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Offline Sir Valimont

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Re: [Fire]Hyroen 2 -1 [Master of Entropy]Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198500#msg198500
« Reply #24 on: November 11, 2010, 01:42:21 am »
Comments like that are a result of your projected attitude  ::)
Projected or interpreted ? :)

Nothing wrong with being a bit eristic, though, right? Call it ... warlike.

Offline majofa

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Re: [Fire]Hyroen 2 -1 [Master of Entropy]Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198504#msg198504
« Reply #25 on: November 11, 2010, 01:46:56 am »
2) Hyroen gets Wings up instantly. This can only happen with 3 pendulums being drawn out of 9 on the first turn.
Hyroen only needs 2 pendulums to get out Wings.

1st turn: 2 pendulums(FIRE) + MARK = 1 Air
2nd turn: 2 pendulums (AIR) + MARK = 3 Air
3+1=4

What are the calculations of winning with that factored in?

Offline Sir Valimont

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Re: [Fire]Hyroen 2 -1 [Master of Entropy]Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198511#msg198511
« Reply #26 on: November 11, 2010, 01:51:24 am »
Hyroen only needs 2 pendulums to get out Wings.

1st turn: 2 pendulums(FIRE) + MARK = 1 Air
2nd turn: 2 pendulums (AIR) + MARK = 3 Air
3+1=4

What are the calculations of winning with that factored in?
Very good point ... the earliest possible Wings is turn-3. You're clearly right that 2 pendulums does it. Will be interesting to see what this means in tests.

MrBlonde

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Re: [Fire]Hyroen 2 -1 [Master of Entropy]Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198512#msg198512
« Reply #27 on: November 11, 2010, 01:54:07 am »
2) Hyroen gets Wings up instantly. This can only happen with 3 pendulums being drawn out of 9 on the first turn.
Hyroen only needs 2 pendulums to get out Wings.

1st turn: 2 pendulums(FIRE) + MARK = 1 Air
2nd turn: 2 pendulums (AIR) + MARK = 3 Air
3+1=4

What are the calculations of winning with that factored in?
And technically it means he only has to pull 1 pende turn 1 and another one turn 2 since on turn 2 any pendelums pulled will be air quanta. So the equation will have the chances of 2/9 pendelums pulled turn 1 and 1/9 pulled turn 1 and 1/8 pulled round 2. I'm not going to bother doing the math on this though. Have fun Valimont.

Re: [Fire]Hyroen 2 -1 [Master of Entropy]Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198533#msg198533
« Reply #28 on: November 11, 2010, 02:27:21 am »
Hyroen only needs 2 pendulums to get out Wings.

1st turn: 2 pendulums(FIRE) + MARK = 1 Air
2nd turn: 2 pendulums (AIR) + MARK = 3 Air
3+1=4

What are the calculations of winning with that factored in?
No need to flame ... just think about the timing of quanta flow from pendula.
    The fire deck needed to draw 2 pendula by its second turn in order to play a turn 3 wings.The entropy deck needed to draw 3 entropy pendula on its first turn in order to play a turn 2 discord, or else it could not get discord out until turn 4.[/list]
    hi

    kobisjeruk

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198538#msg198538
    « Reply #29 on: November 11, 2010, 02:37:42 am »
    hi
    someone invalidated your post while agreeing with another dude who basically said the same thing
    OHAI2U2

    congratulations hyroen for taking down a general
    welcome to da club

    Offline Boingo

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198546#msg198546
    « Reply #30 on: November 11, 2010, 03:00:18 am »
    So, it seems this thread has boiled down to an argument about probabilities instead of facts.  One deck wasn't "supposed to" beat another deck, yet magically did.  I guess that's why we actually go through the bother of actually playing the matches instead of just posting the decks and declaring a winner.  FWIW, I like it better this way.
    Bring back Holy Cow!

    QuantumT

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198553#msg198553
    « Reply #31 on: November 11, 2010, 03:11:36 am »
    Just so that we have actual numbers to put on things, I ran the odds. Hyroen's chances of getting then wings off turn 3:

    Going first (meaning access to 8 cards):44%
    Going first (meaning access to 9 cards):53%

    Zeru's chance of a turn 2 discord
    Going first:12%
    Going second:15%

    I could go through and calculate the probabilities further, but I think this shows that the odds really aren't that bad for Hyroen.

