How to win Matt Parker's jackpot - finding the median of the following distribution

Question

In a recent video the legendary Matt Parker claimed he kept flipping a two-sided (fair) coin untill he scored a sequence of ten consecutive 'switch flips', i.e. letting $T$ denote a tail and $H$ a head, then a sequence of ten switch flips is defined to be either $THTHTHTHTH$ or $HTHTHTHTHT$. He set up a contest and allowed each viewer to guess once at the exact amount of flips he needed to obtain such a sequence. The ten viewers with the ten closest answers would be awarded a prize.

The contest is over, so there is no incentive to keep a solution to the following problem to yourself. What is the best number to bet? Of course this somehow depends on how other viewers answer: you are more likely to win if your bet is not close to many other bets, so if a large number of viewers is mathematically inclined and bets the same number - say 1023 - then it no longer is the best (profit maximizing) bet. I've therefore simplified to the following question: let $X$ be the stochastic variable representing the number of flips needed untill a sequence of ten consecutive switch flips if obtained, then for which number $a \in \mathbb{N}$ does the expected value of the (absolute value of the) error $$ \mathbb{E}[\vert X - a \vert] $$ reach its minimal value? It is well-known that $a$ is the median of the (distribution of) $X$, but how can one compute it? Numerical approximations are welcome, theoretical (generalizable) results are preferred.

I found a way to compute the expected value of $X$ itself for general $n$ (i.e. the total number of coin flips needed to get a sequence of the form $THTHTHTH...$ or $HTHTHTHT...$ of length $n$). Let $\mathbb{E}_i$ denote the expected number of coin flips needed to get a desired sequence of length $n$, assuming we already have a sequence of length $i \in \mathbb{N}$. We immediately find $$ \mathbb{E}_0 = 1 + \mathbb{E}_1 $$ since we are certain to have a sequence of length $1$ after one flip. Furthermore, for $1 \leq i \leq n-1$ $$ \mathbb{E}_i = \frac{1}{2}\left(\mathbb{E}_1 + 1\right) + \frac{1}{2}\left(\mathbb{E}_{i+1} + 1\right) $$ since, given a sequence of $i$ flips which ends, say, on a tail, we have a $\frac{1}{2}$ chance to increase this to a sequence of $i+1$ flips (if we get, say, a head) and a $\frac{1}{2}$ chance to get back where we started, at $1$ flip. Using that $\mathbb{E}_n = 0$ the above gives us system of $n$ equations in the $n$ variables $\mathbb{E}_0, \ldots, \mathbb{E}_{n-1}$. One can easily check that the unique solution is given by $$ \mathbb{E}_i = 2^{n} - 2^{i} \quad 0 \leq i \leq n $$ Since Matt Parker started at $0$ and wanted to get $10$ flips, the expected value of the number of flips needed is $2^{10} - 1 = 1023$ and this should be a reasonable bet.

Does anyone know how to find the distribution of $X$ (or directly the median of $X$)? Like I said, analytical solutions are of course preferred, but any kind of method - even requiring numerical computations, but preferably not Monte Carlo simulations - would be interesting to me.

EDIT: I found out that the problem can be reduced to a combinatorial problem. Indeed, we have that $$ P(X \leq k) = \frac{\# \lbrace \text{sequences of length $k$ which contain a desired subsequence of length $n$}\rbrace}{2^k} $$ where $2^k$ is the total number of sequences of length $k$, since every sequence of length $k$ is equally likely to occur. Let $S_k$ be the set of sequences of length $k$ of $0$'s and $1$'s (we identify tails with $0$ and heads with $1$). We have a map $$ f: S_k \to S_{k-1} $$ where, for any sequence $s \in S_k$, the $i$th element of $f(s)$ is $1$ if $s(i) \neq s(i+1)$ and $0$ is $s(i) = s(i+1)$. This map makes the desired sequences in $S_k$ correspond bijectively with the sequences in $S_{k-1}$ which contain $n-1$ zeroes in a row. Hence, it suffices to count the number of sequences of a given length $k-1$ which contain $n-1$ zeroes in a row. Any ideas on how to continue?

Answer 1

The probability to get $10$ consecutive switch flips can be modelled as a Markov chain, with the states corresponding to the number of consecutive switch flips in the immediate past. The state in which $10$ consecutive switch flips have been encountered is absorbing. The stationary distribution with eigenvalue $1$ has probability $1$ at this absorbing state. There is a quasi-stationary distribution in which each state has half the probability of the one with one less switch flip and probability "leaks out" into the absorbing state at a rate of roughly $2^{-10}$. The median is the least integer $a$ for which $\textsf P(X\le a)\gt\frac12$. The time it takes for $\frac12$ to leak out is roughly given by $a\cdot2^{-10}=\log2$, or $a=2^{10}\log2\approx 710$.

