Getting Babylon in 4 out of the 5 last games

[SluggerBaloney]

Deleted

[Jellby]

But you've played 1361 games of 7 Wonders, that's 272 independent sets of 5 consecutive games, so that should have happened a few times already. Of course, you didn't come here and cheer every 5th game after you didn't get Babylon 4 times (or all the times that you should have gotten the same non-Babylon wonder 4 times out of 5).

[euklid314]

First of all, SluggerBaloney miscalculated by losing a 7 in the last steps of the calculation. The probability is only 0.001795 = 30/16807 and not 0.0125 =30/2401.

But Jellby's counterargument is also much stronger as Jellby stated, since those 272 independent sets do not include the case that SluggerBalloney did get 4 out of 5 Babylon in, say, his 13th, 14th, 15th, 16th, 17th game. Jellby is considering only the games 1-5, 6-10, 11-15,... when he calculates with the number 272. But the games 13-17 are equally noteworthy as the games 11-15 or 16-20..

The truth is, therefore, that you have to consider all 1357 of the 5-game-long series within the 1361 played games. These 1357 series are not independent anymore, since if you get lots of Babylon in, say, games 13-17, this of couse means that you will have lots of Babylon in games 12-16 and games 14-18 as well. This effect, however, is cancelled out by the fact that if you get few of Babylon in games 13-17 you will have few Babylon in games 12-16 and games 14-18.

SluggerBaloney is therefore expected to already have had 1357×0.001795 = 2.4 instances of "4 Babylon out of 5 games" in the long 7wonder career. Same with every other wonder besides Babylon...

[Remkar]

But you've played 1361 games of 7 Wonders, that's 272 independent sets of 5 consecutive games, so that should have happened a few times already. Of course, you didn't come here and cheer every 5th game after you didn't get Babylon 4 times (or all the times that you should have gotten the same non-Babylon wonder 4 times out of 5).

They don't even have to be completely independent. There are technically 1357 five games runs in there too choose from.

That's not even taking into consideration if they would be upset about getting Babylon 3 out of 3 games, or 6 out of 8 games, etc, which all could have also led to a post like this.

So I'd garner that the odds of sometime playing 1361 games and somewhere in there being upset about how many times they got a wonder in some sort of small grouping of games is just about 100%.

I also think his calculation is wrong, though, % seems too high, I'd have to check. (Edited to add: looks like Euklid was posting at the same time and actually figured it out.)

[Jellby]

But Jellby's counterargument is also much stronger as Jellby stated, since those 272 independent sets do not include the case that SluggerBalloney did get 4 out of 5 Babylon in, say, his 13th, 14th, 15th, 16th, 17th game. Jellby is considering only the games 1-5, 6-10, 11-15,... when he calculates with the number 272. But the games 13-17 are equally noteworthy as the games 11-15 or 16-20..

Yes, I was aware of that, that's why I said "independent sets". I took that as a lower bound for the probability, not wanting to bother computing it properly. I was about to say that it should have happened around 3-4 times, but then I settled for saying just "a few times". Thanks for pointing that out and for fixing the base probability, though.

[Romain672]

I did 100 simulations: https://docs.google.com/spreadsheets/d/ ... l=fr#gid=0 .
0 serie => 22
1 => 22
2 => 18
3 => 15
4 => 5
5 => 8
6 => 3
7 => 3
8 => 3
11 => 1

Note that a potential serie of 5 babylons /5 will count for at least 3 ("xvvvv", "vvvvv", "vvvvx").

1.746%. Really close from euklid calculations of 1.795%. But you can see massive differences between simulations. 5/6/7/8/11 seems to have a lot of results.

[euklid314]

@Jellby: Yes, I noticed that your wording was correct and precise (and I cited you in this way). It is far from obvious that my reasoning is mathematically correct, because of the statistical dependences of all the overlapping 5-game series. I could not even give a mathematical proof - but I knew from similar mathematical problems that my claim of the expected value was correct.

@Romain: Thanks for your simulation that "proves" my claim that I could not have (easily) proven myself. As you pointed out nicely, the distribution clusters - but the expected value E(X) of the random variable X of "4 out of 5" occurrences is still calculated by E(X) = n×p.
In Romain's 100fold simulation there are (on average)
2,37=(0×22+1×22+2×18+3×15+4×5+...+1×11)/100 occurrences which underlines my (mathematical) prediction of 2,4 expected occurrences in SuggerBaloney's career.

[LaidBackAries]

Yup. It's been doing that again to me. It'll be the same wonder for a handful of games in a row. It stopped doing it for awhile and it doing it again. No idea why.

[Jellby]

Yup. It's been doing that again to me. It'll be the same wonder for a handful of games in a row. It stopped doing it for awhile and it doing it again. No idea why.

When things happen with no apparent reason, we tend to call them "random"

[LaidBackAries]

More likely it's a bug or i just didn't notice for awhile because i got the good Wonders! Lol