Clique bei der Verlosung der Tickets

[quchen]

Eine Clique mit N Personen wird mit Gewicht 1/N gezogen. Dadurch hat jedes Individuum ein Gewicht von 1, genau wie wenn man sich alleine anmeldet.

Alle Cliquen in einen Topf, ziehen, wiederholen bis 70000 Leute ein Ticket haben.

[Ch33zy_K]

Würde ja aber bedeuten.
Wenn sich alle Bewerber:innen in ein und der selben Clique anmelden müssten alle ein Ticket bekommen. Oder niemand....

[wellenreiter]

Würde ja aber bedeuten.
Wenn sich alle Bewerber:innen in ein und der selben Clique anmelden müssten alle ein Ticket bekommen. Oder niemand....

Edit: Sorry falsch verstanden haha

[irrelevant]

On a Google expedition once, I came across this semi advanced calculation. As languages, and not math, are my forte, I've NO clue how accurate it might be, but for what it's worth:

https://gist.github.com/robbl/1376571?f ... MTZieXRlcw

[charno]

On a Google expedition once, I came across this semi advanced calculation. As languages, and not math, are my forte, I've NO clue how accurate it might be, but for what it's worth:

https://gist.github.com/robbl/1376571?f ... MTZieXRlcw

These calculations are not correct anymore. They based on an algorithm which worked on the principle "If half or more of a clique gets selected, the whole clique gets selected", and thus favored cliques with an even number of members. This problem was solved by the fusion crew (I'm not sure if after the first raffle, or if they already fixed it before). A thread with (mathematical) discussion can be found at https://forum.kulturkosmos.de/viewtopic ... 6&start=30, I'm not sure if the solution for this problem was also discussed and confirmed in this thread or if this information is elsewhere.

If I remember right (no guarantees!), the solution was that each clique and individual get one ticket in the raffle, and random tickets get selected until the number of people (members of cliques and individuals) is higher than the limit.

[irrelevant]

On a Google expedition once, I came across this semi advanced calculation. As languages, and not math, are my forte, I've NO clue how accurate it might be, but for what it's worth:

https://gist.github.com/robbl/1376571?f ... MTZieXRlcw

These calculations are not correct anymore. They based on an algorithm which worked on the principle "If half or more of a clique gets selected, the whole clique gets selected", and thus favored cliques with an even number of members. This problem was solved by the fusion crew (I'm not sure if after the first raffle, or if they already fixed it before). A thread with (mathematical) discussion can be found at https://forum.kulturkosmos.de/viewtopic ... 6&start=30, I'm not sure if the solution for this problem was also discussed and confirmed in this thread or if this information is elsewhere.

If I remember right (no guarantees!), the solution was that each clique and individual get one ticket in the raffle, and random tickets get selected until the number of people (members of cliques and individuals) is higher than the limit.

Ooh, ok. Thanks for the update! I don't quite follow here though: "random tickets get selected until the number of people (members of cliques and individuals) is higher than the limit." Could you maybe rephrase? (As stated, math is gibberish to me -- think explaining to a 4-year-old. )
A clique only gets ONE ticket though, or every individual in a clique gets one each ...?

[charno]

Let me try explaining. That's just from memory, maybe I misremember or the system changed in the meantime.

Lets assume:
Limit is 75 people coming to fusion
50 People are individually applying
5 Cliques with 10 People in it.

This means, there are 55 cliques (if we assume, that each individual applicant is also a clique with just one member).

Now, repeat until "number of people" is greater than the limit:

• Select one of the cliques

• add how many people are in this clique to "number of people"

In the end, this means AT LEAST as many people as the limit get the chance to buy a ticket after the raffle.

[irrelevant]

Let me try explaining. That's just from memory, maybe I misremember or the system changed in the meantime.

Lets assume:
Limit is 75 people coming to fusion
50 People are individually applying
5 Cliques with 10 People in it.

This means, there are 55 cliques (if we assume, that each individual applicant is also a clique with just one member).

Now, repeat until "number of people" is greater than the limit:

• Select one of the cliques

• add how many people are in this clique to "number of people"

In the end, this means AT LEAST as many people as the limit get the chance to buy a ticket after the raffle.

TYVM, think I could wrap my head around this one. But doesn't this also then mean that cliques (the ones with 10 people each, here) have a LOWER chance of being picked, if they only "count" as a single entry?

[charno]

I don't think so.

In each round of the raffle, each person has the same chance to be selected. If you think about individual applications as "one person clique", and each round one clique gets selected, then the chance for each clique is the same. If at this moment only one or ten people get the ticket doesn't change the chances, it only affects how many future rounds are left.

[irrelevant]

I don't think so.

In each round of the raffle, each person has the same chance to be selected. If you think about individual applications as "one person clique", and each round one clique gets selected, then the chance for each clique is the same. If at this moment only one or ten people get the ticket doesn't change the chances, it only affects how many future rounds are left.

Hmm, ok. Appreciate the time taken to explain.