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

Some more problems on a Deck of cards

We already saw how a standard deck of cards forms a nice setting for studying both Coupon Collector problem without replacement and Birthday problem without replacement. Here we extend those to some more problems for a fun!

The first problem is to ask for the expected number of cards such that we have atleast one matching suit or one matching rank. This problem is relatively easy. We proceed using the same idea as before.

Let $X$ be the random variable denoting the number of cards need to get atleast one matching suit or rank. Then, the probability that we need more than $k$ cards to get this,

$\displaystyle \mathbb{P}(X>k)=\frac{k!\binom{13}{k}\binom{4}{k}}{\binom{52}{k}}$

That is, we are finding the probability that we neither have a mathcing suit nor a matching rank after we have drawn $k$ cards without replacement. Therefore,

$\displaystyle \mathbb{E}(X)=\sum_{k>0}\mathbb{P}(X>k)$

So, we need, on average, $\approx 3.0799$ cards to get either a matching suit or a matching rank when drawing cards without replacement from a randomly shuffled standard deck of cards.

To simplify text, let $S_k$ denote the event of having atleast one matching suit after having drawn $k$ cards without replacement and $R_k$ denote the same event for rank. Then, we know $\mathbb{P}(S^c_k)$, $\mathbb{P}(R^c_k)$, and $\mathbb{P}(S^c_k\text{ and }R^c_k)$.

Using Inclusion-Exclusion principle, we have

$\displaystyle \mathbb{P}(S^c_k \text{ or } R^c_k)=\mathbb{P}(S^c_k)+\mathbb{P}(R^c_k)-\mathbb{P}(S^c_k\text{ and }R^c_k)$

This allows to calculate the expected number of cards need to get both a matching suit and a matching rank when drawing cards without replacement. If $X$ denotes this random variable, then

$\mathbb{P}(X>k)=1-\mathbb{P}(S_k\text{ and }R_k)=\mathbb{P}(S^c_k \text{ or } R^c_k)$

This shows that

$\displaystyle \mathbb{E}(S \text{ and } R)=\mathbb{E}(S)+\mathbb{E}(R)-\mathbb{E}(S\text{ or }R)\approx 3.2678 + 5.6965 - 3.0799=5.8844$

Therefore, we need, on average, $\approx 5.8844$ cards to get both a matching suit and a matching rank when drawing cards without replacement from a randomly shuffled standard deck of cards.

Now we ask a similar question in the Coupon collector problem. That is, we seek the expected number of cards needed to collect all suits and all ranks when drawing cards without replacement from a well shuffled pack of cards.

Like we did before, we will be using idea of negative Geometric random variable and the Min-Max identity. Let $X_k$ ($k \in \{\text{Spades}, \text{Clubs}, \text{Hearts}, \text{Diamonds}\}$)be the random number of draws needed without replacement to get first suit. Similarly, let $Y_k$ for $1\leq k \leq 13$ be the corresponding random variables for ranks.

If we let $Z$ denote the random number of draws needed to collect all suits and ranks, and $A$ be the set $\{X_S, X_C,\cdots,Y_1,Y_2,\cdots,Y_{13}\}$, then by the Min-Max identity, we know

$\displaystyle \mathbb{E}(Z)=\mathbb{E}(\text{max }A)=\sum_{M \subset_\emptyset A}(-1)^{|M|}\mathbb{E}(\text{min }M)$

where we have used $\subset_\emptyset$ to denote a non-empty subset.

It just remains to calculate the expected values of the subsets. This is not very difficult and we can proceed as we did in the earlier post. For example, let's solve for $M=\{X_S,X_C,Y_5,Y_6,Y_7\}$.

Then $\mathbb{E}(\text{min }M)$ is the expected number of cards needed to get either a Spade or a Clubs or any of ranks $5$, $6$ or $7$. This is again a Negative Geometric variable with population size $N=52$ and "successes" $K=13+13+4+4+4-6=32$. We are subtracting the six cards which are the 5, 6 and 7 of Clubs and Spades to avoid double counting.

Now using the expectation of Negative Geometric variable, we easily see that $\mathbb{E}(\text{min }M)=(52+1)/(32+1)$.

