Expected values with Bertrand's Paradox

Bertrand's paradox is a neat problem that shows what happens when we take the words 'at random' a bit too casually. In this post, we will be looking at some problems in the same setup and how the different methods yield answers for the 'seemingly' same question.

Bertrand's paradox asks for the probability of a 'random' chord in a unit circle being longer than the side of an inscribed triangle. We are interested in the expected length of such a chord. We now see how the expected length varies if we choose the chord according to the three methods in the paradox.

(1) Circumference method

In this method, we choose two points uniformly randomly on the circumference of the circle and connect those points to create the chord. After the first point is chosen, the second point is uniformly distributed in the circumference.

With this method, if we draw a tangent at the first point, then the angle between the chord and the tangent is uniform in the range $(0,\pi)$. If we let that angle be $\theta$, then the length of the chord is simply $2\sin\theta$. Therefore,

$\displaystyle\mathbb{E}(L)=\int_0^{\pi/2}2\sin\theta\,\frac{d\theta}{\pi/2}=\frac{4}{\pi}\approx 1.2732$

(2) Radial method

In this method, we first choose a random radius. We then choose a point uniformly on this radius and draw a chord perpendicular to this line. If we let $x$ be the length of the point from the origin, then the length of the chord is $2\sqrt{1-x^2}$.

Because $x$ is uniform in $(0,1)$, we have

$\displaystyle \mathbb{E}(L)=\int_0^12\sqrt{1-x^2}dx=\frac{\pi}{2}\approx1.5708$

But here's something interesting. If we use $x=\cos\theta$ in the above integral, the integral becomes,

$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}2\sin\theta\cdot\sin\theta\,d\theta$

From what we saw in the 'Circumference method', we know that the $2\sin\theta$ denotes the length of the chord. Hence, whatever remains after that must be the PDF of the variable $\theta$.

Therefore, the PDF, $f(\theta)$, for this case is $\displaystyle f(\theta)=\sin\theta$

(3) Midpoint method

Here, a point is chosen randomly in the circle and the unique chord which has this point as the midpoint is drawn. The expected length is slightly tricky here. The trick is to identify the parameter of interest here is the distance of the point from the origin.

If we let $r$ be that distance, then the length of the chord is $2\sqrt{1-r^2}$. But $r$ is not uniform. It is well known that the density of function of $r$ is $2r$ for a unit circle. We then have

$\displaystyle \mathbb{E}(L)=\int_0^12\sqrt{1-r^2}\cdot 2r\,dr=\frac{4}{3}\approx 1.3334$

If we use the substitution $r=\cos\theta$ in the above integral, we can see that the PDF of $\theta$ in this case is $f(\theta)=\sin2\theta$

(4) Random point and line at random angle method

The video also specifies another method of creating a chord which involves choosing a point randomly in the circle and drawing a line at a random angle through this point. To make things clear, let the $r$ be the distance between the point and other centre of the circle.

Now we draw a line perpendicular to the radial line containing this point. The angle, $t$, that the random line (that we draw to create the chord) makes with this perpendicular line is uniformly distributed between $(0,\pi)$.

It can shown then the distance of the random chord and the centre of the circle is then $r\cos t$. Therefore,

$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}\int_0^12\sqrt{1-(r\cos t)^2}\cdot 2r\,dr\frac{dt}{\pi/2}=\frac{16}{3\pi}\approx 1.69765$

The above integral was solved using Wolfram Alpha.

Here we use the substitutions $u=r\cos t$ and $v=r\sin t$. Then we know that $dudv=rdrdt$. The limits of $r$ and $t$ covers the first quadrant of the unit circle and so should the limits of $u$ and $v$. Therefore,

$\displaystyle \mathbb{E}(L)=\int_0^{1}\int_0^{\sqrt{1-u^2}}2\sqrt{1-u^2}\cdot 2\,\frac{dvdu}{\pi/2}=\int_0^{1}2\sqrt{1-u^2}\cdot 2\sqrt{1-u^2}\,\frac{du}{\pi/2}$

If we now use $u=\cos\theta$, then we see that

$\displaystyle f(\theta)=\frac{2\sin^2\theta}{\pi/2}$

(5) Random point on a radial line and a line at random angle method

Just for the sake of needlessly complicating things, we look at another method where we choose a point randomly in a randomly chosen radial line and draw a at a random angle through this point (the number of randoms in this sentence!). This is almost exactly like the previous case.

$\displaystyle \mathbb{E}(L)=\int_0^{\pi/2}\int_0^12\sqrt{1-(x\cos t)^2}\cdot \,dx\frac{dt}{\pi/2}=\frac{2}{\pi}(2G+1)\approx 1.80286$

where $G$ is the Catalan constant. Note that this integral can be expressed as the integral of the Elliptic integral of the second kind. With that and some known results, this can also be calculated manually.

