Crypt Probabilities. Again. 'Cause I'm bad at math.

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Hey--math people!

Say I put together a crypt of two groups each of 6 vampires (say, 6 with POT
and 6 with pot)--we will call them 6 of group A and 6 of group B.

You'd think that most of the time, you would draw two of each group (i.e. 2
from group A and 2 from group B). It seems, in practice, however, that most
of the time, you actually draw three of one group and one of another (i.e. 3
from group A and 1 from B or 1 from A and 3 from B).

Am I imagining this, or does probabilitiy support this concept?

Thanks,

Peter D Bakija
pdb6@lightlink.com
http://www.lightlink.com/pdb6

"So in conclusion, our business plan is to sell hot,
easily spilled liquids to naked people."
-Brittni Meil

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

Peter D Bakija wrote:
> Hey--math people!
>
> Say I put together a crypt of two groups each of 6 vampires (say, 6 with POT
> and 6 with pot)--we will call them 6 of group A and 6 of group B.
>
> You'd think that most of the time, you would draw two of each group (i.e. 2
> from group A and 2 from group B). It seems, in practice, however, that most
> of the time, you actually draw three of one group and one of another (i.e. 3
> from group A and 1 from B or 1 from A and 3 from B).
>
> Am I imagining this, or does probabilitiy support this concept?

I ain't a big math person but yes, over enough trials you should get
50% from group A and 50% from group B. But that's a lot of trials!
What you're doing is flipping a coin a few times and wigging out
because it came up tails a few times in a row instead of coming up
heads once and then tails once and then heads once.

Then there's the pyschological stuff about only remembering the bad
trials. Nobody remembers the normal draws--only the odd trials stick in
your head.

-Robert

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

"Peter D Bakija" <pdb6@lightlink.com> wrote in message news:BF56108D.22132%pdb6@lightlink.com...
> Hey--math people!
>
> Say I put together a crypt of two groups each of 6 vampires (say, 6 with POT
> and 6 with pot)--we will call them 6 of group A and 6 of group B.
>
> You'd think that most of the time, you would draw two of each group (i.e. 2
> from group A and 2 from group B). It seems, in practice, however, that most
> of the time, you actually draw three of one group and one of another (i.e. 3
> from group A and 1 from B or 1 from A and 3 from B).
>
> Am I imagining this, or does probabilitiy support this concept?

***Raises hand!***

I've just built a couple decks with 50/50s like that. Strangely enough, I was
surprised by the exact same thing. I have one 50/50 deck that I swear I've
played at least a dozen times, at least TWICE drawn four of one of the vampires
and none of the others, and have yet to draw two and two.

Weird!

Fred

 

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Robert Goudie wrote:

> I ain't a big math person but yes, over enough trials you should get
> 50% from group A and 50% from group B. But that's a lot of trials!
> What you're doing is flipping a coin a few times and wigging out
> because it came up tails a few times in a row instead of coming up
> heads once and then tails once and then heads once.

Heh. Mostly I was wondering if my understanding of probability was off, as
that is a perfectly reasonable assumption on my part

Like, if someone came on and posted a nice equation and was like "Yeah, see,
due to the combinatorics or whatever, you actually have a 35% chance of
getting 3A+1B; a 35% chance of 1A+3B; and a 30% chance of 2A+2B." Or
something.

I mean, now if someone posts that, I'll assume they are jerking me around



Peter D Bakija
pdb6@lightlink.com
http://www.lightlink.com/pdb6

"So in conclusion, our business plan is to sell hot,
easily spilled liquids to naked people."
-Brittni Meil

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

On Tue, 20 Sep 2005, Peter D Bakija wrote:

> Like, if someone came on and posted a nice equation and was like "Yeah, see,
> due to the combinatorics or whatever, you actually have a 35% chance of
> getting 3A+1B; a 35% chance of 1A+3B; and a 30% chance of 2A+2B." Or
> something.

