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? – Squirtle 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. – Coin Flipper 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 $${144084523275\over 1099511627776}=0.13104$$

protected by Community♦Oct 7 at 13:01

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?