Again using the subtitution $u=x\cos t$ and $v=x\sin t$ with the Jacobian $dudv=xdxdt$, we see 

$\displaystyle \mathbb{E}(L)=\int_0^{1}\int_0^{\sqrt{1-u^2}}2\sqrt{1-u^2}\cdot \,\frac{dvdu}{x\cdot\pi/2}$

But $x=\sqrt{u^2+v^2}$. Using this, the fact that the integral of the secant function is the inverse hyperbolic sine of the tangent function and the substitution $u=\cos t$, we can see that the PDF in this case as

$\displaystyle f(\theta)=\frac{\sin\theta\sinh^{-1}(\tan\theta)}{\pi/2}$

That sums up the Expected values. But we can do more. There was a recent Numberphile video on similar lines. The question discussed was to find the probability that two points selected at random in a circle is lies on different sides of a 'random' chord. Here's where the PDFs we've found so far will be really useful. We'll see those in the next post.

Hope you enjoyed this.

Until then

Yours Aye

Me

Candles and Cake problem - Probabilities with Bertrand's Paradox

The video I referred to in the earlier post only shows a Monte Carlo solution but we can find the exact values quite easily using the density functions.

Because the two candles (or points) are assumed to be uniformly randomly distributed in the circle (or the cake), an important parameter that keeps repeating in all the cases below is the area of a circular segment which can given in terms of the central angle $t$ as $(t-\sin t)/2$. The probability that a point randomly chosen in a unit circle lies in this segment is then $(t-\sin t)/2/\pi$.

Note that the density we found in the earlier refers are for the random variable $\theta$, the angle the chord makes with the tangent. The angle subtended by the chord is twice this value.

Let $E$ denote the event that two randomly chosen points on the circle lie on the opposite side of the chord chosen according to the following methods.

(1) Circumference method

Like before, if we let $\theta$ be the angle between the tangent and the chord, the central angle becomes $2\theta$. Therefore, the probability that a point randomly chosen in the circle lies in the segment created by this chord is the ratio of the segment's area to that of the circle. Therefore,

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\,\frac{d\theta}{\pi/2}=\frac{1}{3}-\frac{5}{4\pi^2}\approx 0.20668$

Even though we can solve this easily, Wolfram Alpha does the job perfectly.

(2) Radial line method

We know from our earlier post that the density function of the tangent angle in this case is just $\sin\theta$. Therefore the required probaility

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin\theta\,d\theta=\frac{128}{45\pi^2}\approx 0.28820$

This can also be simplified manually, but again Wolfram Alpha minimizes our effort and shows

Note that this is $\pi$ times the expected distance between two points chosen at random inside a circle. Coincidence?

(3) Midpoint method

Again, using the density from the previous post, the probability in this case is,

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin2\theta\,d\theta=\frac{1}{8}+\frac{2}{3\pi^2}\approx 0.19255$

Simple even without Wolfram Alpha.

(4) Random point and line at random angle method

Like the cases above we use the density function for this case that we already found in the previous post.

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot2\sin^2\theta\,\frac{d\theta}{\pi/2}=\frac{1}{3}\approx 0.33334$

Solved with Wolfram Alpha.

(5) Random point on a radial line and a line at random angle method

Diving right into the probability with the density function at our disposal,

$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin\theta\sinh^{-1}(\tan\theta)\,\frac{d\theta}{\pi/2}\approx 0.386408$

This time Wolfram Alpha only gives us a numerical approximation. But with Mathematica, we can confirm that the exact value is

$\displaystyle \mathbb{P}(E)=\frac{427+60\pi^2-480\log 2}{180\pi^2}$

For the sake of completion, let's finish this post with one more method that was used in the video. Here, the cut is created by choosing two random points on the circumference while the two candles were chosen randomly on a randomly chosen radius.

The first point on the circumference can be chosen to be the 'south pole' of the circle. After the second point is chosen on the circumference, let the chord subtend an angle of $2t$ at the centre. Using the symmetry of the problem, we can limit the range of $t$ to be in $(0,\pi/2)$.

We now calculate the probability that the first candle lies to the right of the chord and the second on the left. Let the radii selected for the first candle make an angle $x$ with the radius perpendicular to the chord. We get a a non-zero probability only if $-t \leq x \leq t$. When $x$ lies in this range, the probability of the first candle landing to the right of the cut is $1-\cos t/\cos x$ and $0$ otherwise.

Similarly, let $y$ be the angle between the radius chosen for the second candle and the radius perpendicular to the cut. When $-t \leq y \leq t$, the probability of the second candle ending up to the left of the cut is $\cos t/\cos y$. Else, the probability is $1$.

Therefore probability in this case depends on the following two integrals.

$\displaystyle \mathcal{I}_1=\int_0^{\pi/2} \int_{-t}^t \int_{-t}^t \frac{\cos t}{\cos y}\left(1-\frac{\cos t}{\cos x}\right)\,dy\,dx\,dt$

$\displaystyle \mathcal{I}_2=\int_0^{\pi/2} \int_{-t}^t \int_{t}^{2\pi-t} \left(1-\frac{\cos t}{\cos x}\right)\,dy\,dx\,dt$

These are very hard to evaluate even with Mathematica. Numerically evaluating these shows,

$\displaystyle \mathbb{P}(E)=2\cdot\frac{\mathcal{I}_1+\mathcal{I}_2}{\pi/2\cdot 2\pi \cdot 2\pi}\approx 0.161612$

Hope you enjoyed this.