    Offline Sir Valimont

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198631#msg198631
    « Reply #32 on: November 11, 2010, 07:26:06 am »
    Just so that we have actual numbers to put on things, I ran the odds. Hyroen's chances of getting then wings off turn 3:

    Going first (meaning access to 8 cards):44%
    Going first (meaning access to 9 cards):53%

    Zeru's chance of a turn 2 discord
    Going first:12%
    Going second:15%

    I could go through and calculate the probabilities further, but I think this shows that the odds really aren't that bad for Hyroen.
    More accurately:

    Zeru's chance of at least 1 Discord in first 8 cards: 62.1%
    Zeru's chance of at least 3 Pendulums in first 8 cards: 54.8%
    Total probability: 34.0%

    So, 34% of games Zeru wins. Then, of the next 66% of games:

    Hyroen's chance of at least 1 Wings in first 8 cards: 85.1%
    Hyroen's chance of at least 2 Pendulums in first 8 cards: 74.3%
    Total probability: 63.2%

    63% of those 66%, or 41.5% of total games, Hyroen gets Wings immediately.

    Of the 25% of games where neither player gets their item out in the first 4 turns, Zeru has a Scorpion and Chaos Seed dealing damage around 90% of the time. Point being, Hyroen rarely wins; two hits from Scorpion (a low estimate) is usually around extra 14 damage besides the extra neurotoxin.

    Further, of the 41.5% of games where Wings comes out first, there are 38% of games where Hyroen does not get Fahrenheit in his first 6 turns. These are games he will lose. And there are 24.8% of those games where he will not get a second Wings within his first 6 turns either (those percentages overlap, they don't add of course). That brings the 41.5% winning to 41.5% x 53.5% = 22.2% winning.

    These are rough estimates but the probabilities are all accurately calculated using hypergeometric distribution tables. There are of course a minority of anomalous cases where one player might win not accounted for here but I think this is the majority of cases. I will post statistics of actual victories when I have them; likely I will test the decks against each other 30 times to get an accurate win%.

    QuantumT

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198633#msg198633
    « Reply #33 on: November 11, 2010, 07:31:19 am »
    Note:

    The percentages I gave are exact and not up for debate. I know combinatorics and I can setup the equations myself.

    You can't just find the odds for one and the odds for the other and multiply them because they depend on each other, ie when you have 3 pends in hands you don't get 8 cards to draw the discord.

    Offline Sir Valimont

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198635#msg198635
    « Reply #34 on: November 11, 2010, 07:32:38 am »
    Note:

    The percentages I gave are exact and not up for debate.
    It would be more useful if you showed your math since I disagree with your figures.


    QuantumT

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    Re: [Fire] Hyroen 2 -1 [Master of Entropy] Zeru https://elementscommunity.org/forum/index.php?topic=15485.msg198645#msg198645
    « Reply #35 on: November 11, 2010, 07:49:14 am »
    Sigh... ok I'll try to teach you combinatorics.

    In these formulas, ncr[N,n] means N!/(n!(N-n)!)

    Hyreon formula- odds of drawing at least 2 pends and 1 wings in 8 cards

    Sum[Sum[ncr[9,p]*ncr[6,w]*ncr[17,8-p-w]/ncr[32,8],{w,1,8-p}],{p,1,8}]~44%

    For 9 cards:

    Sum[Sum[ncr[9,p]*ncr[6,w]*ncr[17,9-p-w]/ncr[32,9],{w,1,9-p}],{p,1,9}]~53%

    Note that the mulligan is ignored as it adds well under a percent to the value.

    For Zeru

    In 7 cards:

    Sum[Sum[ncr[10,p]*ncr[3,d]*ncr[17,7-p-d]/ncr[30,7],{d,1,7-p}],{p,1,7}]~12%

    Sum[Sum[ncr[10,p]*ncr[3,d]*ncr[17,8-p-d]/ncr[30,8],{d,1,8-p}],{p,1,8}]~15%

    There I gave you the formula. Hyreon's chance of getting his out first is significantly higher.

     

    anything
    blarg: