This thread is for the discussion of my latest article, Cranial Insertion: Waiting for Moko. We would be grateful if you would let us know what you think, but please keep your comments on topic.
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"Let x equal the quantity of all quantities of x..." -- Proof by David Auburn
Please create an online petition for no questions regarding Mirrorweave and Opalescence. I will sign it. Under 500 alternate names.
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Level 3 Magic Judge
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Q: How exactly do tiebreakers work in Magic tournaments?
A: In a single-player swiss tournament, rankings are based on the following. First, it’s based on match points (you get 3 match points for every match win, 1 for each draw, and 0 for each loss). After that, players with the same number of match points are ranked based on their opponent’s match win percentage. This is the percentage of matches won by their previous opponents in that tournament. After that, ties are broken using your game win percentage. The final tiebreaker is the game win percentage of your previous opponents in that tournament. You can think of the tiebreakers this way:
1. How much you’ve pwnz0rd.
2. How hard your opponents were to pwnz0r.
3. How hard you pwnz0rd them.
4. How hard they pwnz0rd other people.
Note that these get jostled a bit in the pairing process sometimes to avoid having two players face off twice in the swiss portion of the same tournament.
The official explanation of the swiss system can be found in the Tournament Organizer’s Handbook, which can be downloaded here. My explanation is far better, though.
I bolded the part I have a question on... When doing parings for a small tournament (non-sanctioned, less than 8 players) that we didn't have a computer to run, I would write each player's name on the top of a piece of paper, then they're opponent's Name for each round followed by the player's result, and total points. For example:
Tom
1. John 2-0 3
2. Tim 1-2 3
3. Paul 2-0 6
4. Anthony 2-1 9
So for this example, Tom would have 9 points for the first part. Then what I would do to figure out the second part was based on opponent's match win points (as opposed to match win percentage like you stated). So to figure out the final standings, I would write down each player's match points next to their name and add them up, for example:
Tom
1. John 2-0 3 0
2. Tim 1-2 3 6
3. Paul 2-0 6 9
4. Josh 2-1 9 9
21
Josh
1. Tony 2-0 3 3
2. Jim 2-1 6 3
3. Sam 2-0 9 6
4. Tom 1-2 9 9
21
So looking at this, we'd have to go to opponent win percentage so basically added up all all the wins/losses for each person's opponent's and then comparing for that. Now these are just examples and maybe need more than 8 players to actually get these results, but looking at it from just the resulting points, is this a legitimate way to figure it out, or am I doing something wrong?
Now these are just examples and maybe need more than 8 players to actually get these results, but looking at it from just the resulting points, is this a legitimate way to figure it out, or am I doing something wrong?
In an unsanctioned event, you can really use whatever method you want to determine standings for the players. DCI sanctioned events, however, use a very specific system that is described in the Magic Floor Rules, section 118. I think that the Tournament Organizer's Handbook has that same information as well, and also describes a number of other methods as well. But the four described in the article are the ones that are required to be used in sanctioned events.
the Withering Borderguard doesn't make sense... you say the SBE's will check and remove the tokens all at once.. But Wotc says otherwise.. (in fact April 7th Arcana/CardoftheDay) and i recently asked them trying to submit it as a bug on MTGO.. they responded with It should get 9 Kithkin Tokens when it dies)
Also... Just to spite you Moko (but we love you ^.^) If i Cast Mirrorweave on an Animated Howling Mine, Due to March of the Machines. with 8 Tokens on either side.. how many Cards will They Draw (if i play it during thier Upkeep)
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Ravarshi Kashaku, Ancient Dragon of the Darkened Realms;
The Merciless Lord of Torture, Permanently Bound To: ">[THE PACK] 11/5/63 - 11/25/09 Goodbye mom, i'll always love you...
Think of them as different types of data. If your operation is looking for a boolean variable, it won’t accept an integer, even if the integer in question is “1,” for example.
