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Star Wars: The Card Game» Forums » General

Subject: Mathematics behind drawing objectives rss

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David C.
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Okay, I give up...

Although I studied Mathematics for teaching in Elementary school and I had some courses concerning probabilities (like 6 to 7 years ago), I cant wrap my brain around the draw of a specific set of objectives.

Since I never had to use those formulars after the University, I can't remember all the necessary stuff. I'm sure some of you can help me to figure it out.


-----
Questions:
I will use colors instead of objective set numbers, so its easier to identify how the group is put together.

(1)
My objectives contain: 7 blue, 3 red

Its obvious that you will draw one of the blue objectives no matter what, but:
Questions:
1) Whats the probability to draw all 3 red objectives?
2) Whats the probability to draw only blue objectives?
3) What about the possible combinations of objective sets?

(2)
My objectives contain: 3 blue, 3 red, 4 green.

1) Whats the probability to draw at least one of each?
2) Whats the probability to draw only 2 colors?
3) Whats the probability to draw only 4 greens?

-----
Important:
I would really, really like to understand the mathematics and logic behind it, so please dont just hand me the percentages to my questions as they are examples for possible combinations.
I would rather like you to try to explain me the necessary formulars (or at least tell me which ones to use), so I can try to figure the % out on my own in the future.

Thanks in advance,

David

Edits
*edit* I'm sure there are sites out there for Deckbuilding Games which explain the Mathematics behind deckbuilding. If you know such an adress, please let me know. Maybe that would be already help enough.

*edit2* In this thread there is one guy at least explaining some of the stuff he calculates. But as I dont know what theory he uses behind it, I cant translate it to other combinations.

----

Okay... some ideas to get this (and my brain) started.

(1)
If I have 6 blue / 4 red objectives.
Whats the probability to draw only red objectives?

Its: ((4/10)*(3/9)*(2/8)*(1/7)) = 0,47%

(2)
If I have 6 blue / 4 red objectives.
Whats the probability to draw at least 1 blue objective on the start?

Its the probability of 1 minus the probability of getting only red objectives.
So: 1 - ((4/10)*(3/9)*(2/8)*(1/7)) = 99,5%

Is this correct?
 
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Jarrett McBride
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You know how I look at it? I just look at 7:3 as the limit just because by running 6:4 there's a CHANCE that you could start with all four of the wrong affiliation. Just the fact that you have that chance, abet a small one, is enough reason not to run it that way. Who wants to lose a game per tournament because you can't resource match your cards? Better safe than sorry in my eyes. Only way I would run a deck without that ratio is if there was a deck that could completely dominate the metagame and win a huge majority of its game, and in my eyes, there is no deck that can do that.
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PenumbraPenguin
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Torala wrote:

(1)
My objectives contain: 7 blue, 3 red

Its obvious that you will draw one of the blue objectives no matter what, but:
Questions:
1) Whats the probability to draw all 3 red objectives?
2) Whats the probability to draw only blue objectives?
3) What about the possible combinations of objective sets?


1) How many sets of four objectives contain all three red ones? 7.
How many sets of four are there? 10C4 = 10*9*8*7 / 4*3*2*1 = 210.

The chance of getting all three reds is 7/210 = 1/30.

2) How many sets of four objectives are all blue? 7C4 = 35. How many sets are there in total? 10C4 = 210. The probability is 35 / 210 = 1/6.

3) Not sure what you mean here. Is it that there are 10C4 = 210 possible sets of 4, assuming they're all different?


Torala wrote:

(2)
My objectives contain: 3 blue, 3 red, 4 green.

1) Whats the probability to draw at least one of each?
2) Whats the probability to draw only 2 colors?
3) Whats the probability to draw only 4 greens?


1) How many sets contain at least one of each? Let's break it up:

BBRG: 3C2 * 3C1 * 4C1 = 3*3*4 = 36 (the factors are the ways of choosing the blues, reds, greens independently)
BRRG: 3C1 * 3C2 * 4C1 = 3*3*4 = 36.
BRGG: 3C1 * 3C1 * 4C2 = 3*3*6 = 54.
Total: 126.

Total number of sets of 4 is still 10C4 = 210, so probability is 126/210 = 3/5.

2) It's easier to calculate this by calculating the chance that it doesn't happen. If you don't get 2 colours, then you get either one or three. The chance of getting all three is 3/5, from above. The chance of getting only one colour is only 1/210 (the only way of doing this is to get all four greens).

Therefore the chance of getting exactly two colours is 1 - 3/5 - 1/210 = 83/210, which is about 2/5.

3) 1/210, as in the previous part.
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David C.
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Well.. I run mostly 8-2 or 7-3 decks, but still I want to know the math behind it.

Also: not everyone wants to play his/her decks in tournament only. Granted, propably those guys wouldnt be interested in the math then either.
 
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Cameron McKenzie
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Personally, I am comfortable with running as few as four objectives from the same affiliation, without the corresponding affiliation card. This gives a 1/18 chance that you will NOT draw one of those objectives, which is an acceptable risk in my opinion. You are more likely to lose for many other reasons anyway.

The math only takes a simple understanding of combinatorics. You basically need to compute two values, and take the ratio.
How many total ways are there to draw 4 objectives?
How many ways are there to draw 4 objectives which meet some criteria?

