[Fire] Hyroen 2 -1 [Master of Entropy] Zeru

[TheonlyrealBeef]

*Looks at all these numbers and what not*
You just go with your guts and hope your deck wins.
Odds are irrelevant afterwards, outcomes are not, Entropy was defeated this match, and I'm glad to see serious competition take a beating (though we need to watch out for Fire )

[~Napalm]

I'm with beef on this one. Who cares what the odds were? RNG says Hyroen wins. Simple as that.

[QuantumT]

I'm with beef on this one. Who cares what the odds were? RNG says Hyroen wins. Simple as that.

I agree. I just figured if we were going to throw percentages around, they might as well be the right ones.

[xdude]

Seems like Master of Entropy got betrayed by luck.

[Sir Valimont]

Quantum, I can see that you're calculating every possible combination of cards to find the probability - I've seen nCr before and understand how it works. But your notation I find somewhat confusing; specifically the curly-braced parts of your formula.

Perhaps a way to look at this more simply:

Using hypergeometric distribution:

The chance of exactly 1 Discord in 8 cards: 45.5%
The chance of exactly 2 Discords in 8 cards: 15.2%

The chance of exactly 3 Pendulums in 7 cards: 29.8%
The chance of exactly 4 Pendulums in 7 cards: 13.0%
The chance of exactly 5 Pendulums in 7 cards: 2.8%

The chance of exactly 3 Pendulums in 6 cards: 24.5%
The chance of exactly 4 Pendulums in 6 cards: 7.6%
The chance of exactly 5 Pendulums in 6 cards: 1.0%

So estimates of totals:

If you assume 1 in the first 8 is a Discord (45.5%) then you have 7 chances for pends: 29.8% + 13.0% + 2.8% = 45.6%

So about 45.5 x 45.6 (underestimate) = 20.7 % chance of drawing the pends with 1 Discord;

If you assume 2 in the first 8 are Discords (15.2%) then: 24.5% + 7.6% + 1.0% = 33.1%

So about 15.2 x 33.1 = 5.0 % chance of drawing the pends with 2 Discords.

Call the total about 26% (underestimate - my guess is closer to 30) of the time you have success with second-turn Discord. Note that neither my calculations nor yours take into account that in fact one can draw the Discord the same turn one uses it (so it can be the 9th card in the deck, not in the first 8 ) and still be considered successful.

I presume there is a miscalculation somewhere in your formula but I haven't pored over it to try to figure out where ... it would take me a good while to discern exactly what you're doing. Still, given the straightforward logic of the above, there's clearly an inconsistency somewhere. Unless you have an explanation for why my logic here would be wrong?

[MrBlonde]

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%.

You seem to understand math pretty well. So you should know that 30 games is a very, very, small sample size and that it will not get you an accurate win %.

[QuantumT]

*snip*

What you did won't quite work. Why it won't work is rather hard to explain, but basically you just overcount some things. However, it should get you in the ballpark, and when I when back and ran the more likely successful hands, the number was distinctly higher than I gave before (ballparking it as somewhere in the 30s). The formulas I gave are correct, so I must have entered them wrong. Unfortunately, I'm not at a computer where running them again would be easy, so I'll have to get back to you on the exact numbers.

I did also forget about the possibility of drawing the discord on the turn you play it, so that would raise it a bit as well (same goes for wings).

My formulas give the complete chances of getting them, but for most purposes that's overkill (ie it includes hands like 7 pendulums and 1 discord). The way they're setup is to run over all the ranges for pendulums and discords that would be considered successful. The curly brackets are the {variable sum is over, starting value, ending value}.

Just to show one example *so the formulas are a bit easier to understand), let's look at the chances of having 8 cards with exactly 3 pendulums and 1 discord.

ncr[10,3] (number of ways to get exactly 3 pendulums) * ncr[3,1] (number of ways to get exactly 1 discord) * ncr[17,4] (number of ways to get remaining 4 cards that aren't discords or pendulums) / ncr[30,8] (divide by total number of hands)

=14.6% chance to have exactly 3 pendulums and 1 discord.

So you have to go through and calculate the probabilities for all other successful hands (4 pends 1 discord, 3 pends 2 discords, etc) and add them up.

PS- I'm not trying to argue, I just like trying to teach people who are interested new things.

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%.

You seem to understand math pretty well. So you should know that 30 games is a very, very, small sample size and that it will not get you an accurate win %.

It should be enough to get you in the ballpark though. Obviously it won't be exact, but it will most likely be within 10% or so of the right answer.

[~Napalm]

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.

I completely missed this the first time running through that gibberish. Well said

Looking at these numbers I would have to say that probability AND luck were on Hyroens side. But w/e. I'm finished with this. See what happens when you update the results? You get dragged into what's happening on the more interesting threads. Now if you'll excuse me... I need some sleep.

[LongDono]

Hmmmm this seemed like an interesting game.
Hyroen seemed to have a number of outs to dune scorp, but chaos power can change that, it all depends.
Honstly there is too much luck between the 2 decks to honestly say who should have won. Some skill would be needed as well I would think in certain situations that these 2 decks might have against eachother.
I love entropy but they lost today, end of story.

[Kuross]

Maybe it's just me, but am I the only one that sees something and it'd like to point it out.


Option one: Sir Valimont congratulates Hyroen for the win in a good sportsmanlike spirit like Zeru did.

Option two: Sir Valimont finds way to showcase why Zeru lost so as to lessen Hyroen's victory?

Yes, chance was a factor. Yes, both decks were built well. The games seemed challenging and both players played well. In the end, the matches were completed and a winner was determined.

So I ask, why can't Sir valimont simply say "good game" and move on? By dragging this out, Zeru gets indirectly dragged into a debate of statistics when I'm sure he wants to look at moving forward and bringing team Entropy to victory rather than have his match nitpicked over due to statistical anomalies.

I'm sorry if this seems harsh, I'm just at a loss as to why this needs to be gone over and tested to show statistics to explain why Hyroen should of lost when that is not the case and any energy put into that end is a complete waste of time. Round over, games played well and both contestants (Hyroen and Zeru) are mature enough to have left the results as they are and look to the next matches. I get a little smack talking between teams happens, but wasting time to test out WAR decks to show why one should have lost to another doesn't win the next round. Nor is the way you're portraying yourself really helping, SV. Most people have issue with your implications, not because of your arguments but because of the way you state them. It's much more difficult to convince someone of something if you do so in a condescending way. Just my constructive criticism.

[Sir Valimont]

Heh, the trouble with threads is if you aren't following them in real-time you don't really see the progress of the conversation.

I did already congratulate Hyroen, who played brilliantly, and I did already say I would post stats when they are complete. I don't think the tenor of the conversation was too negative either; Quantum ran some stats using combinatorics and we had a brief discussion on them in which we found each of our estimations to be closer to the other's. I will run the tests and post the information for everyone.

[Sir Valimont]

@ QuantumT:

Really interesting. Essentially you are figuring out the odds by calculating how many possible card combinations exist and dividing the number of successful ones into that number. I figured the curly braces were iterations of your other math but I'm a bit old school and I see things a little more clearly with Sigmas (actually when I copied it down the first time I left out the "Sums" at the beginning and that made it a bit more confusing).

It will be interesting to see what results happen in testing - one of the x-factors is going to be those games where anomalous outcomes happen, like if Wings comes out first but then no other Wings are drawn, etc. To the testing chamber!