## Introduction

This weekend I got hooked on a funny little probability puzzle, and have finally found some closure. It started with an off-hand comment by a student, which led me to think more deeply about the problem once I actually worked it out. The problem was simple enough:

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:

I compared *with replacement* to *without replacement*, and got the same answer, much to my surprise. After satisfying my surprise with a numerical simulation, I thought, “there are clearly cases where it *does* make a difference between replacement and no replacement (drawing two jacks, for example), so where does it matter?”

## Another case

So I did another case:

You draw two cards from a deck, and ask what is the probability that the first is a

facecard, and the second is a jack. In math notation, we want:

### With replacement:

### Without replacement

## The General Case

It appeared to me that there was a pattern – some relationship between the number of jacks, the number of cards, and the number of the sub-population (color, face, etc…) such that the replacement and the no-replacement cases would come out the same, because *most* cases don’t. So I looked at it in general, where I define:

### With replacement:

### Without replacement

### Solving

When these two expressions are the same, we have:

which, believe it or not, simplifies to

or, in other words, for the replacement and non-replacement probabilities to be the same in this simple game, the fraction of the subpopulation to the deck has to be the same fraction as the jacks in that subpopulation to the number of jacks. In the case of color, 1/2 the deck is black and 1/2 the jacks are black. However, 3/13 of the deck are face cards and 4/4 of the jacks are face cards. An interesting symmetry.

Essentially, when there exists this symmetry, knowledge of the first draw gives you no information about the second. I imagine there is some fancy math theorem to this effect, but it is still pretty cool.

I commented on your last post before I saw this one. Looks like you worked out mathematically what I was thinking.

what’s a little weird is that drawing a black gives you no extra information about drawing a jack on the second – not even that you draw a card, and thus possibly reduced the deck size. just shows how a well-written simulation trumps intuition any day. 🙂