It's more like the opposite. Basically, the odds of you getting
2A+2B are greater than getting 3A+1B, and greater than getting 1A+3B, but
less then the odds of getting *either* 3A+1B *or* 1A+3B, by a bit. I
believe the relevant numbers are:

2A+2B = 15/33
1A+3B + 3A+1B = 16/33
4A + 4B = 2/33

Marc G.

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

On Tue, 20 Sep 2005, Peter D Bakija wrote:

> Hey--math people!
>
> Say I put together a crypt of two groups each of 6 vampires (say, 6 with POT
> and 6 with pot)--we will call them 6 of group A and 6 of group B.
>
> You'd think that most of the time, you would draw two of each group (i.e. 2
> from group A and 2 from group B). It seems, in practice, however, that most
> of the time, you actually draw three of one group and one of another (i.e. 3
> from group A and 1 from B or 1 from A and 3 from B).
>
> Am I imagining this, or does probabilitiy support this concept?
>

All 4 are A/B: 1/16 x2
3 are A/B and 1 is B/A: 4/16 x2
2 are A and 2 are B: 6/16

It is more likely that you have 2 As than 3 As.

It is more likely that you have 3/1 than 2/2.

You have a 50/50 chance of getting 3/1.

 

[pat]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

"Peter D Bakija" <pdb6@lightlink.com> wrote in message
news:BF56108D.22132%pdb6@lightlink.com...
> Hey--math people!
>
> Say I put together a crypt of two groups each of 6 vampires (say, 6 with
> POT
> and 6 with pot)--we will call them 6 of group A and 6 of group B.
>
> You'd think that most of the time, you would draw two of each group (i.e.
> 2
> from group A and 2 from group B). It seems, in practice, however, that
> most
> of the time, you actually draw three of one group and one of another (i.e.
> 3
> from group A and 1 from B or 1 from A and 3 from B).
>
> Am I imagining this, or does probabilitiy support this concept?
>
> Thanks,
>

I think I have this right, and if I do, you're not imagining things, based
on the way you're asking the question.

Call your two crypt groups A1, A2, ... , A6 and B1, B2, ..., B6.

Let C(n,k) be the number of ways to choose k things from among n things,
where order doesn't matter.

C(n,k) = n! / (k! * (n-k)!)

For example, you have C(12,4) ways of drawing your opening crypt, since
you're picking 4 vampires out of a possible 12.

■ C(12,4) = 12! / (4! * 8!) = 479,001,600 / (24 * 40, 320) = 495

Now, we turn to the subgroups.

Since we care only about A vs. B, and not A1 vs. A2, there are only 5
possible opening draws:

I. 0 A, 4 B
II. 1 A, 3 B
III. 2 A, 2 B
IV. 3 A, 1 B
V. 4 A, 0 B

The number of ways to do I is C(6,0) * C(6,4), because you're picking 0
things out of one group of six, and 4 things out of another group of 6, and
the selections from each subgroup are independent from the other subgroup.

C(6,0) = 1
C(6,4) = 15

So, the result of option I is 15.

Similarly, the rest of the options are counted as follows:

II. C(6,1) * C(6,3) = 6 * 20 = 120
III. C(6,2) * C(,2) = 15 * 15 = 225
IV. C(6,3) * C(6,1) = 20 * 6 = 120
V. C(6,4) * C(6,0) = 15 * 1 = 15

As a check, we add I. through V.: 15 + 120 + 225 + 120 + 15 = 495, which
equals the answer we derived in ■.

And so, now we come back to your original question: you're essentially
asking whether II. + IV. is more than III., and the answer is yes, since 120
+ 120 is slightly greater than 225.

But 2 pot and 2 POT is still the single most likely outcome, as you would
expect.

The more math-empowered can correct me if I'm wrong on any of this.

