Archived from groups: rec.games.trading-cards.jyhad (
More info?)
For the math experts out there, I'd like to ask some help and advice
on this.
I've tested a simple method of deck distribution last year during a
sealed deck tournament and it went quite well (9 players), but I lack
the math skills to calculate the real difference in probabilities. I
thought it would be good to add some choices to players and therefore
diminishing the chances of everybody not being really satisfied with
the deck they get during this type of event, and thought it would be
good to our pre-release events (delayed for one week due to customs
problems).
Basically, the organizer asks for each player to prioritize their
choices given the decks available. So, let's say, I'd like to get the
Baali deck, Alastor second, !Gangrel third.
The organizer polls the results and see which decks had too much
demand based solely on the players' first options. If he has the
available number to give one type of deck or another to all that chose
it as first option, he does it. Say that he has 4 copies of each deck
available and only 2 players chose Alastors. They get that deck.
Conversely, say that 6 players chose Baali. The organizer uses some
random method to assing the 4 copies available between these players.
The two unlucky ones don't get the Baali deck (say I am one of these
two).
Now the organizer compares which players are left with no deck and
which copies are still available. He repeats the process using any
player's second option. In our example there are still 2 Alastors. Say
that me and two other players chose Alastor as second option. We enter
a new round of random assignment. This time I'm in and get the Alastor
deck. The third player is left to a new round based on third options,
and so on, until everyone gets one deck to play.
I assume this method greatly reduces the chance of everyone getting a
third or fourth option, and at least introduces the notion that if one
gets a deck they'd not normally choose, they have to be out of luck in
two or three drawings instead of just one. But besides these pratical
conotations, I'd like to know if I'm right regarding math.
I suspect that a single, simple drawing means that *all* players could
potentially get what would be otherwise third and fourth options, even
if some chose decks that are not all too popular in their metagames,
while the method I've used last year guarantees that at least some get
their first option and maximize the chances to make (almost) everyone
get their first or second options. I suspect that in my city most will
choose Alastor, per example, so a single drawing could frustrate
people who choose Baali or Anathema, even if not that many players
want these and there are plenty of copies to satisfy them. Could
someone calculate the probabilities to confirm if my suspicions are
correct?
best,
Fabio "Sooner" Macedo
V:TES National Coordinator for Brazil
Giovanni Newsletter Editor
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V:tES Brasil Site (only in Portuguese for now)
http://planeta.terra.com.br/lazer/vtesbrasil/