Expanding the idea, the expected value of $Z$ can be seen to be

$\begin{align}\displaystyle \mathbb{E}(Z)&=(52+1)\left[1-\sum_{i=0}^4\sum_{j=0}^{13}\binom{4}{i}\binom{13}{j}\frac{(-1)^{i+j}}{13i+4j-ij+1}\right]\\ &=(52+1)\left[1-(-1)^{4+13}\sum_{i=0}^4\sum_{j=0}^{13}\binom{4}{i}\binom{13}{j}\frac{(-1)^{i+j}}{52-ij+1}\right]\end{align}$

Evaluating this with Wolfram Alpha, we finally see that the number of cards needed, on average, to collect atleast one card from each suit and rank when drawing cards without replacement from a shuffled deck of cards is $\approx 27.9998$, very close to an integer.

Until then
Yours Aye
Me

Estimating the height of a Mountain (or a Building)

 I recently watched Matt Parker's attempt to measure the Earth's radius with Hannah Fry. As always, Matt made it entertaining but there was something missing there which I wanted to share in this post.

I remember seeing the idea used in the video in History of Geodesy and the related idea of finding the height of a mountain by ancient mathematicians. The idea that I find the most problematic in the approach used in the video is the use of sextant in finding the tower's height.

In my opinion, parallax error induced in hand made sextants is difficult to control and can completely throw off estimates. While we can't do away with the sextant in finding the Earth's circumference, we certainly can do better in the other case.

All we need now is a pole of known length and two measurements using that pole as shown in the figure above. That is we first place the pole at some point and move ourselves into a position such that the top of the pole and the top of the coincide in our line of sight. Now repeat the same at a different point.

Let $OH$ be the height we are tying to measure. Let $AC=BD$ be the pole whose length we consider unity to simplify calculations. Now, ssing the similarity of triangles $OA'H$ and $OB'H$, we have

$$\displaystyle \frac{OH}{AC}=\frac{OB'}{BB'}=\frac{OA'}{AA'}=\frac{B'A'}{BB'-AA'}=\frac{B'A'}{B'A'-BA}=\frac{AB+BB'-AA'}{BB'-AA'}$$

which is an expression of length measurements. Not only do we avoid sextants here, we also don't have to worry about measuring lengths from the mountain's (or building's) base.

Note that $A'$ and $B'$ are the only measurement in the expression which is prone to error because of the parallax. If we let $OH=y$ and $BB'-AA'=z$, then we can rewrite the above as

$$OH=y=AC\cdot \left(1+\frac{AB}{z}\right)$$

Differentiating the above expression, we see that

$$\displaystyle \frac{dy}{dz}=-AC\cdot \frac{AB}{z^2}$$

This shows that if $z$ is too small (which means $AA'$ and $BB'$ are almost equal), then even a small change in $z$ would result in a large change in the height. So the best thing is to make the first measurement close enough to the mountain and the second measurement far so as to reduce this effect.

Because we rely on line of sight in finding both $A'$ and $B'$, we are interested in some sort of interval estimate for the height. To this end, we consider any measurements involving points $A'$ and/or $B'$ are independent Normal variables with variance $\sigma^2$. Now,

$$\displaystyle OH=AC \cdot \left(1+\frac{AB}{Z}\right)$$

where $Z =Z_{BB'}-Z_{AA'}\sim \mathcal{N}(BB'-AA',2\sigma^2)$.

Now to attain a confidence level of $\alpha$, we have

$\displaystyle \mathbb{P}(OH^{-} \leq OH \leq OH^{+})=\alpha$ where $\displaystyle OH^{\pm}=AC\cdot\left(1+\frac{AB}{BB'-AA' \mp z_\alpha \sqrt{2}\sigma}\right)$

Note that $z=BB'-AA'$ needs two measurements. We could alternately use a single measurement $A'B'$ which also is random with variance $2\sigma^2$. Practically though, as the two measuring points are far away, $A'B'$ will be long. In this case, we have,

$\displaystyle \mathbb{P}(OH^{-} \leq OH \leq OH^{+})=\alpha$ where $\displaystyle OH^{\pm}=AC\cdot\left(1-\frac{BA}{B'A' \mp z_\alpha \sqrt{2}\sigma}\right)^{-1}$

I plan to use this idea to estimate the height of a temple tower near my place. I'll update if I find anything.

Until then

Yours Aye

Me

Animating the Spherical Tautochrone and Isochrone

I recently made a submission to 3Blue1Brown's Summer of Math Exposition based on the idea of Spherical Tautochrones. It's a short clip and if interested you can watch the video on my channel.

