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Walking Dead #1 Black Label vs White Label - an Answer! UPDATED 2/4/14
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304 posts in this topic

Hello!

 

It took me longer than I thought, but I think I have enough data and re-educated myself enough on statistics to come up with a reasonable answer to the following question:

 

How rare is black label Walking Dead #1?

 

Here is the information used to determine the answer:

 

Data:

white - 112

black - 20

 

 

id - grade - color

1266002 - 9.8 - white

5884001 - 9.8 - white

79182012 - 9.8 - white

113915004 - 9.8 - white

129296015 - 9.8 - white

140573002 - 9.8 - white

144970011 - 9.8 - black

146321020 - 9.8 - black

154026025 - 9.8 - white

154946001 - 9.6 - white

165995005 - 9.8 - white

167232001 - 9.8 - white

178638010 - 9.8 - white

182175002 - 9.6 - white

186798001 - 9.6 - white

191516002 - 9.6 - white

191821001 - 9.6 - black

191970001 - 9.6 - white

192488001 - 9.6 - white

193861001 - 9.4 - white

194079001 - 9.8 - white

195496001 - 9.4 - white

197225001 - 9.6 - white

197449001 - 9.6 - white

197644001 - 9.4 - white

197759001 - 9.6 - black

197855002 - 9.4 - black

198629002 - 9.4 - white

198760005 - 9.4 - black

198872001 - 9.8 - white

198923002 - 9.8 - white

199787001 - 9.4 - white

200322007 - 9.6 - white

202176002 - 9.4 - white

202875001 - 9.6 - white

203420002 - 9.8 - white

204153001 - 9.6 - white

204501001 - 9.8 - black

204895001 - 9 - white

205393001 - 9.4 - white

205791001 - 9.6 - white

206059009 - 9 - black

206222001 - 9.4 - white

206563012 - 9.6 - white

206563013 - 9.8 - white

206563015 - 9.8 - white

206563016 - 9.9 - white

207421001 - 9.8 - white

207506001 - 9.4 - white

208527001 - 9.4 - white

209169001 - 9.8 - white

209413001 - 9.6 - white

209503001 - 9.6 - white

209755001 - 8.5 - white

209755002 - 9.6 - white

209786001 - 9.4 - white

210256001 - 9.8 - white

210616001 - 9.6 - white

210873001 - 9.8 - white

211206001 - 9.6 - white

212599001 - 9.6 - white

212759001 - 9.8 - white

628194007 - 9.6 - white

709182012 - 9.8 - white

723222007 - 9.6 - white

797804003 - 9.6 - black

808607002 - 9.8 - white

905937002 - 9.8 - white

938958010 - 9.8 - black

952169001 - 9.8 - white

956195018 - 9.8 - white

957831001 - 9.6 - white

966396001 - 9.9 - black

968161006 - 9.8 - black

975877003 - 9.8 - white

975877004 - 9.6 - white

985729009 - 9.4 - white

991934004 - 9.4 - white

994664002 - 9.9 - white

1006423009 - 9 - white

1011229007 - 9.4 - white

1019293020 - 9.2 - white

1020211001 - 9.9 - white

1024487001 - 8.5 - white

1027451008 - 9.6 - black

1054779004 - 9.8 - white

1055026002 - 9.8 - white

1055107007 - 9.8 - white

1056971002 - 9.8 - white

1072637001 - 9.8 - white

1074787001 - 9.2 - white

1076175004 - 9.6 - white

1076238002 - 9.8 - white

1076401004 - 9.6 - black

1091638001 - 9 - white

1091638002 - 9 - black

1091638006 - 9.8 - white

1094188002 - 9.8 - white

1094188003 - 9.8 - black

1094428002 - 9.8 - white

1096874002 - 9.8 - black

1100551001 - 9.8 - white

1101018001 - 9.8 - white

1103146001 - 9.8 - black

1104801013 - 9.6 - white

1105049002 - 9.8 - white

1107526004 - 9.8 - white

1107956001 - 9.9 - white

1108057001 - 9.4 - white

1108376001 - 9.6 - black

1108586004 - 9.8 - white

1109516002 - 9.8 - white

1109815002 - 9.6 - white

1110547006 - 9.8 - white

1110580001 - 9.8 - white

1115011001 - 9.4 - white

1125635001 - 9.9 - white

1126235001 - 9.8 - white

1126506001 - 9.8 - black

1127144001 - 8.5 - white

1127144002 - 9.2 - white

1127646001 - 9.6 - white

1129133001 - 9.2 - white

1129425003 - 9.4 - white

1136103001 - 9.4 - white

1136174005 - 9.8 - white

1136548001 - 9.8 - white

1159072001 - 9.8 - white

1165456001 - 9.8 - white

1165603001 - 9 - white

1196966001 - 9.8 - white

1980335001 - 9.8 - white

 

 

 

Here is how I came about the answer:

 

 

To solve this problem, I had to freshen up on Binomial Distributions. This happens when data can go one of two ways, for example: yes or no, right or wrong, and in this case black or white. I used the follow two sites to re-educate myself:

 

http://www.sigmazone.com/binomial_confidence_interval.htm

http://books.google.com/books?id=m8rYUEWQx00C&pg=PA360&lpg=PA360&dq=binomial+distribution+margin+of+error&source=bl&ots=qG0eP3IPrb&sig=LbOZI4bceo6Pdoou8x9FBiIvdGQ&hl=en&sa=X&ei=SJ9_Upr2G8WqkAf6pICYDQ&ved=0CDkQ6AEwAg#v=onepage&q=binomial%20distribution%20margin%20of%20error&f=false

 

One of the things you can do with statistics is make statements about very large populations of data with a comparatively small amount of data. Businesses do this everyday when they do things like quality control. For example, instead of testing every item off of a production line, they grab random samples and test those. If they grab enough of them, they can be reasonable certain of the quality of all they items they produce.

