Probability on a standard deck of cards

I'm a great fan of Probability and Combinatorics. The question that we are going to solve in this post remains as one of my most favorite question in this area. I've been meaning to solve this one for so long and finally I'm glad I did in recently.

Consider a standard deck of 52 cards that is randomly shuffled. What is the probability that we do not have any King adjacent to any Queen after the shuffle?

Like I said, there were multiple instances where I used to think about this problem for long and then get distracted elsewhere. This time Possibly Wrong brought it back again to my attention. There is already a closed form solution there but I couldn't understand it. So, I decided to solve it Generating functions.

Stripping the problem down of all labels, the equivalent problem is to consider the number of words that we can form from the alphabet $\mathcal{A}=\{a,b,c\}$ with no occurrence of the sub-word $ab$ or $ba$. The $a$'s correspond to the Kings in the deck, $b$'s to the Queens and $c$'s to the rest of the cards.

Two important points here: First, We have to specifically find the number of words with 4 $a$'s, 4 $b$'s and 44 $c$'s. Second, convince yourself that both the problems are equivalent.

Pattern avoidance problems are easily tacked with the idea presented in Page 60 of 'amazing' Analytic Combinatorics.

Following the same, let $\mathcal{S}$ be the language words with no occurrence of $ab$ or $ba$, $\mathcal{T}_{ab}$ be those words that end with $ab$ but have no other occurrence of $ab$ and likewise for $\mathcal{T}_{ba}$ be those words that end with $ba$ but have no other occurrence of $ba$.

Appending a letter from $\mathcal{A}$ to $\mathcal{S}$, we find a non-empty word either in $\mathcal{S}$, $\mathcal{T}_{ab}$ or $\mathcal{T}_{ba}$. Therefore,

$\mathcal{S}+\mathcal{T}_{ab}+\mathcal{T}_{ba}=\{\epsilon\}+\mathcal{S}\times\mathcal{A}$

Appending $ab$ to $\mathcal{S}$, we either get a word from $\mathcal{T}_{ab}$ or a word from $\mathcal{T}_{ba}$ appending with $b$. Therefore,

$\mathcal{S}\times ab=\mathcal{T}_{ab}+\mathcal{T}_{ba}b$

Similarly, we have, $\mathcal{S}\times ba=\mathcal{T}_{ba}+\mathcal{T}_{ab}a$

The three equations in terms of OGFs become,

$S+T_{ab}+T_{ba}=1+S\cdot(a+b+c)$

$S\cdot ab=T_{ab}+T_{ba}\cdot b$

$S\cdot ba=T_{ba}+T_{ab}\cdot a$

We have three equations in three unknowns. Solving for $S$, we get,

$\displaystyle S=\frac{1-ab}{(1-a)(1-b)-(1-ab)c}$

which should be regarded as the generating function in terms of variables $a$, $b$ and $c$. So the coefficient of $a^ib^jc^k$ gives the number of words that avoid both $ab$ and $ba$ using $i$ $a$'s, $j$ $b$'s and $k$ $c$'s.

The coefficient of $c^k$ in this generating function is elementary. We get,

$\displaystyle [c^k]S=\frac{(1-ab)^{k+1}}{(1-a)^{k+1}(1-b)^{k+1}}=\left(\frac{1-ab}{(1-a)(1-b)}\right)^{k+1}$

Something interesting happens here. We note that

$\displaystyle \frac{1-ab}{(1-a)(1-b)}=1+a+b+a^2+b^2+a^3+b^3+a^4+b^4+\cdots$

Therefore the coefficient of $a^ib^j$ in the expansion $[c^k]S$ above has the following interpretation: It is the number of ways that a bipartite number $(i,j)$ can be written as sum of $k+1$ bipartite numbers of the form $(u,0)$ and $(0,v)$ with $u,v\geq0$.

This can be achieved with simple combinatorics. Of the $k+1$ numbers, choose $m$ numbers for $i$. Distribute $i$ in those $m$ numbers such that each numbers gets at least $1$. Distribute $j$ in the remaining $k+1-m$ numbers.

Therefore,

$\displaystyle [a^ib^jc^k]S=\sum_{m=1}^i \binom{k+1}{m}\binom{i-1}{m-1}\binom{j+k-m}{j}$

There we have it!

But, wait a minute.. The final answer gives us a very simple combinatorial way of arriving at the answer. Just lay down all the $k$ $c$-cards. Now there will be a total of $k+1$ gaps between those cards. Choose $m$ among them and distribute the $i$ $a$-cards in those $m$ spaces such that each space gets at least one card. Now distribute the $j$ $b$-cards in the remaining $k+1-m$ spaces.

Knowing this, I feel stupid to have gone through the generating function route to get the solution. Grrr...

