## A funny little probability problem

(Note: this problem has an error…can you find it? I will post the correction soon here, which itself has some interesting properties)

We often make the important distinction between drawing cards with or without replacement when determining the results of card games. Clearly, if you draw a Jack, then the probability of drawing another Jack is different whether you replace and shuffle, or you leave the drawn Jack out. In an extreme case, imagine a deck with only one Jack.

After going through an example in class, and applying replacement, a student asked about the same example without replacement. I didn’t have time to go over the example in class, but I sketched how the calculation would be different, and would yield a different answer. After class, going through the calculation, I was in for a surprise.

# The problem

You draw two cards from a deck, and ask what is the probability that the first is a black card, and the second is a jack. In math notation, we want:

$P(B1,J2)$

## With replacement

In replacement, we replace the first card after drawing it, reshuffle, and then draw the second. Thus the two events are independent.

$\begin{array}{rcl} P(B1,J2|{\rm replace}) &=& P(B1) \times P(J2) \\ &=&\frac{26}{52} \times \frac{4}{52} = 0.0385 \end{array}$

## Without replacement

Without replacement is a little trickier to set up, because the second draw depends on the first.

$\begin{array}{rcl} P(B1,J2|{\rm no-replace}) &=& P(B1) \times P(J2|B1) \\ &=&\frac{26}{52} \times \left(P(J2|B1,J1)P(J1)+P(J2|B1,\overline{J1})P(\overline{J1}) \right)\\ &=&\frac{26}{52} \times \left(\frac{3}{51} \cdot \frac{4}{52}+\frac{4}{51} \cdot \frac{48}{52} \right)\\ &=&\frac{26}{52} \times \left(\frac{204}{51\cdot 52} \right)\\ &=&0.0385 \end{array}$

(Note: this problem has an error…can you find it? I will post the correction soon here, which itself has some interesting properties)

## Conclusion

The only conclusions are

1. My intuition fails me sometimes, on even simple problems
2. Drawing a black card on the first draw tells you no more information about drawing a jack on the second (or any other number, for that matter).

Cool!

(Note: this problem has an error…can you find it? I will post the correction soon here, which itself has some interesting properties)