# For the math experts (random draw of decks)

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Anonymous

February 22, 2005 3:19:17 PM

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

-----------------------------------------------------

V:tES Brasil Site (only in Portuguese for now)

http://planeta.terra.com.br/lazer/vtesbrasil/

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

-----------------------------------------------------

V:tES Brasil Site (only in Portuguese for now)

http://planeta.terra.com.br/lazer/vtesbrasil/

More about : math experts random draw decks

Anonymous

February 22, 2005 3:36:05 PM

Fabio "Sooner" Macedo wrote:

> 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?

(I assume by "simple drawing" people draw numbers, and then pick a

remaining deck in order of number drawn.)

Under your method, at least as many, and probably more, people will get

their first pick (exact probabilities depend on how many people want

each starter, of course). Conversely, less people will get their second

or third pick. So you'll have more people who have their first pick,

and more people who have their fourth pick.

Whether this is preferable or not is up to you. It punishes people who

pick the most popular starter, and rewards the people who pick the

second or third most popular starter. People who pick the least popular

starter are unaffected -- there will never be a situation where they

won't be able to get it.

--Colin McGuigan

Anonymous

February 22, 2005 3:36:29 PM

Fabio "Sooner" Macedo wrote:

> 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?

(I assume by "simple drawing" people draw numbers, and then pick a

remaining deck in order of number drawn.)

Under your method, at least as many, and probably more, people will get

their first pick (exact probabilities depend on how many people want

each starter, of course). Conversely, less people will get their second

or third pick. So you'll have more people who have their first pick,

and more people who have their fourth pick.

Whether this is preferable or not is up to you. It punishes people who

pick the most popular starter, and rewards the people who pick the

second or third most popular starter. People who pick the least popular

starter are unaffected -- there will never be a situation where they

won't be able to get it.

--Colin McGuigan

Related resources

Anonymous

February 22, 2005 5:17:16 PM

> 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.

There are a few too many permutations to consider to get paydata, I'd

suggest giving the problem to a college student to do an Independent Study

Project on.

> 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.

>

I like your idea, good luck with it.

> 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).

>

I'd actually offer to give people their second choices before using a

random system. People with unpopular second choices can get their second

choice and be relatively happy. Then, if the First Choice deck is still

too much in demand, you roll off.

> I assume this method greatly reduces the chance of everyone getting a

> third or fourth option

It does, but at some cost. Let's say we have 6 (AN) - 6 (B) - 2 (G) - 2

(AL). Worst case scenario is someone lists Baali and Anathema as their top

2. He has a 2/3 chance of getting his top choice - versus 1/4 - but a 0

chance of getting his second choice - versus 1/4.

If the numbers are 6 (B) - 4 (AN) - 4 (G) - 2 (AL), the two players who

miss out on Baali get Alastors, regardless of where it lies in their

priorities.

> I suspect that a single, simple drawing means that *all* players could

> potentially get what would be otherwise third and fourth options

That's why all of the sealed play I've seen allowed you to trade your

deck. An even number of each deck is drawn for the group, and everyone

gets assigned a random starter, allowed to trade before opening it. Even

if you do implement your system, I'd still give people the chance to trade

decks, on the off chance that Person A's Third Choice is more personally

satisfying than Person B's Second Choice.

Anonymous

February 22, 2005 6:01:44 PM

LSJ wrote:

> You might try adapting one of the methods for assigning

> powers in Diplomacy, since quite a lot of noses have been

> poked into that bag.

>

> (In fact, it sounds like the method you adopted was

> a parallel to method 2A here

> http://www.diplom.org/Zine/F1998M/Tarzan/assign.html

> )

>

> It seems that method 2C is the fairest, but then

> you also have to come up with an accurate

> "satisfaction" weighting.

> It's also the most complicated to compute, IMO.

Or, you could simply debate which satisfaction table best suits your

group among the seven players. Meta meta meta.

Jeff

Anonymous

February 22, 2005 6:56:53 PM

You might try adapting one of the methods for assigning

powers in Diplomacy, since quite a lot of noses have been

poked into that bag.

(In fact, it sounds like the method you adopted was

a parallel to method 2A here

http://www.diplom.org/Zine/F1998M/Tarzan/assign.html

)

It seems that method 2C is the fairest, but then

you also have to come up with an accurate

"satisfaction" weighting.

It's also the most complicated to compute, IMO.

--

LSJ (vtesrep@white-wolf.com) V:TES Net.Rep for White Wolf, Inc.