Until then

Yours Aye

Me

The Problem of points - Without Replacement

Consider an Urn with $A$ green balls and $B$ red balls. Two players A and B play a game where they draw balls from this Urn one at a time (without replacement). If the drawn ball is green, Player A gets a point. Else, Player B gets a point.

If Player A needs $a$ points to win the game whereas Player B needs $b$ points, the question that we address in this post concerns the probability that Player A wins the match.

As the title says, this is just the Problem of points but without replacement. We've already encountered the problem in our blog in the post titled A note on Partial fractions where we saw how Pascal and Fermat independently solved the problem with different reasoning.

Interestingly, Pascal reasoning involves the (CDF of) Neg. Binomial distribution whereas Fermat's idea involves the (CDF of) Binomial distribution. The central insight in both their ideas is in realising that we'll need a maximum of $a+b-1$ games to conclude the match.

Let's consider the original Problem for points for a moment. If we let $p$ and $q$ represent the probability of A and B winning a single game, then using Pascal's idea,

$\displaystyle \mathbb{P}(\text{A wins the match})=\sum_{k=0}^{b-1}\binom{k+a-1}{k}p^aq^k$

where the individual terms are probabilities from the Neg. Binomial distribution, $NB(a,p)$.

Using Fermat's idea,

$\displaystyle \mathbb{P}(\text{B wins the match})=\sum_{k=b}^{a+b-1}\binom{a+b-1}{k}q^kp^{a+b-1-k}$

where the individual terms are probabilities from the Binomial distribution, $B(a+b-1,p)$.

Now using our discussion about the Neg. Hypergeometric distribution in the previous case and Pascal's idea, we could solve the Without-Replacement case of the Problem of points. Or we could go for simplicity and use the Hypergeometric distribution and Fermat's idea.

Either way, let $P(A,a,B,b)$ be the probability Player A winning the game in the setup described at the start of the post and $M$ be a very large number. It should be clear the classical case can be derived out of this setup as $P(pM,a,qM,b)$ with $p+q=1$.

Obviously, when $A=B$ and $a=b$, either player has an equal chance of winning because of symmetry. Similarly, when $A=B$ and $a<b$, Player A has an advantage. Same goes when $a=b$ and $A>B$.

But consider $p(50,25,40,20)$. Player A has 10 more balls than Player B but needs 5 more points to win the game. It isn't immediately clear whether this is advantageous to Player A or not? In this case, the probability of Player A winning the match is only (approx.) 49.07%.

Naturally, we might think as we break symmetry, the match skews towards one way. Perhaps the most surprising result about the Without-Replacement case of Problem of points is the one family of parameters that form an exception.

Consider $p(21, 11, 19, 10)$. Player A has two more balls than Player B but needs one extra point to win the match. Surprisingly, this match is evenly poised. Not just this, for positive integer $n$,

$\displaystyle p(2n+1,n+1,2n-1,n)=\frac{1}{2}$

Clearly, the parameters that define the game lacks symmetry. Still the game is evenly poised between the two players. I could neither come up with a reason nor a simple explanation of why this should be the case? If you can, please post it in the comments below.

Clear["Global`*"];
pp[A_, a_, B_, b_] := (A / (A + B)) Binomial[A - 1, a - 1] Sum[Binomial[B, j] / Binomial[A + B - 1, j + a - 1], {j, 0, b - 1}];
p[A_, a_, B_, b_] := Sum[Binomial[j + a - 1, j] Binomial[A - a + B - j, B - j], {j, 0, b - 1}] / Binomial[A + B, B];
k = 2;
ListPlot[Table[p[2n + k, n + k, 2n - k, n], {n, 100}]]
p[50, 25, 40, 20]
N[%]
p[21, 11, 19, 10]
N[%]
p[26, 13, 22, 12]
N[%]

Until then
Yours Aye
Me

Expected distance between two points inside a Sphere

This post picks up where we left earlier. As discussed before, we want to find the expected distance between two points randomly selected in an $n$-sphere.

Let the two points be $P$ and $Q$, and $O$ be the centre. The first step is to apply the technique of projective Reduction with $O$ as the scaling point. This shows that the expected length between two points inside a $n$-sphere is $2n/(2n+1)$ times the expected length between a point on the surface of $n$-sphere and another point inside the $n$-sphere.

WLOG, let $P$ be the point on the surface and let's orient the sphere so that point $P$ becomes the south pole. The next step is apply another Projective Reduction with the south pole as the scaling point. This step gives another factor of $n/(n+1)$ for the expected length. Note that while the first reduction preserves the uniformity of the points on the respective 'surfaces', it is not the case with the second reduction. This is what makes the problem tricky.

To find the density of $Q$, consider a random chord from point $P$ (the south pole) along with the axis on which $P$ lies. Let $\theta$ be the angle between the chord and the axis.

