nandbit

Determining lootbox chances

Introduction

Imagine you have a lootbox with $N$ possible item tiers. For example:

Common
Uncommon
Rare
Legendary

Now let's say we would like to have the probability of rarer items to be exponentially less likely to be drawn from a lootbox than the tier above. What are the assigned probabilities to each tier?

Math

My initial solution was formulated as:

\sum_{i = 1}^{n} e^{x i} = 1

e^{x} + e^{2 x} + . . . + e^{n x} = 1

Let's try a concrete example with 4 tiers

\sum_{i = 1}^{4} e^{x i} = 1

x \approx - 0.65626

So the tier drop chances would be

Tier	Expression	Probability
Common	$e^{1 (- 0.65626)}$	$0.518788...$
Uncommon	$e^{2 (- 0.65626)}$	$0.269141...$
Rare	$e^{3 (- 0.65626)}$	$0.139627...$
Legendary	$e^{4 (- 0.65626)}$	$0.072437...$

Generalization (it's more math)

My formulation is okay but it's not generalizeable. What if we want to use another base besides $e$ ? What if we would like to have a specific lowest/highest probability? A friend of mine helped me come up with a generalized formula to answer these questions

1 = k^{0} p + k^{1} p + . . . + k^{n} p

Here, $k$ is essentially the scaling factor for the chosen set probability. If we choose the common tier to have the probability of $0.5$ , and $n = 4$ , we get

1 = k^{0} (0.5) + k^{1} (0.5) + k^{2} (0.5) + k^{3} (0.5)

Solving for $k$ yields

k \approx 0.54369

Plugging it back into the equation gives us the following probabilities

Tier	Expression	Probability
Common	${0.54369}^{0} (0.5)$	$0.5$
Uncommon	${0.54369}^{1} (0.5)$	$0.271845$
Rare	${0.54369}^{2} (0.5)$	$0.147799$
Legendary	${0.54369}^{3} (0.5)$	$0.080357$

We can actually choose any one tier to have a specific probability to calculate the rest off of. Let's say we want the rarest tier to have the probability of 1%. What we need to do is to "isolate" it in our equation

1 = k^{0} p + k^{1} p + k^{2} p + k^{3} p

To do that we divide the probabilities of the other elements by the prefix of the probability we want to keep constant

1 = \frac{k^{0}}{k^{3}} (0.01) + \frac{k^{1}}{k^{3}} (0.01) + \frac{k^{2}}{k^{3}} (0.01) + 0.01

Solving for

k

yields

k \approx 0.23536...

And one more time plugging the values...

Tier	Expression	Probability
Common	$\frac{{0.23536}^{0}}{{0.23536}^{3}} (0.01)$	$0.76701$
Uncommon	$\frac{{0.23536}^{1}}{{0.23536}^{3}} (0.01)$	$0.18052$
Rare	$\frac{{0.23536}^{2}}{{0.23536}^{3}} (0.01)$	$0.04249$
Legendary	$0.01$	$0.01$

Notes

My original formulation is a special case of the generalized formula where $k = e^{x}$ and $p = e^{x}$