How many different 2v2 setups are possible...

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trickser
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How many different 2v2 setups are possible...

Postby trickser » Thu Apr 02, 2009 11:53 am

and have you seen/played them all?

We have hit the 200 games mark in the current 2v2 ladder season, that are more games then previous ladder.
I wondered if I have seen or played all possible setups and did some math.

There are 6*5*4*3=360 different 4 player setups. But from the players view, its not important if the other team is russia and asia or asia and russia.
So we have 180 unique setups to play.

From the specs view, you would neither care if Team A switches its territories or if Team B does, or if Team A and B switch roles altogether.
So we have 360/2/2/2=45 unique setups to watch.

So how many games we have to watch to have seen them all? Well at least 45 or if we are unlucky we never see all. But how many in average?
The 1st time we watch a 2v2 we certainly see a new unique setup, but the 2nd time we have a 1/45 chance to see the very same setup again and just a 44/45 chance to see a new one. After we have seen 2 unique setups, we have a chance of 43/45 to see a new one and so on, until we have a 1/45 chance to see the last one.

Thats the probability, but how many games we need to play for probability=1? (44/45)*(45/44)=1
So we need to watch 1+45/44+45/43+...+45/2+45/1 games.

I have no clue how to calculate that other then brute force, but i know how to let my pc brute force it. :)


And the results are (please imagin a fanfare):
We need to watch 197.8 games to see the 45 unique 2v2 setups
We need to play 1039.1 games to encounter all 180 unique 2v2 situation


If you watched every ladder game, you probably have seen them all, but no way you played them all, even The Stiffmeister has not.

I am known to myself to include errors in stuff like this, please dont nail me to the results.
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Postby bert_the_turtle » Thu Apr 02, 2009 12:27 pm

Well done. To my knowledge, there's no simple closed formula for the sum you're facing. You can express it in terms of an obscure special function, but that's cheating. The usual approach is to approximate such a sum by an integral; in your case, the integral would be over 1/x and yield a logarithm as result, so the total number of games you need to watch would rise with n * log n, where n is the number of unique combinations you're interested in.
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Postby tllotpfkamvpe » Fri Apr 03, 2009 8:20 am

.
Last edited by tllotpfkamvpe on Fri Feb 24, 2023 5:30 am, edited 1 time in total.
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Postby trickser » Fri Apr 03, 2009 3:08 pm

games you need to watch to see the 1st unique setup: 1
probability to see 2nd unique match in the next game you watch: 44/45
games you need to watch to have a chance of 1(100%) to see 2nd unique match: 45/44
games you need to watch to see 1st AND 2nd unique setup: 1+45/44 = sligtly more then 2

games you need to watch to see 3rd unique match: 45/43
games you need to watch for 100% chance to see 1st,2nd and 3rd = 1 + 45/44 + 45/43


and so on
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Postby bert_the_turtle » Fri Apr 03, 2009 3:15 pm

trickser wrote:average number of games you need to watch to see 2nd unique match: 45/44
fix'd :) There's no such thing as 100% probability for any finite number of watched matches here.
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Postby trickser » Fri Apr 03, 2009 7:44 pm

Hehe, good point.
Its 1/probability, the reversed dimension. Thats frequency , i believe. Nah, but something like this.

2nd thought. initially it was:
probability to see new unique setup
----------------------------
number of games


where i only meaned 1 game and therefor only spoke about probability

reversed its:
number of games
----------------------------
probability to see new unique setup


where i meaned probability=1, assuming its the average

Frequency usually means quantity of repeat per time or per lenght. So its indeed some kind of frequency, but not in the usual meaning.

3rd thought.
probability is already
chance
--------
number of games


Ah well, not so easy to fit dimensions concept into language usage, but truth must be near.
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Postby Nuke this » Tue Apr 07, 2009 2:31 am

Your probability will never be 1, it will approach 1.

Instead you could work out the probability of seeing all 45 games in...

10 games (0)
45 games (miniscule)
90 games (slightly better)
...
...
100,000 (high 90's)
1^20 (Very high 99's)
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Postby trickser » Wed Apr 08, 2009 5:07 pm

While developing the calculation, i substitute chance (eg. 44/45) by number of games (eg 45/44) by seting the product to 1 ( (44/45)*(45/44)=1 ). I did it intuitively and mistook it for probability. But i still think its correct, i just dont know why.

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