- Pat

 

[pat]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

"Marc Gabriele" <gootmu@dementia.org> wrote in message
newsine.LNX.4.62.0509201955060.24138@lily.dementia.org...
> On Tue, 20 Sep 2005, Peter D Bakija wrote:
>
>> Like, if someone came on and posted a nice equation and was like "Yeah,
>> see,
>> due to the combinatorics or whatever, you actually have a 35% chance of
>> getting 3A+1B; a 35% chance of 1A+3B; and a 30% chance of 2A+2B." Or
>> something.
>
> It's more like the opposite. Basically, the odds of you getting 2A+2B are
> greater than getting 3A+1B, and greater than getting 1A+3B, but less then
> the odds of getting *either* 3A+1B *or* 1A+3B, by a bit. I believe the
> relevant numbers are:
>
> 2A+2B = 15/33
> 1A+3B + 3A+1B = 16/33
> 4A + 4B = 2/33
>
> Marc G.

This agrees with my more pedantic calculations, elsewhere in the thread.

- Pat

 

[pat]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

"Gregory Stuart Pettigrew" <etherial@sidehack.gweep.net> wrote in message
news:20050920195608.D12568@sidehack.gweep.net...
> On Tue, 20 Sep 2005, Peter D Bakija wrote:
>
>> Hey--math people!
>>
>> Say I put together a crypt of two groups each of 6 vampires (say, 6 with
>> POT
>> and 6 with pot)--we will call them 6 of group A and 6 of group B.
>>
>> You'd think that most of the time, you would draw two of each group (i.e.
>> 2
>> from group A and 2 from group B). It seems, in practice, however, that
>> most
>> of the time, you actually draw three of one group and one of another
>> (i.e. 3
>> from group A and 1 from B or 1 from A and 3 from B).
>>
>> Am I imagining this, or does probabilitiy support this concept?
>>
>
> All 4 are A/B: 1/16 x2
> 3 are A/B and 1 is B/A: 4/16 x2
> 2 are A and 2 are B: 6/16
>

You're a bit off with the exact probabilities, I think. The denominator
should be 33, not 32, so I think maybe you missed one possibility when
counting.

> It is more likely that you have 2 As than 3 As.
>

True.

> It is more likely that you have 3/1 than 2/2.
>

True, if you consider 3/1 to be equivalent to 1/3, as Peter did in
formulating his question.

> You have a 50/50 chance of getting 3/1.

Again, not quite.

- Pat

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

Howdy Peter,

> So looking at all the math floating around (and thank you all for the cool
> math--I always like math, I'm just not real good at it), it looks like:
>
> -The most likely single outcome of 5 specific outcomes (0A/4B; 1A/3B; 2A/2B;
> 3A/1B; 4A/0B) is 2A/2B.
>
> -This being said, you are more likely by some amount to get one of the other
> 4 outcomes than you are the most likely single outcome (from what I can
> tell, it is something like a 45% chance of getting 2A/2B and a 55% chance of
> getting something other than 2A/2B).
>
> Interesting.

The relevant line from the crypt draw tables I once posted:

6: 3.03, 24.24, 45.45...

Where the '6' means you have six copies of the relevant vampire in your
12 vampire crypt (from which you are drawing 4 vampires). The 3.03(%)
is the chance of drawing exactly 0 copies, the 24.24(%) is the chance
of drawing exactly 1 copy, and the 45.45(%) is the chance of drawing
exactly 2 copies.

If you consider the remaining six vampires to all be the same, then you
have the case you asked about, and you see that:

pr(0A + 4B) = pr(4A + 0B) = 3.03% (=1/33),
pr(1A + 3B) = pr(3A + 1B) = 24.24% (= 8/33), and
pr(2A + 2B) = 45.45% (=15/33)

So, yes, your observation is correct. 18 times out of 33 (54.54% of
the time) you will not get 2 and 2.

Hope that helps,
Alex

 

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Frederick Scott wrote:

> I've just built a couple decks with 50/50s like that. Strangely enough, I was
> surprised by the exact same thing. I have one 50/50 deck that I swear I've
> played at least a dozen times, at least TWICE drawn four of one of the
> vampires
> and none of the others, and have yet to draw two and two.

So looking at all the math floating around (and thank you all for the cool
math--I always like math, I'm just not real good at it), it looks like:

-The most likely single outcome of 5 specific outcomes (0A/4B; 1A/3B; 2A/2B;
3A/1B; 4A/0B) is 2A/2B.