Here's Java code that follows the evolution of the distribution of the process until the probability for $10$ switch flips exceeds $\frac12$. The median turns out to be $712$. The precision of the above estimate is in part gratuitous, since the process takes $10$ steps to equilibrate before probability starts leaking into the absorbing state. Nevertheless, the agreement shows that the quasi-stationary model is quite good, so $P(X\le x)\approx1-\left(1-2^{-10}\right)^t\approx1-\mathrm e^{-2^{-10}t}$ should be a good approximation.

Regarding the game-theoretic aspect of the problem, this is related to the ice-cream vendor problem, where the flip numbers are mapped to the beach using the cumulative distribution function $P(X\le x)$. However, the conclusion in the continuous case that there is no equilibrium for $3$ players doesn't go through in the discrete case.

Answer 2

Let $a^k_{n}$ denote the number of zero-one sequences of length $n$ with longest zero run non-exceeding $k$, and $a^k_{n,m}$ denote the number of such sequences with $m$ trailing zeros, $m=0,\dots,k$.

Then $$ a^{k}_{n,m} = a^{k}_{n-m,0},\ m = 1,\dots,k, n\ge m, $$ and $a^k_{n,0} = a^k_{n-1}$, $n\ge 0$. Therefore, $$ a_n^k = \sum_{m=0}^k a^k_{n,m} = a^k_{n-1} + \sum_{m=1}^{k} a^k_{n-m,0} = \sum_{j=1}^{k+1} a^k_{n-j} \tag{1} $$ for $n\ge k+1$. (In other words, take a sequence and remove all trailing zeroes and preceding one; you'll end up with a sequence whose longest zero run does not exceed $k$.)

The initial values are $a_n^k = 2^n, n\le k$. Therefore, it easy to see that the generating function $A^k(z) = \sum_{n=0}^\infty a_n^k z^n$ is equal to $$ A^k(z) = \frac{\sum_{j=0}^k z^j}{1-\sum_{j=1}^{k+1} z^j} = \frac{1-z^{k+1}}{1-2z+z^{k+2}}. $$

Writing the partial fraction expansion for $A^k(z)$, it is possible to derive an explicit formula for $a_n^k$.

Now, denoting $T_m$ the number of throws to get $m$ heads, we have $$ P(T_m>n) = \frac{1}{2^n}P(\text{at most $m-1$ consecutive heads in the first $n$ throws}) = \frac{a^{m-1}_n}{2^n}. $$

However, for $m=9$ (which corresponds to $10$ alternating flips, as you've explained) there is nothing very exciting, as the roots can be computed only approximately. Nevertheless, using the recurrent formula (1) gives an efficient way to compute the required number of sequences.

A good rough approximation is $a_n^{8} \sim C_\tau \tau^{-n-1}$, where $\tau = 0.500493118286\dots$ is the unique positive root of the polynomial $f(\tau) = \sum_{j=1}^{9}z^{j} -1$ (alternatively, the root of $z^{10}-2z+1$ inside $(0,1)$), and $$ C_\tau = \frac{\sum_{j=0}^{8} \tau^j}{f'(\tau)} = 0.503980275733\dots $$ Indeed, since $A^8(z)$ is rational, and $f$ does not have double roots, it has a partial fraction expansion $A^8(z) = \sum_{\zeta: f(\zeta) = 0} \frac{C_\zeta}{\zeta - z}$. Therefore, the sequence is of the form $a_n^8 = \sum_{\zeta: f(\zeta) = 0} C_\zeta \zeta^{-n-1}$. In particular, since $\tau$ is the root with the smallest absolute value, $a_n^8\sim C_\tau \tau^{-n-1}$. Moreover, the norms of other roots are bigger than $1.11$, so the relative error of this approximation is of order $r^{-n}$ with $r>2$. This is especially good for our problem, since we will be interested with very large values of $n$.

So the probability we need is approximately $$ P(T_9>n)\sim \frac{C_\tau}{\tau (2\tau)^n}, $$ and this is very sharp (up to some exponentially small relative error). In order to find the median, one should find the smallest $n$ such that $P(T_9>n)< 1/2$. This gives the value $$n= \lceil- \log(4C_\tau)/\log(2\tau)\rceil-1 = 711$$ for the median, which agrees with joriki's calculations.

Answer 3

Intuitively it seems like you should bet on this event occurring in the first 10 flips.

Reasoning: If you pick any particular 10 flips you have a $\frac{1}{2^9}$ probability of those particular ten flips being a sequence of interest.

However, we specifically want the first such sequence, which means that there will be constraints on the earlier flips (to avoid an earlier sequence of this type). Therefore your probability of some sequence of 10 being the first such sequence is less than or equal to $\frac{1}{2^9}$, with equality only for the first ten flips.