Probability of certain rarities in a trading card game booster box

So I've previously had a discrete math course that included very basic combinatorics, but this is over my head. I'm in a discussion/argument on Facebook over rarities and "pull rates" in Pokémon TCG booster boxes. (This is not a homework question.)

Given:

• 1 out of every 72 booster packs contains a "secret rare" card,
• a pack cannot contain more than one secret rare,
• each booster box contains 36 packs, and
• packs in each box are completely random, with no guarantees or restrictions of any kind on particular rarities per box,

I can calculate the odds of having two or more secret rares in a box as follows:

• Probability of zero secret rares: $$\left(\frac{71}{72}\right)^{36}=0.60440866...$$
• Probability of exactly one secret rare...
• in the first pack: $$\left(\frac{1}{72}\right)\times\left(\frac{71}{72}\right)^{35}=0.00851279...$$
• anywhere in the box: $$0.00851279...\times36=0.30646073...$$
• Probability of two or more: $$1-0.60440866...-0.30646073...=0.08913060...$$, or approximately 1 in 11.2 boxes.

However, I'm lost on the bigger issue in the argument: what are the odds of a box having 3 or more secret rares? (In particular, I don't know how to figure the probability of exactly two secret rares.)

• What have you tried so far? As a hint, can you calculate the probability of there being exactly 0 secret rares? What about exactly 1? – Sidd Singal Mar 25 at 4:17
• @SiddSingal I can get the probability of zero secret rares pretty easily: (71/72)^36 = 0.6044... Then for exactly one, (1/72)(71/72)^35 = 0.0085... for the secret rare in the first pack, times 36 = 0.3064... for a secret rare in any one pack. And I see now that I can subtract those from 1.0000 for the chances of two or more, so that's solved. But for three or more, I'm lost. – jcgoble3 Mar 25 at 4:36
• @SiddSingal Question updated with that information. – jcgoble3 Mar 25 at 4:55
• Note that in practice, booster packs in booster boxes are not random, and are less likely than chance to contain consecutive rares. – eyeballfrog Mar 25 at 5:23
• @eyeballfrog I've seen some crazy stuff with Pokémon TCG booster boxes, and the company that makes them is silent on rarity distribution, leading many people to think they are totally random. However, per the result below, it seems that they aren't completely random after all, so I can add that info to the argument on Facebook. – jcgoble3 Mar 25 at 5:57

To compute the chance of exactly two secret rares in one box, you have $${36 \choose 2}$$ ways to choose the packs that contain the secret rares, so the chance is $${36 \choose 2}\left(\frac 1{72}\right)^2\left(\frac {71}{72}\right)^{34}$$. Similarly for three it is $${36 \choose 3}\left(\frac 1{72}\right)^3\left(\frac {71}{72}\right)^{33}$$. Your computation for one is $${36 \choose 1}\left(\frac 1{72}\right)^1\left(\frac {71}{72}\right)^{35}$$ for comparison.