# Tossing the coin problem: We toss symmetrical coin $10$ times. Calculate probability of tails appearing at least five times in a row.

We toss symmetrical coin $$10$$ times. Calculate probability of tails appearing at least five times in a row.

I tried by dividing by cases (the row TTTTT I observe as one object); first case exactly $$5$$ tails, second exacty $$6$$ tails etc. For the first case I decided to separate cases when the tails appear from the first toss to the fifth toss and when they appear from $$i$$th toss to $$(i+5)$$th toss but each time I get stuck on that case.

• You're going to want to use the binomial expansions Mar 17, 2019 at 9:27
• Define $A_i$ to be the set of outcomes in which $5$ consecutive tails occur, beginning in position $i$, $1 \leq i \leq 6$, then use the Inclusion-Exclusion Principle. Mar 17, 2019 at 10:29

This is easier than it looks! There are six mutually exclusive cases:

1. First five throws are Tails: probability = $$\frac{1}{32}$$
2. First throw is a Head, and next five throws are Tails: probability = $$\frac{1}{64}$$
3. Second throw is a Head, and next five throws are Tails: probability = $$\frac{1}{64}$$
4. Third throw is a Head, and next five throws are Tails: probability = $$\frac{1}{64}$$
5. Fourth throw is a Head, and next five throws are Tails: probability = $$\frac{1}{64}$$
6. Fifth throw is a Head, and next five throws are Tails: probability = $$\frac{1}{64}$$

So the total probability is $$\frac{7}{64} = 0.109375$$.

Note that this method doesn't work if there are more than ten throws, because then you have to guard against double counting (e.g. T T T T T H T T T T T).

I recognize that this probably isn't the desired way of approaching this, but this will nonetheless give someone else a check of their work. This problem is identical. We have that, in general, with a sequence of $$n$$ Bernoulli trials each with probability of success being $$p$$, the probability of obtaining at least $$m$$ consecutive successes in the sequence is given by$$\mathbb{P}(\ell_n \geq m)=\sum_{j=1}^{\lfloor n/m\rfloor} (-1)^{j+1}\left(p+\left({n-jm+1\over j}\right)(1-p)\right){n-jm\choose j-1}p^{jm}(1-p)^{j-1}.$$

With $$n=10$$, $$m=5$$, and $$p=0.5$$ we get

$$\mathbb{P}(\ell_n \geq m)=0.109375$$

As a perhaps cumbersome way of simulating this using R statistical software:

coin <- c("H","T")
count=0
for(i in seq(1,10^6,1)){
u<-sample(coin,10,repl=T)
if(u[1]=="T"&u[2]=="T"&u[3]=="T"&u[4]=="T"&u[5]=="T"){count=count+1}
else if(u[2]=="T"&u[3]=="T"&u[4]=="T"&u[5]=="T"&u[6]=="T"){count=count+1}
else if(u[3]=="T"&u[4]=="T"&u[5]=="T"&u[6]=="T"&u[7]=="T"){count=count+1}
else if(u[4]=="T"&u[5]=="T"&u[6]=="T"&u[7]=="T"&u[8]=="T"){count=count+1}
else if(u[5]=="T"&u[6]=="T"&u[7]=="T"&u[8]=="T"&u[9]=="T"){count=count+1}
else if(u[6]=="T"&u[7]=="T"&u[8]=="T"&u[9]=="T"&u[10]=="T"){count=count+1}
}

count/10^6

> 0.109379


which is accurate up to $$5$$ decimal places.

It may be easier to find the probability that a sequence of five tails does not occur.

There are $$2^{10}$$ possible sequences of coin tosses, all of which we assume are equally likely. We would like to count the number of sequences which does not contain a sequence of five tails in a row.

One way to do this is with a set of recursive equations. Define $$a_i(n)$$ to be the number of sequences of $$n$$ coin tosses which do not contain a run of five tails and which end in $$i$$ tails, for $$i=0,1,2,3,4$$. Clearly $$a_0(1) = a_1(1) = 1$$ and $$a_i(1) = 0$$ for $$i=2,3,4$$. Any sequence of $$n$$ tosses followed by a head results in a sequence of $$n+1$$ tosses ending in zero tails; so $$a_0(n+1) = a_0(n) + a_1(n) + a_2(n) + a_3(n) + a_4(n)$$ for $$n > 1$$. The only way to get $$n+1$$ tosses ending in a sequence of $$i$$ tails for $$i > 0$$ is to start with a sequence of $$n$$ tosses ending in $$i-1$$ tails and then get a tail on the $$(n+1)$$th toss; so $$a_i(n+1) = a_{i-1}(n)$$ for $$i=1,2,3,4$$ and $$n>1$$.

Using these equations and the given initial conditions, we can calculate $$a_i(n)$$ for $$0 \le i \le 4$$ and $$n$$ as large as we want. (It's easy to do on a spreadsheet; see below.) The total number of sequences of length $$n$$ not containing any run of five tails is then $$\sum_{i=0}^4 a_i(n)$$, and the probability of such a sequence is $$\frac{\sum_{i=0}^4 a_i(n)}{2^n}$$ We are interested in the case $$n=10$$. By calculation, we find $$\sum_{i=0}^4 a_i(10) = 912$$, so the probability that a sequence of $$10$$ coin tosses does not include a sequence of five tails is $$912/2^{10}$$. The answer to the original question, the probability that a sequence of $$10$$ tosses contains at least one sequence of five tails, is $$\boxed{1-\frac{912}{2^{10}}}$$

In case it's not clear how to calculate the $$a_i(n)$$ values, here are the first few rows of my spreadsheet, containing $$a_i(n)$$ for $$0 \le i \le 4$$ and $$1 \le n \le 5$$, just to get you started. The first row is for $$n=1$$, the second for $$n=2$$, etc. Not shown are the spreadsheet formulas which perform the calculations. $$\begin{matrix} 1 &1 &0 &0 &0\\ 2 &1 &1 &0 &0\\ 4 &2 &1 &1 &0\\ 8 &4 &2 &1 &1\\ 16 &8 &4 &2 &1\\ \end{matrix}$$

To complement the other answers, you can do backtracking to at least know what the solution is going to be. It's been some time since I've programmed in C++, so I've found this problem a good excuse to remember how to do some things. The code is

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

typedef vector<int> VE;

VE c;
int n, count;

void f(int i, int consec, bool b){
if (i == 10) {
if (consec == n) b = true;
if (consec <= n and b){
++count;
//for (int j = 0; j < 10; ++j) cout << c[j] << ' ';
//cout << endl;
}
return;
}
if (consec > n) return;
if (consec == n) b = true;
c[i] = 1;
f(i+1, consec+1, b);
c[i] = 0;
f(i+1, 0, b);
return;
}

int main(){
int total = 0;
for (int i = 0; i <= 4; ++i){
c = VE(10);
n = i; count = 0;
f(0,0, false);
total += count;
}
cout << 1 - total/pow(2,10) << endl;
}


The output is $$0.109375$$

The function finds how many sequences have $$n$$ tails appearing in a row (and no more).

In general, we find $$1 - P$$(a sequence of five or more tails does not occur), so we'll call the function for $$n=1,2,3,4$$.