Uniform Open pack: 1 random sticker (uniform) per pack out of 20 possible, priced 1 unit of currency. Unlimited availability of packs. Size of pack: N Non-uniform What if we lose the “unlimited availability”? As a kid I lived through several collecting manias in the midst of my peers. Hockey cards, Pogs, Cs… Most of them managed to mostly pass me. I had owned some, but I was never really into the thing to the point of spending my own money on it. The only exception were stickers that came with Resanka (filled croissant in package with a drawing of a girl). If one collected a full set, one could enter a draw for prizes or something. (I haven’t managed to find more information about it beyond stickers being numbered, there being about 16 or 20 different ones, number one being “Kamoš Loptoš”, and number sixteen “Hadica” (very childish pun on snake female and hose). I think I managed to fill about three of the albums which shows how many I must have eaten. 😅) Primary mechanism of acquiring these collectible items is buying an opague collection which contains a random selection of given size. Common names are “Pack” or “Loot box” or similar. However, often times people start trading these cards, either for other cards, or through established value-exchange media such as money (and so secondary markets emerge). CG20 We’ll start with a very simple card collecting game CG20 (Collecting Game 20). It will be our playground to explore some basic principles guiding the price of these cards. CGA works as follows: There are 20 different cards. Cards are sold in individual packs (one card per pack). Pack costs ¤1. Cards have uniform probability. (So each card has the same probability to be found in each pack. For this particular setup that is probability that is sometimes expressed as 1/20, 0.05, 5%, or 1:19.) There is no limit on how many packs can be bought (so scarcity is not a part of our problem in this game). Question #1: How much should we expect to pay to get a specific card? (Maybe that is the last one we need to get the dopamine hit from completing our collection?) As each pack costs ¤1 this is the same as asking what is the expected number of packs to open to get a specific card. Probability of us getting it on the first try is 0.05. Getting it on the second try means we did not get it on the first try so that is 0.95 * 0.05. Getting it on the third try means we did not get it on the first try (0.95) and on the second try (0.95 or remaining probability) and got it on the third time: 0.95 * 0.95 * 0.05. And so on. Now we make weighted average of that: ∑i∈ℕ+i⋅19i−120i=20 \sum_{i \in \mathbb{N}^{+}} \frac{i\cdot{}19^{i-1}}{20^i} = 20 If you don’t know how to get the value of geometric series, infinite sums might look a bit intimidating. However we have previously seen with generating functions, it is possible to turn the infinity to eat itself. x=1⋅19020+2⋅19120+3⋅19220+4⋅19320+⋯1920x=0⋅190201+1⋅191202+2⋅192203+3⋅193204+⋯x20=1⋅190201+1⋅191202+1⋅192203+1⋅193204+⋯19x400=0⋅190201+1⋅191202+1⋅192203+1⋅193204+⋯20x−19x400=120x=40020=20 \begin{align} x &= 1\cdot{}\frac{19^0}{20} + 2\cdot{}\frac{19^1}{20} + 3\cdot{}\frac{19^2}{20} + 4\cdot{}\frac{19^3}{20} + \cdots{} \\ \frac{19}{20}x &= 0\cdot{}\frac{19^0}{20^1} + 1\cdot{}\frac{19^1}{20^2} + 2\cdot{}\frac{19^2}{20^3} + 3\cdot{}\frac{19^3}{20^4} + \cdots{} \\ \frac{x}{20} &= 1\cdot{}\frac{19^0}{20^1} + 1\cdot{}\frac{19^1}{20^2} + 1\cdot{}\frac{19^2}{20^3} + 1\cdot{}\frac{19^3}{20^4} + \cdots{} \\ \frac{19x}{400} &= 0\cdot{}\frac{19^0}{20^1} + 1\cdot{}\frac{19^1}{20^2} + 1\cdot{}\frac{19^2}{20^3} + 1\cdot{}\frac{19^3}{20^4} + \cdots{} \\ \frac{20x - 19x}{400} &= \frac{1}{20} \\ x &= \frac{400}{20} = 20 \\ \end{align} So if we want a specific card and see an offer for less than ¤20, it is better than buying individual packs. And it works the other way too: we should not be willing to pay more than ¤20. This gives us a bound on what a fair price of a card might be. As aspiring finance bros we might be tempted to slap an order book on the problem, and we might be right to let the market discover a fair price. But it is good to know that if such a market goes above ¤20, it is time to sell… (Or maybe something happened that we don’t know about? 🤔) On the other side it is obvious that if price goes under ¤1 it is time to buy like there is no tomorrow. (Or not, could be a sign of collapsing market 🤷.) (Impatient reader likely already noticed that if we keep buying packs, we will also accumulate cards we can try selling ourselves, and that way offset our cost. Let’s not think about that just yet.) Question #2: How much should we expect to pay for two specific cards? Trying to make infinity eat itself will start soon turn to hydroboros. Time for pictures! Key here is to notice that grapph without the red part is the same as graph with it. So we turn that edge on itsel! (Think of a value at the node as an average lenght of path from that node if we were taking random walks with probabilities as labelled.) x=0.05⋅1+0.95*(1+x)x=0.05+0.95+0.95x0.05x=1x=20 \begin{align} x &= 0.05 \cdot 1 + 0.95 * (1 + x) \\ x &= 0.05 + 0.95 + 0.95x \\ 0.05x &= 1 \\ x &= 20 \\ \end{align} From there we can draw diagram for 2 cards Let’s merge duplicate parts. x=0.1(1+20)+0.9(1+x)x=2.1+0.9+0.9x0.1x=3x=30 \begin{align} x &= 0.1 (1+20) + 0.9 (1+x) \\ x &= 2.1 + 0.9 + 0.9x \\ 0.1x &= 3 \\ x &= 30 \\ \end{align} One can go to three and beyond but things escalate rather quickly… Question #3: What is expected spend to get m given cards? ∑i∈{1,⋯,m}20i \sum_{i\in{}\{1,\cdots{},m\}} \frac{20}{i} CGN Let’s bump the things a notch and make our game have N cards. (This way we work with a family of games, one for each N.) To translate previous results without going into details: Individual card’s probability is p=1Np = \frac{1}{N}. Expected number of opened packs to find a given card is ∑i∈ℕ+i(N−1)i−1Ni=ip(1−p)i−1=N \sum_{i\in\mathbb{N^{+}}} \frac{i(N-1)^{i-1}}{N^i} = i p (1-p)^{i-1} = N