While doing the video, I wanted to know exact location of the video at any given time to make the animation scientifically correct. But rather than make this about the spherical case, let's make it more general. Then, using the Conservation of energy we first have,

$\displaystyle\frac{ds}{dt}=\sqrt{2g(y_0-y)}$

where $y_0$ denotes the y-coordinate of the sliding particle at time $t=0$.

Using the condition we derived for the Tautochrone curve in the video,

$\displaystyle \frac{ds}{dy}=\frac{T_0}{\pi}\sqrt{\frac{2g}{y}}$

Using the above two equations, we get

$\displaystyle \frac{dy}{dt}=\frac{\pi}{T_0}\sqrt{y(y_0-y)}$

In my python code for the video, I used python's odeint to numerically solve this equation. But thinking about it, there was a much more easier way to do this.

Using the pendulum analogy we saw in the video, we can multiply the 'force equation' of the pendulum by $l$, the length of the pendulum to get

$\displaystyle \frac{d^2s}{dt^2}+\frac{g}{l}s=0$

which is just the equation of the Simple Harmonic motion as noted in the video. Solving this for the initial condition that $s=s_0$ at $t=0$ and $s=0$ at $t=T_0$, we easily see that

$\displaystyle s=s_0\cos\left(\frac{\pi}{2}\frac{t}{T_0}\right)$

where we have used the expression $T_0=(\pi/2)\sqrt{l/g}$ to eliminate the $g/l$ term.

Also from the condition that we derived for tautochrones, we know that $s$ is directly proportional to $\sqrt{y}$. This immediately gives

$\displaystyle y=y_0\cos^2\left(\frac{\pi}{2}\frac{t}{T_0}\right)$

which gives the position of the particle at time any time $t$. For the spherical co-ordinate, this reduces to

$\displaystyle \cos\left(\frac{\theta}{2}\right)=\cos\left(\frac{\theta_0}{2}\right)\cos\left(\frac{\pi}{2}\frac{t}{T_0}\right)$

This means, we could have simply used this explicit expression rather than numerically solving a differential equation in the animation!!

Suddenly then, I realized that this will not work when animating the (polar) Isochrone. But then it dawned on me that the polar velocity is constant by definition of the polar isochrone. Great!!

Until then

Yours always

Me

Birds on a Wire - a Probability puzzle

This post is about birds on a wire - a nice probability puzzle that I recently saw on cut-the-knot website. The site also gives five solutions all of which are somewhat informal.

While all the solutions were pretty, I was particularly struck by the elegant solution by Stuart Anderson. Studying his solution made me realize that the condition that '$n$ is large' is not very significant. In fact, we'll use his idea to solve the question for the finite version of the problem.

Let the length of the wire be $1$ unit. Let $X_k$ ($1\leq k \leq n$) denote the position of the $k$th bird to sit on the wire. That is, all the $X_k$'s are independent standard uniform random variables.

Let $X_{(k)}$ be the random variable denoting the $k$th Order statistic. Now let's define new random variables $Y_k$ for $0 < k \leq n+1$ such that

$Y_k=X_{(k)}-X_{(k-1)}$ with $X_{(0)}=0$ and $X_{(n+1)}=1$.

Note that the $n$ birds divide the wire into $n+1$ segments and the random variable $Y_k$ denote the length of the $k$th segment.

Using the joint distribution of the Order statistics of the Uniform distribution, we can see that the $Y_k$s are identical (but not necessarily independent) $Beta(1,n)$ variables.

Therefore, the cumulative distribution for any $Y_k$ is $F(y)=\mathbb{P}(Y_k \leq y)=1-(1-y)^n$ and the expected value is $\mathbb{E}(Y_k)=1/(n+1)$.

Now if $n$ were large, we can use the idea that the above expression for CDF is approximated by $1-e^{-ny}$ and the expected value by $1/n$. But as noted earlier, we don't have to assume $n$ is large and proceed directly.

Let's define new variables $C_k$ such that

$\displaystyle C_k=\begin{cases}0, & \text{if }k \text{th segment is not colored}\\  Y_k, & \text{otherwise} \end{cases}$

It is straight forward to see that

(i) $\mathbb{E}(C_1)=\mathbb{E}(C_{n+1})=0$ because those segments never get coloured and

(ii) $\mathbb{E}(C_2)=\mathbb{E}(Y_2)=\mathbb{E}(C_n)=\mathbb{E}(Y_n)=1/(n+1)$ because those segments always gets coloured.

