A funny little probability problem - some closure

In #wordpress_migration

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:

P(B1,J2) P(B1,J2)

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

P(F1,J2) P(F1,J2)

With replacement:

P(F1,J2∣replace)=P(F1)×P(J2)=1252×452=482704=0.0178 \begin{aligned} P(F1,J2|{\rm replace}) &= P(F1) \times P(J2) \\ &=\frac{12}{52} \times \frac{4}{52} = \frac{48}{2704}=0.0178 \end{aligned}

Without replacement

P(F1,J2∣no−replace)=P(F1)×P(J2∣F1)=1252×(P(J2∣F1,J1)P(J1∣F1)+P(J2∣F1,J1‾)P(J1‾∣F1))=1252×(351⋅412+451⋅812)=442652=0.0166 \begin{array}{rcl} P(F1,J2|{\rm no-replace}) &=& P(F1) \times P(J2|F1) \\ &=&\frac{12}{52} \times \left(P(J2|F1,J1)P(J1|F1)+P(J2|F1,\overline{J1})P(\overline{J1}|F1)\right)\\ &=&\frac{12}{52} \times \left(\frac{3}{51} \cdot \frac{4}{12}+\frac{4}{51} \cdot \frac{8}{12} \right)\\ &=&\frac{44}{2652}\\ &=&0.0166 \end{array}

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:

NC=\mboxnumberofcardsNJ=\mboxnumberofjacksNF=\mboxnumberofthesub−population(faces,color,etc...)NJF=\mboxnumberofjacksinthesub−population \begin{array}{rcl} N_C &=& \mbox{number of cards} \\ N_J &=& \mbox{number of jacks} \\ N_F &=& \mbox{number of the sub-population (faces, color, etc...)}\\ N_{JF} &=& \mbox{number of jacks in the sub-population} \end{array}

With replacement:

P(F1,J2∣replace)=NFNC⋅NJNC \begin{array}{rcl} P(F1,J2|{\rm replace}) &=& \frac{N_F}{N_C} \cdot \frac{N_J}{N_C} \end{array}

Without replacement

P(F1,J2∣no−replace)=P(F1)×P(J2∣F1)=NFNC×(P(J2∣F1,J1)P(J1∣F1)+P(J2∣F1,J1‾)P(J1‾∣F1))=NFNC×(NJ−1NC−1⋅NJFNF+NJNC−1⋅NF−NJFNF) \begin{array}{rcl} P(F1,J2|{\rm no-replace}) &=& P(F1) \times P(J2|F1) \\ &=&\frac{N_F}{N_C} \times \left(P(J2|F1,J1)P(J1|F1)+P(J2|F1,\overline{J1})P(\overline{J1}|F1)\right)\\ &=&\frac{N_F}{N_C} \times \left( \frac{N_J-1}{N_C-1}\cdot \frac{N_{JF}}{N_F} + \frac{N_J}{N_C-1}\cdot \frac{N_F-N_{JF}}{N_F} \right) \end{array}

Solving

When these two expressions are the same, we have:

NFNC⋅NJNC=NFNC×(NJ−1NC−1⋅NJFNF+NJNC−1⋅NF−NJFNF) \begin{array}{rcl} \frac{N_F}{N_C} \cdot \frac{N_J}{N_C}&=&\frac{N_F}{N_C} \times \left( \frac{N_J-1}{N_C-1}\cdot \frac{N_{JF}}{N_F} + \frac{N_J}{N_C-1}\cdot \frac{N_F-N_{JF}}{N_F} \right) \end{array}

which, believe it or not, simplifies to

NF/NC=NJF/NJ \begin{array}{rcl} N_F/N_C &=& N_{JF}/N_J \end{array}

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.