Mail-In Rebate Math—Redux

published 2026-May-30 19:21:01

Let $\beta$ be the number of cards which can be provided to Mail-In Rebate and $\alpha$ be the number of cards which cannot. Assuming our deck of 52 cards has four targets from Mail-In Rebate, then the probability of $m$ targets being provided to Mail-In Rebate is

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

This works because we compute the number of ways for $m$ targets to be in the $\beta$ -portion of the deck and $4-m$ targets to be in the $\alpha$ -portion of the deck, and divide that by the total number of ways for the targets to be distributed within the entire deck.

This is mathematically equivalent to the previous result:

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

$p(m) = \frac{4!}{m!\left(4 - m\right)!} \frac{\beta!}{\left(\beta - m\right)!} \frac{\alpha!}{\left(\alpha - \left(4 - m\right)\right)!} \frac{\left(52-4\right)!}{52!}$

$p(m) = \frac{\beta!}{m!\left(\beta - m\right)!} \left(\frac{\alpha!}{\left(4 - m\right)!\left(\alpha - \left(4 - m\right)\right)!}\right) \frac{4!\left(52-4\right)!}{52!}$

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

This second formulation is easier to justify, uses more standard notation, and is more closely related to the derivation. Someone with experience with combinatorics and probability could likely reverse engineer the derivation simply from looking at the formula, so I much prefer this formulation.

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