# What are the odds of getting heads 7 times in a row in 40 tries of flipping a coin?

I know if you flip a coin $$7$$ times, the odds of getting $$7$$ heads in a row is $$1$$ in $$2^7$$ or $$1$$ in $$128$$.

But if you flip a coin $$40$$ times, what are the odds of getting $$7$$ heads in a row in those $$40$$ tries? I only want to know the first time there are $$7$$ heads in a row and not count duplicates. Thanks.

• Coin Flipper, welcome to S.E. mathematics. Do you think that you could provide a few more details and try an attempt at the problem? Dec 11 '13 at 1:32
• Well, I would think it would be more than 1/128 but I never took probability theory in college. I don't understand what you mean about more details. I think the question is clear. Dec 11 '13 at 1:45

Let's count the number of ways not to get $7$ heads in a row. We will put together atoms that consist of $0$ to $6$ heads followed by a tail. Any arrangement of heads and tails without $7$ heads in a row, appended with a tail, can be uniquely made up of a number of such atoms.

All arrangements of such atoms appear once somewhere in the sum $$\sum_{k=0}^\infty(x+x^2+x^3+x^4+x^5+x^6+x^7)^k$$ where
$x$ represents $T$
$x^2$ represents $HT$
$x^3$ represents $HHT$
$\vdots$
$x^7$ represents $HHHHHHT$

For example, if we are looking for $HTTHHTTH$, append a $T$ and we get the term for $k=5$ where in the first factor, the $x^2$ ($HT$) was chosen, in the second factor, the $x$ ($T$) was chosen, then $x^3$ ($HHT$), then $x$ ($T$), then $x^2$ (HT), to get $HTTHHTTHT$. Note that the exponent of $x$ matches the number of tosses. To count the number of sequences of $40$ flips that do not contain $7$ consecutive heads, we look at the coefficient of $x^{41}$ in \begin{align} \sum_{k=0}^\infty(x+x^2+x^3+x^4+x^5+x^6+x^7)^k &=\frac1{1-x\frac{x^7-1}{x-1}}\\ &=\frac{1-x}{1-2x+x^8} \end{align} The coefficient of $x^{41}$ is $955427104501$. There is a degree $8$ recursion to compute this without dividing polynomials: $c_n=2c_{n-1}-c_{n-8}$, where $c_n$ starts $$1,1,2,4,8,16,32,64,\dots$$

The number of sequences of $40$ flips is $2^{40}$. Therefore, the probability of getting a sequence of $7$ heads in a row in $40$ flips is $$1-\frac{955427104501}{2^{40}}=0.131044110526$$

You can find the formula in my answer here. Setting $p=1/2$, $n=40$ and $\ell=7$ we find that, using a fair coin, the chance of a run of heads of length seven (or longer) is $${144084523275\over 1099511627776}=0.13104$$