Rotating the chord along this axis cuts out a (for a lack of a better word) spherical-cone which contains a cone with its apex at the south pole and a spherical cap at the top (Imaging the other end of the chord to be on the opposite side of $P$). With simple geometry, it is also apparent that the angle between $OQ$ and the axis is $2\theta$.

If we let $V_{\text{sp. cone}}(\theta, r)$ be the volume of the spherical cone from a $n$-sphere of radius $r$ and apex angle $\theta$, then we have,

$\begin{align} \displaystyle V_{\text{sph. cone}}(\theta, r) &= V_{\text{cone}} + V_{\text{sph. cap}}\\&=  \frac{1}{n}V_{n-1}(r \sin 2\theta)(r+r\cos2\theta)+\int_{0}^{2\theta}V_{n-1}(r \sin t) \,d(-r\cos t)\\ &= r^n\frac{v_{n-1}}{n}\sin^{n-1}2\theta \cdot 2  \cos^2\theta + r^nv_{n-1}\int_0^{2\theta}\sin^nt\,dt \\ &=\frac{2^nr^nv_{n-1}}{n}\sin^{n-1}\theta\cos^{n+1}\theta + + r^nv_{n-1}\int_0^{2\theta}\sin^nt\,dt \end{align}$

We have used the fact that the area of an $n$-dimensional cone is $1/n$ times the volume of the circumscribing 'cylinder'. For the volume of the cap, we have used that the volume is the sum of the areas of $(n-1)$ dimensional 'circles' with infinitesimal thickness (see here for more). 

Now, $\frac{\partial V_{\text{sph. cone}}}{\partial r \partial \theta}$ gives volume between $V_{\text{sph. cone}}(\theta + d\theta, r+dr)$ and $V_{\text{sph. cone}}(\theta, r)$. When $Q$ lies in this infinitesimal volume, the distance between $P$ and $Q$ is just $2\cos\theta$.

Because we are only interested in distribution of $Q$ on the surface of the $n$-sphere, we can integrate out the $r$. Therefore, the density is given by,

$\displaystyle f(\theta)=\frac{1}{v_n}\int_{0}^1\frac{\partial V_{\text{sph. cone}}}{\partial r \partial \theta}\,dr$

We now get,

$\displaystyle f(\theta)=\frac{2^nv_{n-1}}{nv_n}((n-1)\cos^{n+2}\theta\sin^{n-2}\theta-(n+1)\sin^n\theta\cos^n\theta)+\frac{v_{n-1}}{v_n}\cdot 2\sin^n2\theta$

where we have used Leibniz differential under integral sign for the last term. Further simplification then gives,

$\displaystyle f(\theta)=\frac{2^nv_{n-1}}{nv_n}(n-1)\sin^{n-2}\theta\cos^n\theta$

This nice simplification allows us to express the expected length $e_n$ in terms of Beta function.

$\displaystyle e_n=\int_0^{\pi/2}2\cos\theta\cdot f(\theta)\,d\theta=\frac{2^n}{n}\frac{v_{n-1}}{v_n}(n-1)B(n/2-1/2, n/2+1)$

Finally, we see that the expected length $L$ between two points randomly selected inside an $n$-sphere is,

$\displaystyle \mathbb{E}(L)=\frac{2n}{2n+1}\frac{n}{n+1}\frac{2^n(n-1)}{n}\frac{B(n/2-1/2,n/2+1)}{B(n/2+1/2,1/2)}$

Interestingly, at $n\to \infty$, irrespective of whether we choose points on the surface or the inside, the expected length goes to $\sqrt{2}$. I can't clearly see why but if you have an idea, please do share it with us.

UPDATE 3 Jul 2023: Generalising this, we can see that

$\displaystyle \mathbb{E}(L^k)=\frac{2n}{2n+k}\frac{n}{n+k}\frac{2^{n+k-1}(n-1)}{n}\frac{B(n/2-1/2,n/2+k/2+1/2)}{B(n/2+1/2,1/2)}$

Using this, we can see, using WA, that

$\displaystyle \lim_{n \to \infty}\mathbb{E}(L^k)=\sqrt{2}^k$

Also, using the limit definition of the logarithm function,

$\displaystyle \mathbb{E}(\ln L)=-\frac{1}{n}+\frac{n}{2}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k(n+k)}$

This shows us the geometric mean of distance between two points chosen uniformly randomly from a unit circle is $e^{-1/4}$ and that of the sphere is $2e^{-3/4}$

Until then

Yours Aye

Me

Expected distance between two points on the Surface of a sphere

Problems in Geometric Probability are always fascinating. In this post, we going to see two of the well known and challenging problems in this topic.

The first problem is about the expected distance between two points selected at random from a surface of an $n$-sphere. To clarify, I'm considering a circle to be a $2$-sphere, a sphere to be a $3$-sphere and so on.

This problem is relatively easy as can seen from this stackexchange post. Let $S_n(R)$ and $V_n(R)$ denote the surface area and the volume of an $n$-sphere with radius $R$. We know that,

$S_n(R)=s_nR^{n-1}$ and $V_n(R)=v_nR^n$

where $s_n=\dfrac{2\pi^{n/2}}{\Gamma(n/2)}$ and $v_n=\dfrac{2}{n}\dfrac{\pi^{n/2}}{\Gamma(n/2)}$

For the problem under consideration, we can choose one of the points to be on the south pole of the $n$-sphere. Consider one of the axis that connects this point to the origin. A hyperplane perpendicular to this axis cuts the surface of the $n$-sphere to give a $(n-1)$-spherical surface. 