For the cases $2<k<n$, consider the segment $Y_k$ along with its neighbours $Y_{k-1}$ and $Y_{k+1}$. A minute's thought shows that the $k$th segment is not coloured if and only if it is the biggest among the three variables.

If we let $Z_k=\max(Y_{k-1},Y_k,Y_{k+1})$, we can rewrite $C_k$ as

$\displaystyle C_k=\begin{cases}0, & Y_k=Z_k\\  Y_k, & \text{otherwise} \end{cases}$

But $\displaystyle \mathbb{P}(Y_k=Z_k)=\frac{1}{3}$ by symmetry. Therefore,

$\displaystyle \mathbb{E}(C_k)=\mathbb{P}(Y_k=Z_k)\cdot 0+\mathbb{P}(Y_k \ne Z_k) \mathbb{E}(Y_k|Y_k \ne Z_k)=\frac{2}{3}\mathbb{E}(Y_k|Y_k \ne Z_k)$

To do this, we condition on $Y_k$. This gives,

$\displaystyle \mathbb{E}(Y_k)=\frac{1}{3}\mathbb{E}(Y_k|Y_k=Z_k)+\frac{2}{3}\mathbb{E}(Y_k|Y_k \ne Z_k)$

Rearranging this, we have

$\displaystyle \mathbb{E}(C_k)=\frac{2}{3}\mathbb{E}(Y_k|Y_k \ne Z_k)=\mathbb{E}(Y_k)-\frac{1}{3}\mathbb{E}(Y_k|Y_k=Z_k)=\mathbb{E}(Y_k)-\frac{1}{3}\mathbb{E}(Z_k)$

To compute the expected value of $Z_k$, we proceed in two parts.

First, consider any distribution with CDF $F(v)$ and the order statistics $V_{(i)}$, $V_{(j)}$ and $V_{(k)}$ with $i<j<k$.

It can be seen intuitively that the conditional distribution of $V_{(j)}$ given $V_{(i)}$ and $V_{(k)}$ is the same as the $(j-i)$th order statistic obtained from a sample of size $k-i-1$ whose distribution is $F(v)$ truncated on the left at $V_{(i)}$ and on the right at $V_{(k)}$.

For example, if we know that $V_{(5)}=0.4$ and $V_{(10)}=0.9$ from the standard uniform distribution $U(0,1)$, then $V_{(7)}$ has the same distribution as $W_{(2)}$ from a sample of size $4(=10-5+1)$ from the distribution $W\sim U(0.4,0.9)$.

This result can be obtained by using Theorems 2 and 3 given here.

Second, it is a well known result in probability (cuttheknot, here and here to quote a few) that if a line of length 1 unit is broken at $n-1$ points giving $n$ segments, then the expected length of the $k$th largest segment is $(1/k+\cdots+1/n)/n$. Specifically, the length of the largest segment in a 3 segment case is $11/18$.

Therefore,

$\displaystyle \mathbb{E}(Z_k)=\mathbb{E}(\mathbb{E}(Z_k|X_{(k-2)},X_{(k+1)}))=\mathbb{E}\left(\frac{11}{18}(X_{(k+1)}-X_{(k-2)})\right)=\frac{11}{18}\frac{3}{n+1}$

Using the above and substituting the known values, we get

$\displaystyle \mathbb{E}(C_k)=\frac{1}{n+1}-\frac{1}{3}\frac{11}{18}\frac{3}{n+1}=\frac{7}{18n+18}$

If we now let $C=C_1+C_2+\cdots+C_n+C_{n+1}$, then using the linearity of expectations, we get

$\displaystyle \mathbb{E}(C)=\frac{2}{n+1}+(n-3)\frac{7}{18n+18} =\frac{7n+15}{18n+18}$

Note that this applies only when $n\geq 3$ (cases $n<3$ are trivial to calculate). More importantly, this shows that the expected value tends to $7/18$ for large $n$.

I would like to thank Abhilash Unnikrishan for pointing out the error (I assumed that $Y_k$s are independent) in my original solution and to use the 'expected value of largest segment in a line' to simplify calculations in arrive at the final result. 

Hope you enjoyed the discussion.

Until then

Yours Aye

Me