View Full Version : Distribution of Rarities 1-5
In answer to ongoing debates about the rarity of Uniques, we seem to have decided that there is one Unique in every 6 boosters, on average. Each booster has 4 Heroclix, so 1 out of every 6 X4 = 24 Heroclix is a Unique.
My question is: Has anyone quantified the odds of getting Heroclix with rarity levels of 1 through 5? If 1 out of 24 is a Unique, is 2 out of 24 a Level 5? Are 8 out of 24 Level 1? Anyone?
webhead817
02/28/2003, 16:51
Not an answer, but part of the equation I think.
Someone please correct me if I'm wrong...
There are 6 figure rarities for individual figs in HC, 1-6, with 1 being the most common, and 6 being a unique rarity.
Besides uniques, characters are divided into three rarity "spans" for their versions.
The first span are characters whose rookies are lvl 1, exp. lvl 2, and vet lvl 3.
The second span rookies are lvl 2, exp lvl 3, vet lvl 4.
The third span rookies are lvl 3, exp lvl 4, vet lvl 5.
(At this point, the rarity looks like a bell curve, but wait!)
In each booster, there are 4 figures. 2 figures from the first "common" span, one from the second "uncommon" span, and one from the third "rare" span OR a unique.
If all of this is correct (which it may or may not be) then first span figs should be twice as common as second span figs, with third span figs being slightly less common since they share their slot with the occasional (1 out of every 6 or 8 or whatever) unique.
However, I don't know how that accounts for the differences between the lvl 1 figs in the first span and the lvl 3 figs in the first span.
Oh no, I think I've gone cross-eyed.
Anyone else have any ideas?
Originally posted by webhead817
In each booster, there are 4 figures. 2 figures from the first "common" span, one from the second "uncommon" span, and one from the third "rare" span OR a unique.
I had not heard, or observed, that this was true. Can anyone confirm or refute this?
webhead has it dead on. And yes, Uniques account for 1 in 6 of span 3
So:
Span 1 = 12/24 or 1/2
Span 2 = 6/24 or 1/4
Span 3 = 5/24+1/24 or 1/4
Thus:
123 REV = 1/2
234 REV = 1/4
345 REV = 5/24
U = 1/24
Add into this that more Vets are made than experienced of a given span, and more experienced than R, and we could extrapolate a good guess of the "real" statistical distribution.
And I would assume that a Unique is actually probably about equal in rarety in a booster to a 3/4/5 Vet.
Thanks Styrix!
But...suddenly, this is very depressing. This means that fully HALF of the X-Plosion Heoclix we all get will be generics, Typhoid Mary, Destiny, Boom-Boom, and Mystique. Oh well, I guess we just have to be comforted by the thought that the other half of the heroclix we get WON'T be generics, Typhoid Mary, Destiny, Boom-Boom, and Mystique!
I want tons of typhoid marys. I don't have a clue who she is but I want her all the same. Tons and Tons just cause her name is cool.
Ok, I worked it out, and this is what I'd like people to do with their Xplosion cases (I'm getting one, and know a lot of you others are too).
Count how many of each kind of fig you get.
C = 1/2/3 R/E/V
B = 2/3/4 R/E/V
A = 3/4/5/6 R/E/V/U
So this key would say that 8 E B means eight experienced figures of 2/3/4 rarety.
My definately incorrect guess at what a case should hold:
48 R C
32 E C
16 V C
24 R B
16 E B
8 V B
20 R A
12 E A
8 V A
8 U A
This is a guess, only something to work from. I'd appreciate stats on everyone's full case pulls to make this accurate. I stress "full case" because it tends to give a more acurate statistical representation. And if you use the above format, it'll help the data processing too.
I'm such a business geek. :D
By the way, for ease of reference, in Clobbering Time, I believe the following:
C comes in clear plastic
B Comes in Green Plastic
and A comes in White (R/E/V) or Blue plastic (U).