Anyway, to get the desired probability we use $i=j=4$ and $k=44$,

$\displaystyle \mathbb{P}(\text{No King and Queen adjacent})=\frac{4!4!44!\sum_{m=1}^4 \binom{44+1}{m}\binom{4-1}{m-1}\binom{4+44-m}{4}}{52!}$

which matches with the answer given in the quoted page.

One more small note before we conclude. Take any two Aces out of the deck. Now the probability of no King adjacent to any Queen in this pack of 50 cards is given by using $i=j=4$ and $k=42$ which is $\approx 0.499087$, surprisingly close to $1/2$. Very nearly a fair bet!

I'm convinced that the author of the page left the answer in a specific form as hint for someone attempting to solve the problem but it didn't help me. But I'm glad that I was able to derive a simpler form of the solution with some intuition on why the solution looks the way it does. Hope you enjoyed this discussion.

Clear["Global`*"];

SeriesCoefficient[1/(

  1 - a - b - c + (a b (a - 1 + b - 1))/(a b - 1)), {a, 0, 4}, {b, 0, 

   4}, {c, 0, 44}]/Multinomial[4, 4, 44]

1 - N[%]

1 - NSum[Binomial[44 + 1, m] Binomial[4 - 1, m - 1] Binomial[

     4 + 44 - m, 4], {m, 4}]/Multinomial[4, 4, 44]

SeriesCoefficient[(

  1 - a b)/((1 - a) (1 - b) - (1 - a b) c), {a, 0, 4}, {b, 0, 12}, {c,

    0, 36}]/Multinomial[4, 12, 36]

1 - N[%]

1 - NSum[Binomial[36 + 1, m] Binomial[4 - 1, m - 1] Binomial[

     12 + 36 - m, 12], {m, 4}]/Multinomial[4, 12, 36]

SeriesCoefficient[(

  1 - a b)/((1 - a) (1 - b) - (1 - a b) c), {a, 0, 4}, {b, 0, 16}, {c,

    0, 32}]/Multinomial[4, 16, 32]

1 - N[%]

1 - NSum[Binomial[32 + 1, m] Binomial[4 - 1, m - 1] Binomial[

     16 + 32 - m, 16], {m, 4}]/Multinomial[4, 16, 32]

Until then

Yours Aye

Me

Probability on a Contingency Table

Contingency tables are quite common in understanding a classification problems like that of a ML model or new drug tested against a disease. Given that we are just recovering from a pandemic, let's stick to the case of a Machine Learning model. In the context of ML models, it is called the Confusion matrix and we'll use both the terms interchangeably in this post.

A 2x2 contingency table usually has two columns for the binary classification (Win vs. Lose, Apple vs. Orange, White vs. Black etc.) and two rows for whether the prediction was right or wrong. Let's consider the classification as a 'Hypothesis' and the model's prediction as an 'Evidence' supporting it.

Here is how our contingency table would look like. 

Table 1	H	⌐H
E	n(H∩E)	n(⌐H∩E)
⌐E	n(H∩⌐E)	n(⌐H∩⌐E)

where $n(A)$ denotes the number of elements in set $A$.

We can normalize this table by dividing each of the four entries by the total thereby creating a new table.

Table 2	H	⌐H
E	$\mathbb{P}(H\cap E)$	$\mathbb{P}(\neg H\cap E)$
⌐E	$\mathbb{P}(H \cap \neg E)$	$\mathbb{P}(\neg H \cap \neg E)$

where we can view each entry as the probability of a classification falling in that bracket and $\mathbb{P}(A)$ denotes the probability of event $A$. Note that the sum of all the entries of Table 2 is 1.

The Wiki page on Confusion matrix gives a huge list of metrics that can be derived out of this table. In this post, we visit a couple of probability problems created from them.

Three of the important metrics that I learned in the context of ML from these matrices are

Precision, $\displaystyle p=\frac{\mathbb{P}(H\cap E)}{\mathbb{P}(H\cap E)+\mathbb{P}(\neg H\cap E)}=\frac{\mathbb{P}(H\cap E)}{\mathbb{P}(E)}=\mathbb{P}(H|E)$

Recall, $\displaystyle r=\frac{\mathbb{P}(H\cap E)}{\mathbb{P}(H\cap E)+\mathbb{P}(H\cap \neg E)}=\frac{\mathbb{P}(H\cap E)}{\mathbb{P}(H)}=\mathbb{P}(E|H)$

and Accuracy, $a=\mathbb{P}(H \cap E)+\mathbb{P}(\neg H \cap \neg E)$

Suppose, we want to create a random (normalized) confusion matrix. One way to do this would be create random variables all between 0 and 1, that also sum to 1. We can use Dirichlet distribution with four parameters to achieve this.