 

So the first thing I wanted to was see if I had a large enough sample size (n). According to the text I read, you can calculate the sample size needed if you have a preliminary estimate for the proportion (p) that you are testing. This is the formula:

 

n = p (1 - p) (z / E)^2

 

where

p = proportion of interest

n = sample size

E = maximal error of estimate

z = “z value” for desired level of confidence

 

I found 132 different labels, of which 20 were black labels, so I did have an estimate: 15.15%. Using this calculation, I plugged a few numbers to see what maximal error I could have with a sample of only 132, using a desired level of confidence of 95%. Turn out it is 6.12%.

 

n = p (1 - p) (z / E)^2

n = 0.1515 (1 - 0.1515) (1.96 / .0612)

n = 131.8481

 

*z = 1.96 for 95% confidence

 

Now that I felt okay with my sample size, I needed to figure out how to calculate a confidence interval. What is a confidence interval? Well, remember that we testing a small group of data to make a reasonable estimate of the whole population of data. In statistics, this estimate is given as a range with a level of confidence. For example, "I am 95% confident that the true answer falls between 15 and 20." This is the range in which the real answer lies.

 

While not perfect, a formula used to calculate a confidence interval for binomial distributions is below. It makes a few assumptions which I will not go into (since I don't feel I can appropriately explain them), but it is a good approximation according to the text:

 

Confidence Interval =

= p +/- z (sqrt ( p (1 - p) / n ))

= 0.1515 +/- 1.96 (sqrt (0.1515 (0.8485) / 132))

= 0.1515 +/- 0.0612

 

This means that we can be 95% confident that the true proportion of black labels falls between 9.03% and 21.27%. When multiplied by the population (7266), that gives a range of 656 to 1545.

 

 

 

The best answer that statistics can provide is a range, with a level of confidence in that range. With the data and means available, I am 95% confident that the true proportion of black label Walking Dead #1s fall between 9.03% and 21.27% of the print run. When multiplied by the population (7266), that gives a range of 656 to 1545.

 

The range can get tighter as more data is collected. Feel free to comment with any errors you noticed or any additional slab data provided.

 

Jim

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The black label was first. At least that much we know.

 

http://boards.collectors-society.com/ubbthreads.php?ubb=showflat&Number=5970682

 

As far as it being 50/50, or even 40/60, I think the scales are way more out of balance. It seems to me that the price of a graded Walking Dead #1 is high enough to bring any copy to market, regardless of color. A closer proportion of black to whites would be observed if the ratio were 50/50 or 40/60.

 

Bare in mind, I'm not arguing that a black label should make a difference in demand or price; it's just something that I wanted to dig further into.

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I still stick with Kirkman and his observations.

 

A true scientific poll will have at least 2000 results to compare, and with the CPR game, even that number is going to be skewed. How many of those 9.2-9.6 copies got cracked, pressed, and resubbed?

 

This and other reasons are why I stick with Kirkman's statement.

 

 

 

-slym

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Hey slym,

 

I appreciate the input. Of course it would've been better at the start of the project lol but oh well.

 

As far as needing at least 2000 results to be a true scientific poll, I have not seen that referenced anywhere before. Do you have a resource that you can reference? From what I have read, there is not a hard number needed to create a confidence interval; it just depends on the margin of error that you are comfortable with. Using the formula provided by the text, only 150 samples would be needed if you wanted a CI of 95% and an error of 8%, assuming the occurrence is 50/50. Only 144 would be needed if the ratio is 40/60. Even with a margin of error of 5%, only 369 to 384 samples would be needed.

 

I agree, the project is not without it limits. I would prefer to have more data, but only 1727 have been graded. Of course slabs can be cracked and reslabbed, but the chances of that happening to a black label should be equal to that of a white label, especially if Kirkman's assertions are correct. I eliminated any SS for this reason.

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have to have both, no question about it

 

I mean, I'm not a variant chaser-- especially an error variant chaser, but I've sold both, and have gotten about a 5% premium for the black label. There's no doubt it's the "true" first printing, and that others in some cases are willing to pay a little more for it. If I could turn back time, I would have kept my 9.8 black label for good.

 

Which actually reminds me, I've owned six copies, not five!

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How many people participated, the number of raws out there and the number of books graded by "the other guys" are all factors. The number of resubs is definitely a factor as well, especially with a book this hot. Pressing anyone? This book is hard to get in 9.8. I bought about 10 raws over the years (all advertised nm/m) and got back 2 9.8s. The idea of pressing and resubbing definitely crossed my mind and I'm sure there are a lot of people who took advantage of that service especially once CGC acquired CCS.

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