These are solved using combinations, which are expressed as "n choose k". For the first question, this would be "10 choose 4" if you run the minimum number of objectives, which equals 210.

The second question depends on the criteria, but lets take my scenario as an example. If I am running 6 navy 4 sith, how many ways can I pull 4 cards without getting any of the Sith? Rephrasing, how many ways can I pull 4 cards, from only the 6 navy cards available? This is "6 choose 4" which gives 18.

So in this case, there are 15 ways I can draw only Sith cards, out of 210 total ways to draw cards (all of which are equally likely), so the chance is 15/210, which reduces to 1/18.
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Zach S
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First of all, this is a great thread (I've often wondered about the math behind objective-drawing myself), and thanks to everyone who's provided answers so far.

That being said, I have a follow-up question:

If I'm understanding the "n choose k" thing correctly, to find the number of red (Sith) objectives in the first scenario, we'd do 3C4, right? But . . . how does that equation turn out?

I get that 10C4 is 10*9*8*7/4*3*2*1 (though I admit I'm not necessarily clear on why we calculate the denominator that way, but I'm willing to accept it), but I guess I'm just not sure what to do with the numerator when we're doing 3C4. I mean, it obviously can't be 3*2*1*1(?), but I really don't know what else to try.

I suppose it's easy enough to just figure out that there are 7 different combinations of getting 3 red objectives in that draw (you just assume you have all 3 red cards, then realize there are seven other cards that could fill in that last slot), but I'd like to know the actual math behind the equation instead of just using logic, because logic won't always help with bigger numbers. =P

Thanks!
 
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Matt Lernout
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Here's an excellent mathy article that explains how to determine the chances of drawing a card from a deck. It's a Magic: the Gathering article, but the same principle applies to any set where you want to determine the probability of drawing X copies of something within Y draws from a deck of Z size.
http://www.kibble.net/magic/magic10.php

Edit: The chance of not drawing a Sith within your intial setup of four draw with four matching objectives in an objective deck of ten is 7.1429%, which is actually 1/14.

Not terribly bad, but when you consider adding in just one more copy for the affiliation, that the odds are more than three times better, it really makes a strong case for a five card minimum, at least in my opinion. (2.381% percent chance of not starting a matching affiliation, or 1/42 draws)

At six, you'll only be missing it once in 210 shuffles. And well, obviously at seven, it's impossible not to draw one. But 210 is such a large margin for success that in practicality you can count on six being a definite match.

Full odds of not pulling a resource matching objective (based on available objectives, assumed deck size of 10)

1 - 60% failure
2 - 1/3
3 - 1/6
4 - 1/14
5 - 1/42
6 - 1/210
7 - never
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Cameron McKenzie
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Alazzar wrote:
First of all, this is a great thread (I've often wondered about the math behind objective-drawing myself), and thanks to everyone who's provided answers so far.

That being said, I have a follow-up question:

If I'm understanding the "n choose k" thing correctly, to find the number of red (Sith) objectives in the first scenario, we'd do 3C4, right? But . . . how does that equation turn out?

I get that 10C4 is 10*9*8*7/4*3*2*1 (though I admit I'm not necessarily clear on why we calculate the denominator that way, but I'm willing to accept it), but I guess I'm just not sure what to do with the numerator when we're doing 3C4. I mean, it obviously can't be 3*2*1*1(?), but I really don't know what else to try.

I suppose it's easy enough to just figure out that there are 7 different combinations of getting 3 red objectives in that draw (you just assume you have all 3 red cards, then realize there are seven other cards that could fill in that last slot), but I'd like to know the actual math behind the equation instead of just using logic, because logic won't always help with bigger numbers. =P

Thanks!


There are 10 possibilities for the first card, 9 for the second card, 8 for the third, and 7 for the fourth... That gives 10 * 9 * 8 * 7 permutations. Say permutation because in this case, I'm counting different or orde rings as distinct elements. Pulling red then blue is counted differently than pulling blue then red. We want to eliminate those rearrangements,by dividing e total.
No matter how I pull the four cards, I can arrange them a number of ways as follows: 4 choices for the first, 3 for the second, 2 for the third, 1 for the last. So, out of all my permutations, I'm actually getting some number of combinations, each of which is repeated 4 * 3 * 2 times. I don't want to count these separately, so I will just divide my original count by this number.

3C4 is not a valid expression. The question here is, how many ways can I draw a hand that has all 3 reds? Three of my cards are red, drawn from a pool of 3 reds. That's 3C3 which equals 1. One of my cards is non-red, drawn from a pool of 7 non-reds. That's 7C1, which is 7. I multiply these together and I get a total of 1 * 7 possible hands that have exactly three reds and one non-red. These 7 hands are equally as likely as any other hand from the 210 possible, so it's 7/210 chance of happening.
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Cameron McKenzie
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BakaMattSu wrote:


Full odds of not pulling a resource matching objective (based on available objectives, assumed deck size of 10)

1 - 60% failure
2 - 1/3
3 - 1/6
4 - 1/14
5 - 1/42
6 - 1/210
7 - never


Oops, I have no idea how I got 1/18 earlier. 18 does not even divide into 210!
 
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