A funny little probability problem

In #wordpress_migration

(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} $

the same answer!!

(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)