C] (same)
  • T = C * 0.99^n + 1000 * [1 - 0.99^n] (same)
  • Use whichever seems easier

This gives us the total amount of position xp after playing n matches if we start with C.

Finally how many matches are needed to get from the current amount, C, to a new total of T?

  • n = log [1 - (T-C) / (1000-C) ] / log 0.99  

For example, how matches are needed to get from 600 to 800?

  • n = log [1 - 200/400] / log 0.99 = 68.97

The formula gives us 69 matches, which fits with the earlier table.

il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
il Numpty >> 星期四 十二月 1 - 16:34

Incidentally, for those who are observant you will notice that it takes exactly the same amount of time (matches) to go from 600 to 800 as it does from 0 to 500. This is because it's the same percentage (50%) of the remaining amount up to 1000.

The same rule applies to all percentages. So it takes exactly the same time to get from 0-900 as it does from 900 to 990 and from 990 to 999. Which is 230 matches in each case.

il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
eng wonderlust >> 星期一 十二月 5 - 21:51

This is called a geometric progression and can be simplified like so: 



  • Total Position xp (T) = 1000 * [1 - 0.99^n] 


(Ask me if you want to know how I did that!)


How did you do that?  I think I learned this in school, but that was a loooong time ago!

eng wonderlust
小有名气
注册于2014-10-30
eng Forever We Are Wolves
il Numpty >> 星期一 十二月 5 - 22:13, 已编辑 星期一 十二月 5 - 22:39

Lol, I'm genuinely surprised that someone asked. :D

Summation of a Geometric Progression

This is the kind of thing you may have done in post 16 maths education. Typically 'A' levels in the UK. 

The equation we started with was this:

  • T = 10 * [1 + 0.99 + 0.99^2 + 0.99^3 + ... 0.99^(n-1)] 

This series is called a geometric progression, because each subsequent term is multiplied by the same value. In this case each term gets smaller than the previous one using a multiplier of 0.99.

To simplify the summation we multiply both sides by 0.99, which gives this:

  • 0.99 * T = 10 * [0.99 + 0.99^2 + 0.99^3 + ... 0.99^n] 

Note the similarity between the 2 equations - most of the terms are identical. If we subtract the second equation from the first one then all of the intermediate terms cancel out and we are left with:

  • 0.01 * T = 10 * [1 - 0.99^n] 

Which can then be reformatted as: 

  • T = 1000 * [1 - 0.99^n] 

Converting a recurring decimal back into a fraction

Incidentally, the technique for converting a recurring decimal is very similar - and easier. It's best explained with an example.  

To convert 0.2727272727... into a fraction then first we set up an equation:

  • y = 0.2727272727...

Now we need to multiply both sides by 100 because the frequency of the repeated digits is 2. [If only one digit repeats then multiply by 10, for 3 digit repetition multiply by 1,000 and so on.]

  • 100 y = 27.2727272727...

Subtract the first equation from the second and the recurring digits cancel out, leaving:

  • 99 y = 27  

And this simplifies to 

  • y = 3/11 
il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
eng wonderlust >> 星期一 十二月 5 - 23:01

Ah yes, sneaky trick - thanks!

eng wonderlust
小有名气
注册于2014-10-30
eng Forever We Are Wolves
eng holt >> 星期三 十二月 7 - 13:37

Numpty is now teaching maths on RS...

As with all tricks, it only works till a certain degree, for example trying to show that 0.14285714.. = 1/7 will take you some time to do by hand.

Maybe you should create a forforum m thread where you compile all the tools/formulae that you have figured out so far, it would be great to go through all of them.

Also, please tell me if you're interested in recreational mathematics as I have a few puzzles that I don't have the answers for...

eng holt
尚未出师
注册于2013-05-18
il Numpty >> 星期三 十二月 7 - 14:49, 已编辑 星期三 十二月 7 - 14:53

Lol @Holt. I can't disagree. 

Converting 0.142857... back into a decimal by hand would be a bit laborious, but do-able.  In fact most recurring decimals only have a few digits that repeat.  Although others such as 1/13 and 1/23 would be a nightmare. It was about the method, which is similar to solving a GP series. 

As for your puzzles, I'm up for a challenge if you want to start a new thread. 

il Numpty
小有名气
注册于2018-10-19
eng Heath Hornets
eng wonderlust >> 星期四 十二月 8 - 22:46

@Holt It can be done.  Looking at the repeated sequence for 1/7 28 is twice 14, 56 is twice 28 and 112 is twice 56 (the 1 then carries to make it 57). Then 224 is twice 112, so the 2 carries to give you 114 and you get 4 carrying from the next number (448) to give you another 28.  And so ad infinitum...

So you have 0.14 + 0.14*0.02 + 0.14 x 0.02^2...

Do it infinitely and the formula for a geometric progression becomes

a/(1-r)

This is a generalised version of Numpty's formula above, where a is the starting number (0.14) and r is the multiplier (0.02). The r^n part (0.99^n in Numpty's example) cancels out, as for 0<r<1 this becomes zero if n is infinite.

0.14/(1-0.02) = 0.14/0.98 = 1/7

This is about the most complicated bit of maths theory I can do these days!

@Numpty a thread elsewhere would be good, actually!

eng wonderlust
小有名气
注册于2014-10-30
eng Forever We Are Wolves