Let consider an infinite sequence of tosses of a fair coin ($p = 1/2$ to get head or tail).

What is the probability to get, at least once, twice as many head as tails?

In another words, what is the probability that, there is $n, k \in \Bbb N^*$ such as $H_{3n} = 2n, T_{3n} = n$ (where $H_n$ denote the number of heads, and $T_n$ the number for tails at the $n$)?

What I tried: I don't even know how to start... I tried a lot of things but nothing really worked...

  • 2
    $\begingroup$ What does it even mean to have "twice as many heads as tails" in an infinite sequence? Do you mean, for fixed $n$, what is the probability that there are twice as many heads as tails in the first $n$ tosses, and then take the limit as $n\to \infty$? In that case the answer is zero using any number of results, like Chebyshev's inequality. $\endgroup$
    – J.G
    Jun 21, 2019 at 19:20
  • $\begingroup$ To further question the quote in the above comment, does it need to be exactly twice as many or at least twice as many? $\endgroup$
    – WaveX
    Jun 21, 2019 at 19:24
  • $\begingroup$ "What is the probabilty that, there is $n\in \Bbb N^*$ such as $H_{3n} = 2n, T_{3n} = n$ (where $H_n$ denote the number of heads, and $T_n$ the number for tails at the $n$-th toss. $\endgroup$
    – MiKiDe
    Jun 21, 2019 at 19:30
  • $\begingroup$ And it needs to be exactly twice as many. $\endgroup$
    – MiKiDe
    Jun 21, 2019 at 19:40
  • $\begingroup$ There's a lower bound of $\frac 38$ from the first three tosses. The event may happen at 6 tosses with $\frac{15}{64}$. So by inclusion-exclusion, we have a lower bound by the first six tosses of $\frac38+\frac{15}{64}-(\frac38)^2=\frac{15}{32}$, which is already almost $\frac12$ $\endgroup$ Jun 21, 2019 at 19:54

3 Answers 3


Probability of getting $2k$ heads and $k$ tails is $\frac1{2^{3k}}\binom{3k}{k}$. Thus the expected number of successes (getting twice as many heads as tails) is $$ \begin{align} \sum_{k=1}^\infty\frac1{2^{3k}}\binom{3k}{k} &=\frac1{2\pi i}\sum_{k=1}^\infty\oint_{|z|=1^-}\left(\frac{\frac1z+z^2}2\right)^k\frac{\mathrm{d}z}z\\ &=\frac1{2\pi i}\oint_{|z|=1^-}\frac{\frac{\frac1z+z^2}2}{1-\frac{\frac1z+z^2}2}\frac{\mathrm{d}z}z\\ &=-\frac1{2\pi i}\oint_{|z|=1^-}\frac{1+z^3}{(1-z)(1-z-z^2)}\frac{\mathrm{d}z}z\\[6pt] &=\color{#C00}{\frac{\phi^3+1}{(\phi-1)(\phi+2)}}\color{#090}{-1}\\[6pt] &=\frac3{\sqrt5}\tag1 \end{align} $$ Since the sum in the integral converges for $|z|\lt1$ and not for $|z|\gt1$, we integrate just inside the unit circle and therefore avoid the singularities at $z=-\phi$ and $z=1$. We catch the singularities at $\color{#C00}{z=\frac1\phi}$ and $\color{#090}{z=0}$.

Suppose that the probability of at least one success is $p$, then the probability of exactly $k$ successes is $p^k(1-p)$. Thus, the expected number of successes is $$ \sum_{k=1}^\infty kp^k(1-p)=\frac p{1-p}\tag2 $$ Since $\frac p{1-p}=\frac3{\sqrt5}$, the probability of at least one success is $$ \bbox[5px,border:2px solid #C0A000]{p=\frac3{3+\sqrt5}}\tag3 $$