I must have missed that day in Computer Science class where 1 was a boolean value.
the Withering Borderguard doesn't make sense... you say the SBE's will check and remove the tokens all at once.. But Wotc says otherwise.. (in fact April 7th Arcana/CardoftheDay) and i recently asked them trying to submit it as a bug on MTGO.. they responded with It should get 9 Kithkin Tokens when it dies)
That's... exactly what she said would happen. All SBE are checked at the same time, so the counters are removed at the same time as it is put into a graveyard. When the triggered ability resolves, it needs to know how many counters it has. Since it is no longer in play, it uses Last-Known Information. The last time it was in play (before SBE were checked) it had 9 counters, so they will get 9 tokens.
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Hey, you! Yeah, you behind the computer screen! You're unconstitutional.
thnx Sutherlands. they explained it same way in the email WOTC sent me regarding the bug submission.. i didn't think about LKI. so that's why it was a rejected bug submition
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Ravarshi Kashaku, Ancient Dragon of the Darkened Realms;
The Merciless Lord of Torture, Permanently Bound To: ">[THE PACK] 11/5/63 - 11/25/09 Goodbye mom, i'll always love you...
-- Proof by David Auburn
DCI Level 2 Judge Trainer
Please create an online petition for no questions regarding Mirrorweave and Opalescence. I will sign it. Under 500 alternate names.
Do you know any judges who always impress you with their work ethic, knowledge, or attitude? Nominate them to be the next Judge of the Week!
I bolded the part I have a question on... When doing parings for a small tournament (non-sanctioned, less than 8 players) that we didn't have a computer to run, I would write each player's name on the top of a piece of paper, then they're opponent's Name for each round followed by the player's result, and total points. For example:
Tom
1. John 2-0 3
2. Tim 1-2 3
3. Paul 2-0 6
4. Anthony 2-1 9
So for this example, Tom would have 9 points for the first part. Then what I would do to figure out the second part was based on opponent's match win points (as opposed to match win percentage like you stated). So to figure out the final standings, I would write down each player's match points next to their name and add them up, for example:
Tom
1. John 2-0 3 0
2. Tim 1-2 3 6
3. Paul 2-0 6 9
4. Josh 2-1 9 9
21
Josh
1. Tony 2-0 3 3
2. Jim 2-1 6 3
3. Sam 2-0 9 6
4. Tom 1-2 9 9
21
So looking at this, we'd have to go to opponent win percentage so basically added up all all the wins/losses for each person's opponent's and then comparing for that. Now these are just examples and maybe need more than 8 players to actually get these results, but looking at it from just the resulting points, is this a legitimate way to figure it out, or am I doing something wrong?
*Mayreturn*
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In an unsanctioned event, you can really use whatever method you want to determine standings for the players. DCI sanctioned events, however, use a very specific system that is described in the Magic Floor Rules, section 118. I think that the Tournament Organizer's Handbook has that same information as well, and also describes a number of other methods as well. But the four described in the article are the ones that are required to be used in sanctioned events.
I had a 4/4 Boggart Ram-Gang and an Elf Token Blocking thier Kinsbalie Borderguard with 5 +1/+1 tokens. They recieved 9 Kithkin Tokens when it died
Also... Just to spite you Moko (but we love you ^.^) If i Cast Mirrorweave on an Animated Howling Mine, Due to March of the Machines. with 8 Tokens on either side.. how many Cards will They Draw (if i play it during thier Upkeep)
The Merciless Lord of Torture, Permanently Bound To: ">[THE PACK] 11/5/63 - 11/25/09 Goodbye mom, i'll always love you...
Tibalt & His Devils vs. Avacyn's Inquisitors
My EDH decklists
That's... exactly what she said would happen. All SBE are checked at the same time, so the counters are removed at the same time as it is put into a graveyard. When the triggered ability resolves, it needs to know how many counters it has. Since it is no longer in play, it uses Last-Known Information. The last time it was in play (before SBE were checked) it had 9 counters, so they will get 9 tokens.
Hey, you! Yeah, you behind the computer screen! You're unconstitutional.
America == Velociraptor
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The Merciless Lord of Torture, Permanently Bound To: ">[THE PACK] 11/5/63 - 11/25/09 Goodbye mom, i'll always love you...
Tibalt & His Devils vs. Avacyn's Inquisitors
My EDH decklists