-This being said, you are more likely by some amount to get one of the other
4 outcomes than you are the most likely single outcome (from what I can
tell, it is something like a 45% chance of getting 2A/2B and a 55% chance of
getting something other than 2A/2B).

Interesting.


Peter D Bakija
pdb6@lightlink.com
http://www.lightlink.com/pdb6

"So in conclusion, our business plan is to sell hot,
easily spilled liquids to naked people."
-Brittni Meil

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

>> All 4 are A/B: 1/16 x2
>> 3 are A/B and 1 is B/A: 4/16 x2
>> 2 are A and 2 are B: 6/16
>>
>
> You're a bit off with the exact probabilities, I think. The denominator
> should be 33, not 32, so I think maybe you missed one possibility when
> counting.
>

33 is a rather improbable denominator.

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

Howdy,

> The relevant line from the crypt draw tables I once posted:
>
> 6: 3.03, 24.24, 45.45...

Sorry, should have mentioned how to find those tables. In Google
Groups, search r.g.t-c.jyhad using: 'author:wumpus devil'.

Alex

 

[Guest]

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On Tue, 20 Sep 2005, Gregory Stuart Pettigrew wrote:

>>> All 4 are A/B: 1/16 x2
>>> 3 are A/B and 1 is B/A: 4/16 x2
>>> 2 are A and 2 are B: 6/16
>>>
>>
>> You're a bit off with the exact probabilities, I think. The denominator
>> should be 33, not 32, so I think maybe you missed one possibility when
>> counting.
>>
>
> 33 is a rather improbable denominator.
>

Right, I was oversimplifying. I was just turning A/B into coin flips, but
they're not independent.

4A: 1/33 or 3%
3A 1B: 8/33 or 24%
2A 2B: 15/33 or 45%
3B 1A: 8/33 or 24%
4B: 1/33 or 3%

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

Peter D Bakija wrote:
> Hey--math people!
>
> Say I put together a crypt of two groups each of 6 vampires (say, 6 with POT
> and 6 with pot)--we will call them 6 of group A and 6 of group B.
>
> You'd think that most of the time, you would draw two of each group (i.e. 2
> from group A and 2 from group B). It seems, in practice, however, that most
> of the time, you actually draw three of one group and one of another (i.e. 3
> from group A and 1 from B or 1 from A and 3 from B).
>
> Am I imagining this, or does probabilitiy support this concept?
>
> Thanks,
>
> Peter D Bakija
> pdb6@lightlink.com
> http://www.lightlink.com/pdb6
>
> "So in conclusion, our business plan is to sell hot,
> easily spilled liquids to naked people."
> -Brittni Meil

Yeah, but if I have pocket kings in a full table of Texas Hold 'em,
what're the odds someone at the table has pocket aces?

Jeff

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

> Yeah, but if I have pocket kings in a full table of Texas Hold 'em,
> what're the odds someone at the table has pocket aces?
>

Pretty high. And if I have pocket aces, the chances of someone having a
higher hand are approximately 1.

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

This is correct.
And if you would go with more than 6 of each, the % of times you don't
get 2 + 2 would decrease.
For example 7 of each gives a 53.85% chance, while 8 of each reduces
the chance of getting an unwanted combination to 53.33 %
....I think..

 

[Guest]

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Gregory Stuart Pettigrew wrote:

> Yeah. As Crypt Size approaches infinity, the probabilities approach
>
> 4A 1/6
> 3A 1B 4/16
> 2A 2B 6/18
> 1A 3B 4/16
> 4B 1/16

Doesn't that mean that using more vampires makes it *more* likely to
get an unwanted draw?

That math looked like with a crypt of 12 vampires (6A/6B), you had a
~45% chance of drawing (2A/2B) and a ~55% chance of drawing something
other than (2A/2B).

Looking at the numbers that Greg posted, it looks like as you approach
infinity, you end up with an ~33% chance of drawing (2A/2B) and an ~66%
of drawing not (2A/2B).

