Wednesday, September 26, 2012

Un-Perfect Collation

Many hobbyists will grumble on and on about manufacturer’s poor card collation. ‘Card collation’ refers to the mix of cards in any given pack, box, case, etc. An example of poor card collation would be pulling duplicate cards in the same pack or pulling seven Drew Pomeranz rookie cards and one Yu Darvish rookie cards from the same case of Topps Jumbo.

However, I posit that this incessant critiquing has led to an environment with even worse outcomes for the average sports card collector, and an unprecedented boon for the sports card manufacturer.

I recently busted a hobby box of 2012 Topps Chrome; a standard hobby box of 2012 Topps Chrome has 24 packs of 4 cards (a total of 96 cards). Here are the results of the box (ratio in parentheses):
  • 78 - Base
  • 1 - Die-Cut (1:24)
  • 2 - Rookie Auto (2/hobby box)
  • 8 - Refractor (1:3)
  • 4 - X-Fractor (1:8)
  • 1 - Blue Refractor (1:21)
  • 1 - Gold Refractor (1:50)
  • 1 - Atomic Refractor (1:383)

With the exception of the Gold and Atomic Refractors, I pulled exactly what the stated ratios indicated I would pull. And I pulled no duplicate base cards. And the first and last cards in the packs were always base cards. There were 18 insert cards, and no pack had more than one insert card (so six of the packs had no inserts). And, to be fair, I am very pleased with the pulls.

Some might say, “No duplicate base cards, wow, what great collation!” Or, “Topps finally stopped screwing the collectors!” Let’s not get ahead of ourselves.

Had those 78 been chosen by perfectly random draw, what is the likelihood that I would have pulled 78 unique cards out of the 220 card base set? (Spoiler alert: it’s a very, very, very, small number).

Let’s do a quick refresh on probability; the product of all of the independent events’ is the cumulative probability of a series of events occurring. This cumulative probability is represented in the formula: y(1)*y(2)*y(3)*y(4)….*y(n), where y(n) is the last probability in the series. For example, the probability of pulling a flush (five cards of the same suit) from a standard deck of 52 playing cards in one five-card hand would be calculated as: (52/52)*(12/51)*(11/50)*(10/49)*(9/48)=.001981, or 0.1981%, or odds of 1:504.85.

If each pack of Topps Chrome had three base cards and one insert card, then the probability of those base cards all being unique would be calculated as: (220/220)*(219/220)*(218/220)=.986. 99% - not bad!

What about that second pack, what are the odds of pulling three more unique base cards? 94.6% - well that’s still pretty good, right?

But what are the odds of pulling two consecutive packs that have no duplicates? Let’s do the math: .946*.986=.933. Dang, those odds are great!

By the time we get to our last pack to find cards 76, 77, and 78, there is only a 28.04% chance that those three cards will all be unique; some would argue that those are pretty good odds. But here’s the kicker, the odds pulling 78 unique cards from a base set of 220 cards is 0.0000166% - or 1:6,038,332

And what if the Gold and Atomic Refractor had not been in the box and I pulled two more base cards, what are the odds that those two would also be unique, along with the other 78? Those odds are 0.0000069%, or 1:14,596,708.

The perceived need for great collation diminishes the likelihood that what we receive is random or ‘special’. My two favorite cards from the box were the Gold and Atomic Refractor because there was no good probable chance I would pull either, let alone both.

But this lack of randomness is great for folks like me. I’m devising an algorithm to predict which exact cards are in a pack based on just the first few cards on the top or bottom of the pack.

The cherry on top of all of this? Although I pulled no duplicate cards, I pulled the same player three times on three different cards: a base card, an X-Fractor, and the Atomic Refractor. What are the odds of that? 

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