If we consider multiple such hyperplanes parallel to one another along the axis, the $n$-spherical surface is sliced into multiple rings. The picture below shows this in case of a $3$-sphere (taken from this Stack Exchange post).

Because second point is uniformly distributed in the $n$-spherical surface, it is equally likely to be present in any of these rings.

Let $\theta$ be the angle subtended between the radius that contains the second point and radius containing the first point. Then the distance between the points is $2\sin(\theta/2)$.

If we consider these rings to be infinitesimally small, then the density function of the second point lying in any of these rings is equal to the area of the ring. But this is easy. This is just the surface area of the  sliced $(n-1)$-hypersphere times $rd\theta$ (obviously $r=1$ for the unit sphere; used here for illustrative purposes). Therefore,

$\displaystyle \mathbb{E}(L)=\frac{1}{s_n}\int_0^\pi 2\sin(\theta/2)\cdot S_{n-1}(r\sin\theta)\cdot r\,d\theta=\frac{s_{n-1}}{s_n}\int_0^\pi 2\sin(\theta/2)\cdot \sin^{n-2}\theta \,d\theta$

Using Wolfram Alpha, we then see that

$\displaystyle \mathbb{E}(L)=\frac{\Gamma(n/2)}{\sqrt{\pi}\Gamma((n-1)/2)}\cdot \frac{2\sqrt{\pi}\Gamma(n-1)}{\Gamma(n-1/2)}=2\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\cdot \frac{\Gamma(n-1)}{\Gamma(n-1/2)}$

Now, the second problem asks for the expected distance between a point randomly selected on the $n$-spherical surface and a point selected on the $n$-spherical volume. This problem is more involved and requires a little more trickery. We'll see this in the next post.

Until then

Yours Aye

Me

Notes on the Hypergeometric Distribution and its Negative

I wanted to create a post about the Hypergeometric distribution, one of the most important distributions in probability theory, as a prelude to my next post.

The probability mass function of a Hypergeometric distribution with parameters $N$, $K$ and $n$ is given by

$\displaystyle f(k;N,K,n)=\mathbb{P}(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$

The distribution has several symmetries which will be the key content of this post.

To start with, the role of successes and failures could be swapped. In terms of probability, this means that $f(k;N,K,n)=f(n-k;N,N-K,n)$. The binomial coefficients in the numerator of $f(k;N,K,n)$ gets swapped with this transformation.

In the next setup, the role of drawn and not drawn elements could be swapped. In terms of probability, this gives $f(k;N,K,n)=f(K-k;N,K,N-n)$. This amounts to invoking the symmetry of the binomial coefficient $^nC_k=^nC_{n-k}$.

Finally, the role of drawn elements and successes could be swapped as well. In probability terms, this becomes $f(k;N,K,n)=f(k;N,n,K)$. In other words,

$\displaystyle \mathbb{P}(X=k)=\binom{n}{k}\frac{\binom{N-n}{K-k}}{\binom{N}{K}}$

I find this pretty for two reasons. Firstly, there are two 'nCk' terms with the lowercase pair in the numerator and the uppercase in the denominator which gives it a nice structure. Secondly, and most importantly, this brings out the parallel with the Binomial distribution.

Comparing the probability of $k$ successes which we have above with that of a $\text{Bin}(n,k)$, we have,

$\displaystyle \frac{\binom{N-n}{K-k}}{\binom{N}{K}} \approx \left(\frac{K}{N}\right)^k\left(1-\frac{K}{N}\right)^{n-k}$

where $N \gg n$ and ratio of $K$ and $N$ is finite. This can be seen as an asymptotic expression for the ratio of two binomial coefficients.

To make better use of this asymptotic expression, let $K=pN$ and $p+q=1$. Then, the above can be rewritten as

$\displaystyle p^iq^j \approx \frac{\binom{N-i-j}{K-i}}{\binom{N}{K}}$

Lastly, all the above symmetries can be brought under a single umbrella by using beautiful idea from Duality and Symmetry in the Hypergeometric distribution. Surprisingly, this idea is relatively unknown. The authors show that the probability associated with the Hypergeometric distribution can also be written as,

$\displaystyle \mathbb{P}(X=k)=\binom{N}{k,n-k,K-k}\bigg/ \binom{N}{n}\binom{N}{K}$

where the numerator should be understood as the multinomial coefficient.

Finally, we come to visit the lesser known relative of the Hypergeometric distribution - the Negative Hypergeometric distribution. Just like the Hypergeometric parallels the Binomial, the Neg. Hypergeometric parallels the Neg. Binomial distribution.

We could use the exact idea of Neg. Binomial to derive the probability expression for Neg. Hypergeometric. But considering what we've seen so far, we could do better.