But there may be instances where we want to create a confusion matrix with a given precision, recall and accuracy. There are four entries in the table. Given three metrics and the fact that the entries should add upto 1 would seem to suggest, that these completely define the table.

But not all such values can create a valid table. For example, its impossible to create a valid table with a precision of 66.7%, recall 90.0% and accuracy 60%. So our first question is, given that precision, recall and accuracy are all uniformly distributed random variables, what is the probability that we will end up with a valid table.

To produce a valid table, the three variables need to satisfy the condition

$\displaystyle a \geq \frac{pr}{1-(1-p)(1-r)}$

Let $T$ be event of getting a valid table. Using the above we have,

$\displaystyle \mathbb{P}(T)=\int\limits_0^1\int\limits_0^1\int\limits_0^1\left[ a \geq \frac{pr}{1-(1-p)(1-r)} \right]\,dp\,dr\,da$

For a moment, let's assume we assume that $a$ is given. Then we first solve

$\displaystyle F(a)=\int\limits_0^1\int\limits_0^1\left[a \geq \frac{pr}{1-(1-p)(1-r)}\right]\,dp\,dr$

We can simplify the expression inside the Iverson bracket as a curve in the $r-p$ plane. The equation of the curve is given by

$\displaystyle r=\frac{ap}{(1+a)p-a}$

Plotting the region for $a=1/4$, we get the following graph.

The region to be integrated lies between the curve and the two axes. We can divide this region along the $p=r$ line. This line intersects the graph at $\left(\frac{2a}{1+a},\frac{2a}{1+a}\right)$. Therefore,

$\displaystyle F(a)=\frac{4a^2}{(1+a)^2}+2\int\limits_{\frac{2a}{1+a}}^1\frac{a p}{(1+a)p-a}\,dp$

Solving the second integral with Wolfram Alpha, we get,

$\displaystyle F(a)=\frac{4a^2}{(1+a)^2}+2\frac{a(1-a)}{(1+a)^2}-2\frac{a^2\log{a}}{(1+a)^2}=\frac{2a}{1+a}-\frac{2a^2\log{a}}{(1+a)^2}$

Plugging this back in our original equation and integrating, we see that,

$\displaystyle \mathbb{P}(T)=\int\limits_0^1\left(\frac{2a}{1+a}-\frac{2a^2\log{a}}{(1+a)^2}\right)\,da=4-\frac{\pi^2}{3}\approx 0.710132$

Thus we see that the just about 29% of the tables will not be valid. Something that truly surprised me here is the fact that $\pi$ makes an appearance here. There are no circles (not even something close to that) in this problem!!

The second problem is we see is also quite similar. Now, assume that we create tables (like that of Table 2) such that the values are uniformly distributed and sum to 1. If we want the precision and recall of our random tables to be greater than some threshold, what would be expected accuracy of the table?

For clarity, let $\mathbb{P}(H \cap E)=X$, $\mathbb{P}(\neg H \cap E)=Y$, $\mathbb{P}(H \cap \neg E)=Z$ and $\mathbb{P}(\neg H \cap \neg E)=W$, then $(X,Y,Z,W)\sim \text{Dir}(1,1,1,1)$

$\displaystyle \mathbb{E}(1-Y-Z|\mathbb{P}(E|H)\geq r,\mathbb{P}(H|E)\geq p)=\frac{1}{V}\int\limits_Q 1-y-z \,dx\,dy\,dz$

where $Q$ is the region such that

$Q=\{(x,y,z):(1-p)x\geq py \land (1-r)x\geq rz \land x+y+z \leq 1 \land x,y,z\geq 0\}$

and $V$ is the volume enclosed by $Q$.

Evaluating this integral is not so easy. The region integration depends on the value of $p$ and $r$ and it kind of ends in a mess of equations. But with some luck and a lot of Mathematica, we can see

$\displaystyle \mathbb{E}(\mathbb{P}(H \cap E)+\mathbb{P}(\neg H \cap \neg E)\text{  }|\text{  }\mathbb{P}(E|H)\geq r,\mathbb{P}(H|E)\geq p)=\frac{2(p+r)^2+p^3+r^3-pr(p^2+r^2)}{4(p+r)(1-(1-p)(1-r))}$

I have no way of making sense of that expression but, hey, we have an expectation on probabilities!

Hope you enjoyed this discussion.

