In my previous post on collation, I said that a box with good collation (minimal duplicates) is not the same as a randomly packed box. To make this statement, I need to be able to show what a randomly packed box should contain.

One way to show what a random box should contain is through a direct calculation of an outcome. To keep things as simple as possible, we’ll look at a very small set: a four-card insert set – 1994 Topps Stadium Club Dugout Dirt Series 1. These cards were inserted into 1 in 6 packs. The 1994 TSC boxes had 24 packs, so that’s four Dugout Dirt cards per box. With perfect collation, each box should contain a complete set of the Series 1 Dugout Dirt cards. How about a random box?

In a random box, the first Dugout Dirt card will certainly be one that we don’t have. That first card has a 100% chance of not being a duplicate. Let’s say we pull #1 (Piazza). The second card would need to be #2, #3, or #4 to keep the no duplicate street alive. That’s a 75% of no getting a duplicate. Let’s say we pull #2 (Winfield). For the third card, we need to get #3 or #4 – only a 50% chance of no doubles at this point. Let’s say we pull #3 (Kruk). The last card must be #4 – only a 25% chance of this happening. Cross your fingers for #4 (Ripken).

What are the odds of pulling the four different cards? The odds are just the product of all the percentages. So…

odds of getting a set = 1.00 x 0.75 x 0.50 x 0.25 = 0.09375 or 9.375%

Less than 10%. Those aren’t very good odds. Regardless, most people who open a box of 1994 TSC Series 1 probably expect to get all four Dugout Dirt cards. They might even be disappointed if they don’t get a full set. With less than a 10% chance, it should really be a pleasant surprise if it happens.

This method of calculating odds is pretty easy, but it’s limited. What are the odds of getting one (and only one) duplicate? Two duplicates? I’m sure there are those out there who are more math inclined and can answer these question easily. Still, the thinking is a little more involved.

Is there a better way? I’ll cover one option next week.

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