Let $Y$ be a Neg. Binomial variable that counts number of failures encountered until the $r$th success and $W$ be the corresponding Neg. Hypergeometric variable. We know that

$\displaystyle \mathbb{P}(Y=k)=\binom{k+r-1}{k}p^rq^k$

Now we just have to make use of the asymptotic expression above to derive the same for the Neg. Hypergeometric case. We have,

$\displaystyle \mathbb{P}(W=k)=\binom{k+r-1}{k}\frac{\binom{N-k-r}{K-r}}{\binom{N}{K}}$

Neat, isn't it?

Until then
Yours Aye
Me

Matt Parker and the London Lottery

In his (aptly) named video 'Matt Explains: The Lottery', Matt explains the probabilities associated with the London lottery. In the first video he made on this topic, he worked out the probability of winning the Jackpot with the new system where we have a chance to win an extra lottery.

To make things clear, here are the winnings that I understood from his video. Six numbers are drawn (without repetition) from a pool of 59 numbers. If all 6 numbers match, then we win the Jackpot. If we match 5 numbers, then we get a bonus number whose match win us 50000 pounds and 1000 pounds otherwise. If we match 4 numbers, we win a 100 pounds and 3 matches wins us 25 pounds. Finally, under the new system, matching 2 numbers gives us no cash prize but gives us a new lottery which we can use the next time.

The number of matches are given by the Hypergeometric distribution as explained (and calculated) by Matt in the video. In his first video, Matt ignores all cash prizes and focuses only on the Jackpot and the extra lottery.

If we let $p_k$ denote the probability of $k$ matches and $p$ denote the probability of ultimately winning the Jackpot, we have

$p=p_6+p_2p$ which upon simplification gives $\displaystyle p = \frac{p_6}{1-p_2}$

Substituting the known values, gives a probability of $p=2.459111 \times 10^{-8}$ (or) an Odds of 1 in 40665098.

As we will be dealing with extremely small probabilities, let's define 1 micro-bps as a millionth of a basis point. In this sense, the probability in the above case is approx. 245.91 micro-bps.

At this point, as per a Subscriber's suggestion, Matt tries to attack the problem where we repeatedly buy lotteries using all our winnings until we win the Jackpot. Here Matt calculates the expected number of lotteries won with a given lottery.

If we let $l$ denote the expected number of lotteries won per lottery, then

$l=p_2 \cdot 1 + p_3 \cdot 12.5 + p_4 \cdot 50 + p_{5,1} \cdot 500 + p_{5,2} \cdot 25000 \approx 0.257191$

where we have used the fact that each lottery cost 2 pounds. Matt now uses the following equation (indirectly), to get the probability of winning the Jackpot.

$p=p_6+l \cdot p \tag{1} \label{wrongeqn}$ which gives $\displaystyle p = p_6/(1-l)$.

Using the known values, we get $p=298.7830655$ micro-bps (or) an Odds of about 1 in 33469098.

But something is definitely amiss in $\eqref{wrongeqn}$. For example, we can easily construct a case where $l > 1$ which will then give negative probabilities. What then is the error, if any, in Matt's reasoning? Or is there a hidden assumption that was not explicitly stated here?

To understand whats going on here, we should use idea that the probability of getting at least one win in $n$ trials is $1-(1-p)^n$ and interpret $p$ as the probability of winning atleast one Jackpot. Using this, a better expression is given by,

$p=p_6+(1 - p_6)(1 - (1 - p)^\bar{l}) \tag{2} \label{crudeeqn}$

where $\bar{l}=l/(1-p_6)$ is the expected number of lotteries won conditional on not winning the Jackpot directly. The above equation then means that we either win the Jackpot or get some lotteries to continue our pursuit of winning the Jackpot. Not so surprisingly, this equation can be solved in closed form. We have,

$\displaystyle p = 1 - (1 - p_6)^{1 / (1 - \bar{l})} \tag{3} \label{cruderes}$

This gives a probability of $p=298.7830666$ micro-bps (or) an Odds of about 1 in 33469097.

If we assume $p \approx 0$ in $\eqref{crudeeqn}$, then using the approximation $(1+x)^n \approx nx$ for small $x$, we end up with Matt's result. This clearly shows the hidden assumption in result derived by Matt.

But even $\eqref{crudeeqn}$ is still not exactly correct. This can still result in a value that is less than 0 or greater than 1 in certain cases which doesn't make any sense.

This problem occurs because, in the RHS of $\eqref{crudeeqn}$, we casually make an assumption that $\mathbb{E}(1-(1-p)^L)=1-(1-p)^{\mathbb{E}(L)}$ where $L$ is the random number of lotteries won (conditional on not winning the Jackpot). But this is clearly wrong because of Jensen's Inequality.

Therefore, the correct way to do this would be to solve the following equation.

$p=p_6+p_{5,2}(1-(1-p)^{25000})+p_{5,1}(1-(1-p)^{500})+p_4(1-(1-p)^{50})+p_3(1-(1-p)^{12.5})+p_2(1-(1-p))$

where we explicitly condition on the resulting coupons. Solving this, we get $p=298.782539$ micro-bps (or) an Odds of about 1 in 33469156. 