Clear["Global`*"];

p = 200/1000; r = 250/1000;

ImpReg[a_, p_, r_] := 

  ImplicitRegion[

   y + z <= 1 - a && (1 - p) x >= p y && (1 - r) x >= r z && 0 <= x &&

     0 <= y && 0 <= z && x + y + z <= 1, {x, y, z}];

ImpRegap[a_, p_] := 

  ImplicitRegion[

   y + z <= 1 - a && (1 - p) x >= p y && 0 <= x && 0 <= y && 0 <= z &&

     x + y + z <= 1, {x, y, z}];

ImpRegpr[p_, r_] := 

  ImplicitRegion[(1 - p) x >= p y && (1 - r) x >= r z && 0 <= x && 

    0 <= y && 0 <= z && x + y + z <= 1, {x, y, z}];

ImpRegra[r_, a_] := 

  ImplicitRegion[

   y + z <= 1 - a && (1 - r) x >= r z && 0 <= x && 0 <= y && 0 <= z &&

     x + y + z <= 1, {x, y, z}];

ExpecAcc[p_, r_] := (-2 p^2 - p^3 - 4 p r + p^3 r - 2 r^2 - r^3 + 

   p r^3)/(4 (p + r) (-p - r + p r));

ExpecAcc[p, r]

N[%]

lim = 1000000;

cnt = 0;

val = 0;

Do[

  x = -Log[RandomReal[]]; y = -Log[RandomReal[]];

  z = -Log[RandomReal[]]; w = -Log[RandomReal[]];

  t = w + x + y + z;

  x /= t; w /= t; y /= t; z /= t;

  If[And[x/(x + y) >= p, x/(x + z) >= r],

   cnt += 1; val += 1 - y - z;

   ];

  , {i, lim}];

N[val/cnt]

Until then

Yours Aye

Me

Isochrone on a Spherical surface

The Cycloid in all its grandeur scooped all the glory for itself by way of being both the brachistochrone curve and the tautochrone curve.

But, seemingly out of some dark magic, the semi-cubical parabola makes its way as the (vertical) isochrone curve, a curve on which a bead sliding without friction covers equal vertical distances in equal intervals of time.

In my last post, we found the differential equation of a tautochrone curve on a spherical surface and used it to find the curve. In this post, we kind of continue the same discussion.

Our aim in this post is to find a curve on the surface of a sphere such that a bead sliding (without friction) under the influence of gravity covers equal polar angles at equal intervals of time. In other words, we are trying to find the (polar) Isochrone curve on the spherical surface.

For reasons that'll be apparent later, we modify our problem construct a little. Let's assume that the bead starts at the north pole and slides (without friction) slowly along the $\phi=0$ plane from $\theta=0$ to $\theta=\theta_0$ where the bead seamlessly enters the curve.

Let $\omega=d\theta/dt$ be the constant polar velocity as the bead enters the isochrone curve. Also, note that, because of the way the bead travels before entering the curve, the azimuthal velocity will be zero at the entry point.

Using the differential arclength of a curve on a  (unit) spherical surface, we have,

$\displaystyle \left(\frac{ds}{dt}\right)^2 =\left(\frac{d\theta}{dt}\right)^2+\sin^2\theta\left(\frac{d\phi}{dt}\right)^2=\left(\frac{d\theta}{dt}\right)^2\left(1+\sin^2\theta\left(\frac{d\phi}{d\theta}\right)^2\right)$

Let the plane tangent to the sphere at the south pole be the 'base' from which we measure the gravitational potential energy. Then, using the law of conservation of energy at $\theta=\theta_0$ (just after the bead entered the isochrone) and any point beyond,

$\displaystyle \omega^2\left(1+\sin^2\theta\left(\frac{d\phi}{d\theta}\right)^2\right)+2g(1+\cos\theta)=\omega^2+2g(1+\cos\theta_0)$

Simplifying this,

$\displaystyle \frac{d\phi}{d\theta}=\frac{\sqrt{2g}}{\omega}\frac{\sqrt{\cos\theta_0-\cos\theta}}{\sin\theta}$

It is easy to express $\omega$ in terms of $\theta_0$. Using the law of conservation of energy at $\theta=0$ and $\theta=\theta_0$ (just before the bead enters the isochrone), it is easy to see,

$\omega=\sqrt{2g(1-\cos\theta_0)}$

Using this in the expression above, we finally have,

$\displaystyle \frac{d\phi}{d\theta}=\frac{1}{\sin\theta}\sqrt{\frac{\cos\theta_0-\cos\theta}{1-\cos\theta_0}}$

(WARNING: Now is the time where I give you a fair warning to keep your mind in a secured place because it is about to be blown.)

Looking at the above differential equation, it is clear that it is the same equation we found in our previous post for the tautochrone problem. The same curve solves both the problems just like the cycloid does in plane geometry!!!

This says, if you place the bead at any point on our curve, the time it takes to reach the south pole is the same and hence becomes tautochrone. But if you place the bead at the north pole and let it slide itself into the curve, then it covers equal polar angles at equal intervals of times and hence becomes (polar) isochrone.

Truly, some dark magic stuff going on. I would have never expected in any way that such a weird coincidence would happen in the spherical case. I truly enjoyed this. Hope you did too. See you in the next post.

Until Then

Yours Aye

Me