Mail-In Rebate Math

published 2026-May-28 17:27:44

last edited on 2026-May-28 22:45:08

Drspectred uses the term “Value generators” to refer to a category of Jokers in Balatro which provide the player resources instead of scoring points. Of these Jokers, the most powerful is probably Mail-In Rebate.

A description of the Joker Mail-In Rebate from the Balatro Wiki. The Joker earns five dollars for every discarded card of a target rank. The target rank changes each round. — A description of the Joker *Mail-In Rebate* from the Balatro Wiki. The Joker earns five dollars for every discarded card of a target rank. The target rank changes each round.

While explaining why Mail-In Rebate is my favorite value generator would make for a good a blog post, our purposes for today are to share some math I worked through last night. Specifically, I would like to know how much profit I can expect to make each round from Mail-In Rebate.

To start, we will solve the problem of the probability of seeing the “target” of a Mail-In Rebate twice. We will assume that you are playing gold stake, have a hand size of eight, have 4 hands, and have 2 discards.

To solve this problem, we can recognize that only some fraction of the total 52 cards in our deck can be provided to Mail-In Rebate. There are the eight cards we see initially, and we can play three of our four hands, making sure to play five cards with each of these hands. (We can’t play the fourth hand because that will end the round immediately.) We can also discard our first discard to see five additional cards. The five additional cards we can see from our final discard will be ignored since we cannot discard them for Mail-In Rebate value.

This means we will see $8 + 5 \times \left(\left(4 - 1\right) + \left(2 - 1\right)\right) = 28$ cards. For convenience, I let the variable $\beta = 28$ represent the number of cards which can be provided to Mail-In Rebate, and $\alpha = 24$ be the number of cards which cannot.

Then the probability of seeing the target of a Mail-In Rebate twice, assuming there are four targets in the deck, is the chance of two of the targets being in the $\beta$ portion of the deck and two of these cards being the $\alpha$ portion of the deck. To do this, we pick a spot for each of the four targets. The first target is one of 28 of the 52 cards that can be provided to Mail-In Rebate, and then second is one of the 27 of the remaining $\beta$ -type cards of the 51 remaining cards, the third is the 24 of the remaining 50 cards, and the fourth is in the 23 of the remaining 49 cards. The probability of this occurring is

$\left(\frac{28}{52} \cdot \frac{27}{51}\right) \left(\frac{24}{50} \cdot \frac{23}{49}\right).$

However, we could order things differently. Maybe we the first target goes in the $\alpha$ portion of the deck and the third target goes in the $\beta$ portion of the deck:

$\left(\frac{24}{52}\right) \left(\frac{28}{51}\right) \left(\frac{27}{50}\right) \left(\frac{23}{49}\right).$

But this is the exact same number as the first expression (we can just rearrange numerators and denominators due to the algebra of fractional multiplication). So we have to account for all of the ways we can pick two of the four targets to go in the $\beta$ section of the deck. Then the total probability of exactly two targets can be provided to Mail-In Rebate is

${4 \choose 2}\left(\frac{28}{52} \cdot \frac{27}{51}\right) \left(\frac{24}{50} \cdot \frac{23}{49}\right)$ .

Let me introduce some notation that will ease generalization. Let $(5)_{2} = 5 \times 4$ , $(10)_{3} = 10 \times 9 \times 8$ , etc. That is,

$\left(a\right)_{b} = \frac{a!}{\left(a - b\right)!}.$

Then the probability $p(2)$ of exactly two cards being provided to Mail-In Rebate is

$p(2) = {4 \choose 2} \cdot \frac{\left(\beta\right)_{2}\left(\alpha\right)_{2}}{\left(52\right)_{4}}.$

With some thought, you may recognize that the general expression for having exactly $m$ cards appearing in the deck is

$p(m) = {4 \choose m} \cdot \frac{\left(\beta\right)_{m}\left(\alpha\right)_{4 - m}}{\left(52\right)_{4}}.$

As a reminder, $\alpha + \beta = 52$ and $\beta = 8 + 5 \times \left(\left(4 - 1\right) + \left(2 - 1\right)\right) = 28$ .

(As an aside, this is a ceiling on our probability. We are assuming we always discard five cards and then we are happy to play hands specifically for the purpose of removing cards. In an actual game of Balatro, this will not always be the case. We may need to play hands containing a target card in order to score enough to survive, and we may need to discard a target card in order to play a flush or straight or some other five-card hand, and we may not play/discard four cards instead of five because we are holding onto a blue seal or gold card.)

The average profit then, is simply

$\sum_{i = 0}^{4}{5i \cdot p(i)} = \frac{140}{13} = 10.\overline{769230},$

or approximately 11 dollars.

While this is an optimistic calculation, this does align with my experience. I get two of the four target cards more often than not, and I almost always get at least one target card.

It is worth mentioning that

$p(0) = \frac{1518}{38675} \approx 0.039,$

or that there is about a 4% chance that Mail-In Rebate can’t be given any target cards at all. These odds are approximately $\frac{1}{25}$ .

I made an interactive tool for viewing these results in Desmos. As a bonus, you can change various variables such as hand size, the number of hands and discards, the number of target cards, the number of cards in the deck, etc, and see how they change the resulting probabilities.

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