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I'm trying to calculate the odds of a box of trading cards containing less than $100$ unique cards. From what I can gather, there's $11$ common cards per pack of $159$ unique common cards, $1$ item card of $13$, $1$ rare of $79$, one card that may be rare or higher (very rare - $1$ in $12$ packs, out of $31$ v.rare cards, or super rare - $1$ in $96$ packs, out of $6$ s.rare cards) and one special edition that can be of any of the common cards.

There are $16$ packs to a box. I've not done statistics much since high school some $15$ years ago so I'm a little overwhelmed as to where to start.

Thanks!

David Thiessen
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WhatsANumber
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    Someone can math out the naive model, but worth noting that in reality (without knowing the details of whatever tradings cards you are referencing): probably the cards of the same rarity in the same pack are not independent; probably packs from the same box are not independent; possibly cards of the same (nominal) rarity/type are not equally probable. – tehtmi Jul 12 '22 at 04:35
  • They probably aren't, I'm just assuming that they are completely randomised. – WhatsANumber Jul 12 '22 at 04:37
  • Thank you so much for making the effort to explain the composition of cards. Unfortunately, it's been ages since I've played with trading cards (Pokemon cards , Duel Masters, you name it) , so I'm still not fully comfortable with the composition you've written down. Could you give me an online link with examples, so that I can get comfortable with your description? – Sarvesh Ravichandran Iyer Jul 12 '22 at 04:55
  • How many v.rare and s.rare cards are there? Are you considering a special edition to be different than the base version of the card? Are special edition equally likely to be any of the cards, or are they more likely to be certain rarities? – David Thiessen Jul 12 '22 at 18:51
  • Yep, special edition is unique to the common variant. There is 31 v. Rare and 6 super rare. – WhatsANumber Jul 13 '22 at 03:44

1 Answers1

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This type of question is similar to the coupon collector's problem. You can find many similar types of questions and approaches on this website and online, but once you start making adjustments it becomes very difficult to find exact analytic expressions for the result. The selection of very rare or super rare cards makes this question fairly difficult, for example.

Instead, it's easier to use a computer to simulate the situation many times and find approximate probabilities from that. You can find many examples of this sort of simulation online as well. For example, here's a simulation in R of drawing a box and calculating the number of unique cards in it. I repeat the simulation 1000 times and plot a histogram of the number of unique cards.

# number of cards in each set
num_common <- 159
num_item <- 13
num_rare <- 79
num_vrare <- 31
num_srare <- 6
num_special <- num_common

make card names to sample from


commons <- 
  vapply(1:num_common, FUN = function(x)
    paste0("C", formatC(x, width = 3, format = "d", flag = 0)), "A")
items <- 
  vapply(1:num_item,  FUN = function(x)
    paste0("I", formatC(x, width = 2, format = "d", flag = 0)), "A")
rares <- 
  vapply(1:num_rare, FUN = function(x) 
    paste0("R", formatC(x, width = 2, format = "d", flag = 0)), "A")
vrares <- 
  vapply(1:num_vrare, FUN = function(x)
    paste0("VR", formatC(x, width = 2, format = "d", flag = 0)), "A")
srares <- vapply(1:num_srare, FUN = function(x) paste0("SR", x), "A")
specials <- 
  vapply(1:num_special, FUN = function(x)
    paste0("SP", formatC(x, width = 3, format = "d", flag = 0)), "A")


functions to generate a random pack and random box


generate_pack <- function() {
  common <- sample(commons, 11, replace = TRUE)
  item <- sample(items, 1)
  rare <- sample(rares, 1)
  rare_draw <- sample(1:96,1)  # randomly select type of other rare card
  if(rare_draw == 96) {

    other_rare <- sample(srares, 1)
  } else if(rare_draw >= 88) {
    other_rare <- sample(vrares, 1)
  } else {
    other_rare <- sample(rares, 1)
  }
  special <- sample(specials, 1)
  return(c(common, item, rare, other_rare, special))
}
generate_box <- function() {
  allcards <- c()
  for(i in 1:16) {
    allcards <- c(allcards, generate_pack())
  }
  return(allcards)
}


simulate 1000 times and plot histogram.


set.seed(1111) # for reproducibility
num_sims <- 1000
unique_cards <- rep(0, num_sims) 
for(i in 1:num_sims) {
  unique_cards[[i]] <- length(unique(generate_box()))
}
summary(unique_cards)
#>    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
#>   144.0   155.0   158.0   158.3   161.0   172.0
hist(unique_cards)

Created on 2022-07-13 by the reprex package (v2.0.1)

From this we can see that the odds of getting fewer than 100 unique cards is essentially 0. Out of 1000 simulations the fewest unique cards we saw was 144.

You can try running it online if you don't have R installed, and you can make various adjustments to see how the number of cards change. (This code could be simplified, for example by removing formatting from the card names, but I like it this way. If you want to sort and print the simulated cards from each box or pack it's relatively easy to do that as well.)

David Thiessen
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  • That's so cool! Thank you so much for that, it answered my question and finalized the argument. The outcome was exactly as I (very generally) theorised. Seriously, I really appreciate this. Will definitely check out the website – WhatsANumber Jul 13 '22 at 21:50