### Use of unconditional probability

Recently I had a discussion with some friends about the probability that a certain card was among the first *n* cards dealt from a shuffled deck of *m* cards. Here I’m going to only cover the specific case of a deck consisting of the 13 spades, and the chance that the ace is one of the first two cards dealt — *n*=2 and *m*=13. We came up with four ways to perform this elementary probability calculation.

**Combinatorics:** There are _{12}C_{1} ways to have the ace and one other spade, without regard to order. There are _{13}C_{2} ways to deal two spades. So the probability is _{12}C_{1}/_{13}C_{2} = 12/78 = 2/13.

**Trickery:** The chance the first card *isn’t* the ace is 12/13. The chance that the second card, dealt from the remaining 12 cards, also isn’t the ace is 11/12. So the chance that neither is the ace is (12/13)⋅(11/12) = 11/13. Thus the chance that one of the first two cards *is* the ace is 1 − 11/13 = 2/13.

**Conditional probability:** The chance that the first card is the ace is 1/13 and the chance it isn’t is 12/13. If we represent the event of the first card being the ace as **1** and the event the second card is the ace as **2**, then that is *P*(**1**) = 1/13 and *P*(~**1**) = 12/13. Now, *P*(**2**|~**1**) = 1/12. ∴ *P*(**1** or **2**) = *P*(**1**) + *P*(**2**|~**1**)⋅*P*(~**1**) = 1/13 + (1/12)⋅(12/13) = 1/13 + 1/13 = 2/13.

**Unconditional probability:** *P*(**1** or **2**) = *P*(**1**) + *P*(**2**) − *P*(**1** and **2**) = 1/13 + 1/13 − 0 = 2/13.

They all generalize into dealing *n* cards out of *m*, but the one using unconditional probability not only is the simplest, it generalizes the most cleanly. But most of us found it the least satisfying solution. Maybe because it was *too* simple?