Hack-n-slash
02/28/2003, 18:12
From what I've seen, the breakdown *cannot* be:
1x 1/2/3
1x 2/3/4
1x 3/4/5/6
1x 1/2/3
... as I have seen the following drawn:
Elektra(U) = 6
Captain America(V) = 5
Annihilus(V) = 4
Some shmuck guy = 1/2/3
... from a single booster.
I've seen (elsewhere on HCRealms) the following breakdown:
1x 1/2/3
1x 2/3/4
1x 3/4/5 or 6
1x Random figure (1/2/3/4/5/6)
The booster *would* fit *that* pattern.
Your odds of getting two uniques or even two level 5 vets would still be rather small. Fully half the packs would still fit the 2x 1/2/3 structure, with more than that *appearing* to do so at first glance.
A 2/2/4/4 mix, for example, would have 1V, 2E, 1R.
quixotequest
02/28/2003, 19:21
First this is wholly conjecture...
What we know:
a) uniques are distributed 1:24 (1 per 6 boosters)
b) lesser rarity figures are produced at a factor greater than the previous level
For Infinity Challenge we had:
Rarity 1 -- 10 figs
Rarity 2 -- 22 figs
Rarity 3 -- 46 figs
Rarity 4 -- 36 figs
Rarity 5 -- 24 figs
Rarity 6 -- 12 figs
Now since we know lesser rarities are produced at a quantity greater than their next level we could represent the equation as follows:
10(6) + 22(5) + 46(4) + 36(3) + 24(2) + 12(1) = 510
12% + 22% + 36% + 21% + 9% + 2%
Now if these factors were correct one should only get one Unique every 12 boosters. They are packed at twice this rate: 1 every 6. (If exponential production was non-existant, however, we would get a Unique in every 3 boosters.) So we can confidently say that the exponential factors are not this extreme but they do exist.
Still the important point is we see a noticeable "bell curve," which doesn't significantly alter no matter which factor one chooses to insert in the parenthesis of the equation above.
So while the specific amounts of any rarity besides level-6 Uniques can only be guessed, we can apply some fair "bell curved" statistics. In which case, every six boosters should yield, on average, about:
Rarity 1 -- 3 figs
Rarity 2 -- 6 figs
Rarity 3 -- 7 figs
Rarity 4 -- 5 figs
Rarity 5 -- 2 figs
Rarity 6 -- 1 fig
==========
webhead817
02/28/2003, 22:02
hacknslash - I'm not sure about IC, but I think that HT and CT both followed the pattern I mentioned. Of course, there is always the chance that I'm wrong ;)
Jean_genie
03/01/2003, 01:13
Seems about right. I got a ton of CT, and I know for sure that the clear boxes were generics/'commons' (123), the greens were 'uncommons' (234), and the whites were 'rares'(345). I think every booster was packaged top down: clear, green, white (blue if it had a Unique), and then clear.
shyheim10304
03/01/2003, 19:03
this formula:
1x 1/2/3
1x 2/3/4
1x 3/4/5 or 6
1x Random figure (1/2/3/4/5/6)
....seem to be correct as i just bought 10 clobbin' time boosters to test it
amazingly enough i have got a ct pack with 2 (yes 2) uniques (nick fury and medusa) and have seem it twice before (although this was in the 15 cases bought by the group i play with and all 15 of these were ic)
shyheim10304
03/01/2003, 19:04
oh and the nick fury was in blue plastic and medusa was in green plastic
Great findings.
I ask for case stats because knowing that they're packed by hand, mistakes can happen. I figure the double Unique is a "mistake" more than "policy". Great mistake though!
Just one more month before the XPlosion test. If anyone has any CT case-stats, PM me, or post them here.
When I get my Xplosion case, I'll start a new thread to gather the stats.
WizKids, if you don't want me to "reverse-engineer" your case distribution (and further reverse marketing research would then of course follow) just let me know, and I of course will stop doing it on public forums. And Moderators, dito and some junk, y'know?
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