-Peter

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

p...@lightlink.com wrote:
> Gregory Stuart Pettigrew wrote:
>
> > Yeah. As Crypt Size approaches infinity, the probabilities approach
> >
> > 4A 1/6
> > 3A 1B 4/16
> > 2A 2B 6/18
> > 1A 3B 4/16
> > 4B 1/16
>
> Doesn't that mean that using more vampires makes it *more* likely to
> get an unwanted draw?
>
> That math looked like with a crypt of 12 vampires (6A/6B), you had a
> ~45% chance of drawing (2A/2B) and a ~55% chance of drawing something
> other than (2A/2B).
>
> Looking at the numbers that Greg posted, it looks like as you approach
> infinity, you end up with an ~33% chance of drawing (2A/2B) and an ~66%
> of drawing not (2A/2B).
>
> -Peter

Damn Peter, you beat me to the post I was going to post. Well, since I
also did the work of checking the numbers, I'm going to post anyway.

When I first read the post, I thought, whoa, a larger crypt is
better??? That doesn't seem right. Then I read Greg's post and the math
didn't seem to support it, so I decided to work it out myself. At first
I tried to be fancy and use probabilities, but then I realized even
when I was in college, I wasn't all that good at doing probabilities
correctly, so I did it with brute force.

Greg typed a little too fast and meant to list the probability with an
infinite sized crypt as:

4A 0B 1/16
3A 1B 4/16
2A 2B 6/16
1A 3B 4/16
0A 4B 1/16

Here are my brute force results, proving the above probabilities

AAAA BAAA
AAAB BAAB
AABA BABA
AABB BABB
ABAA BBAA
ABAB BBAB
ABBA BBBA
ABBB BBBB

With an infinite crypt you have an equal chance of A or B for each
position, so brute force is easy/quick to do with a 4 card opening
crypt.

Thus, the chance of getting what you want (2 of each) is 37.5% (6/16)
and getting something else is 62.5% (10/16), which is worse than the
~45% / ~55% split in a 12 card crypt.

So a smaller crypt is better.

Now wasn't that fun?

Later,
~Rehlow

 

[Guest]

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On Wed, 28 Sep 2005, ruffe wrote:

> This is correct.
> And if you would go with more than 6 of each, the % of times you don't
> get 2 + 2 would decrease.
> For example 7 of each gives a 53.85% chance, while 8 of each reduces
> the chance of getting an unwanted combination to 53.33 %
> ...I think..
>
>

Yeah. As Crypt Size approaches infinity, the probabilities approach

4A 1/6
3A 1B 4/16
2A 2B 6/18
1A 3B 4/16
4B 1/16

 

[Guest]

Archived from groups: rec.games.trading-cards.jyhad (More info?)

Howdy,

<snip elegant computation of probability>

> With an infinite crypt you have an equal chance of A or B for each
> position, so brute force is easy/quick to do with a 4 card opening
> crypt.
>
> Thus, the chance of getting what you want (2 of each) is 37.5% (6/16)
> and getting something else is 62.5% (10/16), which is worse than the
> ~45% / ~55% split in a 12 card crypt.

For those who are interested in non-infinite, non-12-card cases, the
probability of drawing exactly 2A and 2B for an arbitrary (even) size
crypt with 50% A and 50% B is:

3 n (n-1)
pr(2A+2B) = f = -----------------
2 (2n-1) (2n-3)

Where n is the number of copies of either vampire A or B (and thus the
total crypt size is 2n).

The limit of the probability as n goes to infinity is 3/8, which
matches both Gregory's corrected values and Rehlow's. (The latter's
method of computing is much more efficient than mine, though it doesn't
help you for non-infinite cases.) Also note that f(2) = 100%; smaller
crypt sizes help.

For those who are interested in making such computations themselves,
grep 'distinguishability' in Google Groups (in r.g.t-c.jyhad, of
course), and you'll find the ancient post in which I discuss the
derivation of the formulae.

Hope that helps,
Alex