Curse

sonofthort · Feb 17, 2011, 03:58 AM // 03:58

I think it would be interest to expand on your idea of monitoring patterns by attempting to record the build of each player in each team and record the outcome of each match. People have gotten degrees in Simpsonology (The Simspons tv show), this sounds like something you could make a dissertation out of if you really wanted, if not under the guise of game theory.

Skye Marin · Feb 17, 2011, 04:13 AM // 04:13

The only thing this confirms is that teams are more likely to streak with a monk (due to healing, protection, your own team's confidence, and the enemy team's frustration) than not.

We already knew that.

Ben-A-BoO · Feb 17, 2011, 06:35 AM // 06:35

Quote:

Originally Posted by Tharg

Simple statistics... You mention that you don't play a monk. So you have 3 chances to have a monk on your team. The other team has 4 chances to have a monk on their team - therefore you have a disadvantage.

^ This,
Plus your number of games (N=100) is still to small to minimize the possibility that chance biases your data significantly. Or in other words "bad luck" can still be a factor in your test.

SmokingHotImolation · Feb 17, 2011, 12:59 PM // 12:59

Quote:

Originally Posted by I Jonas I

http://en.wikipedia.org/wiki/Sample_error
http://en.wikipedia.org/wiki/Confidence_interval
http://en.wikipedia.org/wiki/Confirmation_bias
etc

This. Your observation are just way too random to even make a correlation, and even if it dit it still wouldn't imply causation.

Long story short. This tells us something about your luck and nothing about the pattern (if there is one) of RA match making.

Maver1ck87 · Feb 17, 2011, 01:49 PM // 13:49

This thread documents a fantastic scientific phenomenon! What happens when you give 'lay' people a basic understanding of statistics! I've seen 2 or so answers explaining why you WOULD predict facing more teams with monks in it than your own, terefore can't be arsed to explain it again as u agrued against the first post which explained it quite well! For your sake, either accept that this is what is expected to happen in RA or don't, cos I ain't gonna lose any sleep over it. Also, it's RA dude, just have fun and wait for the upcoming changes which will proba change everything anyways!

Tharg · Feb 17, 2011, 02:38 PM // 14:38

a more complete explanation is as follows:

Scenario A: two complete random teams:
This is when both teams are newly formed. As I mentioned before, if you do not play a monk yourself, the chance you have one monk on your team is significantly lower than a completely random team. Furthermore, the chance you get a monk in your team depends on the percentage of monks present in the pool of players. Using an assumption of an infinitely large pool of players, the chances to have 1 monk in a team are as follows:

%monks-------your team----------other team-----disadvantage

5%------------14.3%--------------18.5%--------4.3%
10%-----------27.1%--------------34.4%--------7.3%
15%-----------38.6%--------------47.8%--------9.2%
20%-----------48.8%--------------59.0%-------10.2%
25%-----------57.8%--------------68.4%-------10.6%

You can see that there is a significant difference. You can go to RA and count the percentage of monks to estimate your actual chance. Let's say if 1 out of 10 are monks, you have a 7.3% lower chance to have 1 monk on your team versus the other team. But not all teams you encounter are newly formed, there are repeat teams. Therefore...

Scenario B: one random team and a repeat team
There is also a chance that the other team is not newly formed and came directly from a previous win. This increases the odds to have one monk on the other team, since one-monk teams have a higher winning percentage than no-monk teams.

Let's say 'monk teams' have an average winning streak of 3. That means monk teams have a winning percentage of 75%. Since the average winning streak of all teams is 1/2, non-monk teams must have a an average winning streak of less than 1/2 and a winning percentage of less than 50%. Let's assume a winning percentage of non-monk teams of 40%. Monk teams have then a factor 1.875 higher winning percentage versus non-monk teams, which means that there is a 65.6% change that a repeat team is a monk team.

How many repeat teams in the queue? If an average match is 3 minutes and you wait an average of 30 seconds for a new match, 1 out of 7 is a repeat team.

Taking into account the repeat teams, the chance that the other team has one monk is now: 6/7*34.4% + 1/7*65.6% = 38.9%.
Your disadvantage taking repeat teams into consideration is about 38.9% - 27.1% = 11.8%.

If you assume longer winning streaks of monk-teams, the winning percentage does not change significantly. For instance an (exaggerated) average monk team winning streak of 10 means a winning percentage of 90%. That would mean that 70% of the repeat teams are monk teams and the total disadvantage would be 12.4%.

And then finally...

Scenario C: one random team and a synch (repeat?) team
Now you are out of luck, since I cannot imagine there would be synch teams without a monk... To quantify this further you need to know how many synch teams there are versus really random teams and how successful they are versus random teams.

I hope this clarifies things further. If I made a mistake with my calculations or assumptions I look forward to your comments / suggestions.

Skyy High · Feb 17, 2011, 03:52 PM // 15:52

I wrote up yet another comment concerning the errors made in this analysis, but it seems that everything that needs to be said has been said.

To make it abundantly clear, however: no one is saying you don't get monks less often than you face them. They're just saying that this is, in fact, to be expected, and not the result of bad luck or bad programming.