This is good but there is still one final piece of hiccup here. In case of 2 matches, we get 25 pounds which we have assumed is equal to 12.5 lotteries. But non-integer tickets doesn't make much sense.

In case 2 matches wins us 24 pounds, then we would have $p=296.70581679$ micro-bps (or) an Odds of about 1 in 33703416. Had it been 26 pounds, then we would have $p=300.888537999$ micro-bps (or) an Odds of about 1 in 33234897.

The precise way to model this is to split the problem into cases: one in which we have just a lottery and the other in which we have the lottery and a pound. If we let $p$ and $p_e$ be the probabilities of winning at least one Jackpot in the respective cases, then we have

$p=p_6+p_{5,2}(1-(1-p)^{25000})+p_{5,1}(1-(1-p)^{500})+p_4(1-(1-p)^{50})+p_3(1-(1-p_e)(1-p)^{11})+p_2(1-(1-p)) \tag{4} \label{firstsimul}$

and

$p_e=p_6+p_{5,2}(1-(1-p_e)(1-p)^{24999})+p_{5,1}(1-(1-p_e)(1-p)^{499})+p_4(1-(1-p_e)(1-p)^{49})+p_3(1-(1-p)^{13})+p_2(1-(1-p_e)) \tag{5} \label{secondsimul} $

Solving this simultaneous equations gives, $p=296.75282794$ micro-bps (or) an Odds of about 1 in 33698077 and $p_e=300.13462527$ micro-bps (or) an Odds of about 1 in 33318380.

These equations have to be solved numerically which may be compute-intensive. A compromise would be let Jensen's inequality slide and consider the following equations.

$p=p_6+(1-p_6)(1-(1-p)^{\bar{l}_{11}}(1-p_e)^{\bar{l}_{12}})$ and $p_e=p_6+(1-p_6)(1-(1-p)^{\bar{l}_{21}}(1-p_e)^{\bar{l}_{22}})$

where $l_{11}=25000p_{5,2}+500p_{5,1}+50p_4+11p_3+p_2$, $l_{12}=p_3$, $l_{21}=24999p_{5,2}+499p_{5,1}+49p_4+13p_3$ and $l_{22}=p_{5,2}+p_{5,1}+p_4+p_2$ are the (unconditional) expected lotteries and lottery-pounds in the respective cases, and $\bar{l}_k=l_k/(1-p_6)$ for $k \in \{11, 12, 21, 22\}$. 

Like before, this has the advantage of being solved in closed form. We have,

$p=1-(1-p_6)^{\bar{m}}$ and $p_e=1-(1-p_6)^{\bar{m}_e}$

where $\displaystyle \bar{m} = \frac{1-\bar{l}_{22}+\bar{l}_{12}}{(1-\bar{l}_{11})(1-\bar{l}_{22})-\bar{l}_{12}\bar{l}_{21}}$ and $\displaystyle \bar{m}_e=\frac{1-\bar{l}_{11}+\bar{l}_{21}}{(1-\bar{l}_{11})(1-\bar{l}_{22})-\bar{l}_{12}\bar{l}_{21}}$

Using the known values, we get $p=296.7533437$ micro-bps and $p_e=300.1351482$ micro-bps. Like before, if we assume $p_6$ is also small, then we get an even simpler linear relation.

Or we could assume both $p$ and $p_e$ are small in $\eqref{firstsimul}$ and $\eqref{secondsimul}$, which will give us a pair of linear simultaneous equations. Solving them would give us,

$\displaystyle p = mp_6$ and $\displaystyle p_e = m_ep_6$

where $m$ (and $m_e$) is the same as $\bar{m}$ (and $\bar{m}_e$) except with unconditional expectations. Either way, these approximations become susceptible to absurd probabilities if not applied carefully.

Now, for the next part, let's say we solve everything exactly. If we have 4 pounds, then we could buy 2 lotteries. If we assume that we don't mix the proceeds from on with the other, then the probability of winning atleast one Jackpot is $1-(1-p)^2$ which with small value approximation gives $2p$. Similarly, With 5 pounds, the probability of winning atleast one Jackpot is $p+p_e$.

But with 6 pounds, we have a choice. We could either play three individual lotteries with winning probability $3p$ or two lottery-pounds with winning probability $2p_e$. With the values at hand, playing three individual lotteries is the clear choice.

This shows that it is better to convert all cash into lotteries to the fullest extent possible. Therefore, for integer $n$, we have,

$p_{2n}=np$ and $p_{2n+1}=(n-1)p+p_e$

All of the above were calculated using the following Mathematica code.