V:TES homepage: http://www.white-wolf.com/vtes/

Though effective, appear to be ineffective -- Sun Tzu

Anonymous

February 22, 2005 10:22:57 PM

On Tue, 22 Feb 2005 14:17:16 -0500, Gregory Stuart Pettigrew

<etherial@sidehack.sat.gweep.net> wrote:

>> 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).

>>

>

>I'd actually offer to give people their second choices before using a

>random system. People with unpopular second choices can get their second

>choice and be relatively happy. Then, if the First Choice deck is still

>too much in demand, you roll off.

That's a good advice. I didn't take into consideration that now we get

4 choices (I tried it with Anarchs decks, only 3 choices for 9 people.

Maybe this changes probabilities, I really don't know).

>It does, but at some cost. Let's say we have 6 (AN) - 6 (B) - 2 (G) - 2

>(AL). Worst case scenario is someone lists Baali and Anathema as their top

>2. He has a 2/3 chance of getting his top choice - versus 1/4 - but a 0

>chance of getting his second choice - versus 1/4.

>If the numbers are 6 (B) - 4 (AN) - 4 (G) - 2 (AL), the two players who

>miss out on Baali get Alastors, regardless of where it lies in their

>priorities.

Oh yeah, I didn't think of that. But I see some point in checking in

with other players to know the risks of going for the most popular

option. It becomes a "no pain, no gain" situation. I like that, seems

like gambling without the dreadful conotations of it.

>> I suspect that a single, simple drawing means that *all* players could

>> potentially get what would be otherwise third and fourth options

>

>That's why all of the sealed play I've seen allowed you to trade your

>deck. An even number of each deck is drawn for the group, and everyone

>gets assigned a random starter, allowed to trade before opening it. Even

>if you do implement your system, I'd still give people the chance to trade

>decks, on the off chance that Person A's Third Choice is more personally

>satisfying than Person B's Second Choice.

It will be allowed, of course. Just forgot to mention it, sorry.

Many thanks for your insights!

best,

Fabio "Sooner" Macedo

V:TES National Coordinator for Brazil

Giovanni Clan Newsletter Editor

-----------------------------------------------------

V:tES Brasil Site (only in Portuguese for now)

http://planeta.terra.com.br/lazer/vtesbrasil/

Anonymous

February 23, 2005 2:43:43 AM

On Tue, 22 Feb 2005 14:17:16 -0500, Gregory Stuart Pettigrew

<etherial@sidehack.sat.gweep.net> wrote:

>> I assume this method greatly reduces the chance of everyone getting a

>> third or fourth option

>

> It does, but at some cost. Let's say we have 6 (AN) - 6 (B) - 2 (G) - 2

> (AL). Worst case scenario is someone lists Baali and Anathema as their

> top

> 2. He has a 2/3 chance of getting his top choice - versus 1/4 - but a 0

> chance of getting his second choice - versus 1/4.

>

> If the numbers are 6 (B) - 4 (AN) - 4 (G) - 2 (AL), the two players who

> miss out on Baali get Alastors, regardless of where it lies in their

> priorities.

You could assign the Alastors to those aspiring Baali players who had

Alastors highest up. Like, if two have it second, the problem is solved.

I made a sum-utility matrix (that was loosely based on the transportation

model) that aims to maximize the total combined utility of all players

participating. From that apporach, giving the most possible number of

players their first picks is the best solution (because in the majority

of cases the majority of players will get their first pick). Shuffling

with deviance-from-group-average-utility is a poor choice, given how few

options there are (If your options are assigning 12 first picks, 2 second

and 2 third picks, or to assign 10 first picks and 6 second picks, it is

probably better to go with the former - that way, 3 out of 4 players get

their first picks).

--

Bye,

Daneel

Anonymous

February 23, 2005 11:53:55 AM

On Tue, 22 Feb 2005 15:56:53 -0500, "LSJ"

<vtesrepSPAM@TRAPwhite-wolf.com> wrote:

>You might try adapting one of the methods for assigning

>powers in Diplomacy, since quite a lot of noses have been

>poked into that bag.

>(In fact, it sounds like the method you adopted was

>a parallel to method 2A here

>http://www.diplom.org/Zine/F1998M/Tarzan/assign.html

>It seems that method 2C is the fairest, but then

>you also have to come up with an accurate

>"satisfaction" weighting.

>It's also the most complicated to compute, IMO.

Thanks! For the sake of brevity (and lack of an available PC/laptop at

the event) I'll leave 2C aside, though I find it's a pity since it

seems really the best method. I'm considering 2B.

best,

Fabio "Sooner" Macedo

V:TES National Coordinator for Brazil

Giovanni Newsletter Editor

-----------------------------------------------------

V:tES Brasil Site (only in Portuguese for now)

http://planeta.terra.com.br/lazer/vtesbrasil/

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