Clear["Global`*"];

e1 = Binomial[6, 1] Binomial[53, 5] / Binomial[59, 6];

e2 = Binomial[6, 2] Binomial[53, 4] / Binomial[59, 6];

e3 = Binomial[6, 3] Binomial[53, 3] / Binomial[59, 6];

e4 = Binomial[6, 4] Binomial[53, 2] / Binomial[59, 6];

e51 = Binomial[6, 5] Binomial[53, 1] / Binomial[59, 6] * 52 / 53;

e52 = Binomial[6, 5] Binomial[53, 1] / Binomial[59, 6] * 1 / 53;

e6 = Binomial[6, 6] Binomial[53, 0] / Binomial[59, 6];

f[p_] := e6 + e52 (1 - Power[1 - p, 25000]) + e51 (1 - Power[1 - p, 500]) + e4 (1 - Power[1 - p, 50]) + e3 (1 - Power[1 - p, 25 / 2]) + e2 p - p;

g[p1_, p2_] := {e6 + e52 (1 - Power[1 - p1, 25000]) + e51 (1 - Power[1 - p1, 500]) + e4 (1 - Power[1 - p1, 50]) + e3 (1 - (1 - p2) Power[1 - p1, 11]) + e2 p1,

e6 + e52 (1 - (1 - p2) Power[1 - p1, 24999]) + e51 (1 - (1 - p2) Power[1 - p1, 499]) + e4 (1 - (1 - p2) Power[1 - p1, 49]) + e3 (1 - Power[1 - p1, 13]) + e2 p2};

(* NSolve[p == e6 + e52 (1 - Power[1 - p, 25000]) + e51 (1 - Power[1 - p, 500]) + e4 (1 - Power[1 - p, 50]) + e3 (1 - Power[1 - p, 12]) + e2 p && 0 < p < 1, p] 

NSolve[p == e6 + e52 (1 - Power[1 - p, 25]) + e51 (1 - Power[1 - p, 5]) + e4 (1 - Power[1 - p, 3]) + e3 (1 - Power[1 - p, 2]) + e2 p && 0 < p < 1, p] *)

(* n = 1; nmax = 1000;

a = N[0, 50]; b = N[1 / 1024 / 1024, 50]; tol = Power[10, -75];

While[n <= nmax,

    c = (a + b) / 2; temp = f[c];

    If[Or[(b - a) / 2 < tol, Abs[temp] < tol tol], Break[];];

    n += 1;

    If[temp > 0, a = c;, b = c;];

];

c *)

(* With only Jackpot and extra lottery *)

p = e6 / (1 - e2);

N[p, 15]

Floor[(1 - p) / p]

(* Matts Solution *)

l = e2 + (25 / 2) e3 + 50 e4 + 500 e51 + 25000 e52;

p = e6 / (1 - l);

N[p, 20]

Floor[(1 - p) / p]

(* Ignoring Jensens *)

lb = l / (1 - e6);

p = 1 - Power[1 - e6, 1 / (1 - lb)];

N[p, 20]

Floor[(1 - p) / p]

(* FindRoot[f[p1], {p1, 1}, WorkingPrecision -> 25] *)

(* With 25 units *)

p = 2.9878253945 Power[10, -8];

N[p, 20]

Floor[(1 - p) / p]

(* With 24 units *)

p = 2.9670581679 Power[10, -8];

N[p, 20]

Floor[(1 - p) / p]

(* With 26 units *)

p = 3.00888537999 Power[10, -8];

N[p, 20]

Floor[(1 - p) / p]

(* With 25 units but with extra lottery*)

(* FindRoot[g[p1, p2] == {p1, p2}, {p1, 1}, {p2, 1}, WorkingPrecision -> 25] *)

p = 2.9675282794 Power[10, -8];

Floor[(1 - p) / p]

p = 3.0013462527 Power[10, -8];

Floor[(1 - p) / p]

(* Ignoring Jensens *)

l11 = 25000 e52 + 500 e51 + 50 e4 + 11 e3 + e2; l12 = e3;

l21 = 24999 e52 + 499 e51 + 49 e4 + 13 e3;l22 = e52 + e51 + e4 + e2;

l11b = l11 / (1 - e6); l12b = l12 / (1 - e6); l21b = l21 / (1 - e6); l22b = l22 / (1 - e6);

bdeno = (1 - l11b) (1 - l22b) - l12b l21b;

mb = (1 - l22b + l12b) / bdeno; meb = (1 - l11b + l21b) / bdeno;

p = 1 - Power[1 - e6, mb];

N[p, 15]

Floor[(1 - p) / p]

pe = 1 - Power[1 - e6, meb];

N[pe, 15]

Floor[(1 - pe) / pe]

(* Small value approximations *)

deno = (1 - l11)(1 - l22) - l12 l21;

N[e6 (1 + (l12 - l22)) / deno, 10];

N[e6 (1 + (l21 - l11)) / deno, 10];

m = (1 - l22 + l12) / deno;

me = (1 - l11 + l21) / deno;

N[e6 m, 10]

N[e6 me, 10]

(* temp[n_] := temp[n] = If[n == 2, p, pe];

Calc[n_] := Module[{res = {}, m = IntegerPartitions[n, All, {2, 3}]},

    If[n < 2, Return[0];];

    Do[

        res = Join[res, {1 - Times @@ (1 - temp /@ k)}];

    , {k, m}];

    Max[res]

];

Power[10, 10] Table[Calc[n], {n, 30}] *)

Hope you enjoyed the discussion. See you in the next post.

Yours Aye

Until then

Me