Here is an alternate computation of $(1)$ that doesn't use contour integration. $$ \begin{align} \sum_{k=1}^\infty\frac1{2^{3k}}\binom{3k}{k} &=\frac1{2\pi}\sum_{k=1}^\infty\int_0^{2\pi}\left(\frac{e^{-ix}+e^{2ix}}2\right)^k\,\mathrm{d}x\\ &=-\frac1{2\pi}\int_0^{2\pi}\frac{e^{3ix}+1}{e^{3ix}-2e^{ix}+1}\,\mathrm{d}x\\ &=-\frac1{2\pi}\int_0^{2\pi}{\scriptsize\left(\frac{10+3\sqrt5}5-\frac{\left(5+3\sqrt5\right)/5}{1-e^{-ix}/\phi}-\frac{\left(-5+3\sqrt5\right)/5}{1+e^{ix}/\phi}-\frac2{1-e^{ix}}\right)}\,\mathrm{d}x\\ &=-\left(\frac{10+3\sqrt5}5-\frac{5+3\sqrt5}5-\frac{-5+3\sqrt5}5-2\right)\\[3pt] &=\frac3{\sqrt5}\tag4 \end{align} $$ Note that for $|r|\lt1$ $$ \begin{align} \frac1{2\pi}\int_0^{2\pi}\frac1{1-re^{\pm ix}}\,\mathrm{d}x &=\frac1{2\pi}\int_0^{2\pi}\sum\limits_{k=0}^\infty r^ke^{\pm ikx}\,\mathrm{d}x\\ &=1\tag5 \end{align} $$ and we take $$ \begin{align} \frac1{2\pi}\int_0^{2\pi}\frac1{1-e^{ix}}\,\mathrm{d}x &=\lim_{r\to1^-}\frac1{2\pi}\int_0^{2\pi}\frac1{1-re^{ix}}\,\mathrm{d}x\\[3pt] &=1\tag6 \end{align} $$

  • $\begingroup$ This is the same $57.2949\%$ that Robert Israel gets. $\endgroup$
    – robjohn
    Jun 22, 2019 at 11:49
  • $\begingroup$ Wow, I never thought like that. Thank a lot! $\endgroup$
    – MiKiDe
    Jun 22, 2019 at 11:53
  • $\begingroup$ one question: could you explain me why $p^k(1-p$ is the probability to get $k$ successes? $\endgroup$
    – MiKiDe
    Jun 22, 2019 at 12:47
  • $\begingroup$ The probability of getting no success is $1-p$. The probability of getting a success at some point is $p$ then getting no more is $1-p$. After getting any success, the probability of getting another is $p$ (since we still need to get twice as many heads as tails from that point) and the probability of getting no more is $1-p$. Thus, the probability of getting exactly $k$ successes is $p^k(1-p)$. Another way to look at it is that after each success, we are back to where we started in terms of the next success. $\endgroup$
    – robjohn
    Jun 22, 2019 at 12:57
  • $\begingroup$ that's totally clear thank's! Do you know how could I show that the sum egals $3/\sqrt5$ only using real analysis/power series? $\endgroup$
    – MiKiDe
    Jun 22, 2019 at 13:24

Let $X_n = H_n - 2 T_n$. This process forms a Markov chain; the conditional probabilities for $X_n$ given $X_{n-1}$ are $1/2$ for $X_n = X_{n-1}+1$ or $X_{n-1}-2$, $0$ otherwise. Let $u(x)$ be the probability of ever being at $0$ for this process starting at $x$ (including the starting point, so $u(0) = 1$). Thus $ u(x) = (u(x+1) + u(x-2))/2$ for $x \ne 0$, and you want $(u(1) + u(-2))/2$. The recurrence $u(x) = (u(x+1) + u(x-2))/2$ has basic solutions $u(x) = r^x$ for $r = 1$, $(1- \sqrt{5})/2$ and $(1+\sqrt{5})/2$. Thus for some constants $a_-, b_-, c_-, a_+, b_+, c_+$ we should have $$ u(x) = \cases{a_- + b_- r_1^x + c_- r_2^x & if $x < 0$\cr a_+ + b_+ r_1^x + c_+ r_2^x & if $x > 0$\cr}$$ Now because $|r_1^x| \to \infty$ for $x \to -\infty$ and $r_2^x \to \infty$ for $x \to +\infty$, we should have $b_- = 0$ and $c_+ = 0$. Moreover we should have $u(x) \to 0$ for $x \to -\infty$, so $a_- = 0$. To fit with $u(0)=1$ we should have $c_1 = 1$. Then $a_+$ and $b_+$ are determined by $u(1) = (u(-1) + u(2))/2$ and $u(2) = (u(0)+u(3))/2$, i.e.

$$ \eqalign{a_+ + b_+ r_1 &= (r_2^{-1} + a_+ + b_+ r_1^2)/2 \cr a_+ + b_+ r_1^2 &= (1 + a_+ + b_+ r_1^3)/2 \cr}$$

whose solutions are $$ \eqalign{a_+ &= \frac{3 \sqrt{5} - 5}{2}\cr b_+ &= \frac{7 - 3 \sqrt{5}}{2}\cr}$$

That makes your answer $$(r_2^{-1} + a_+ + b_+ r_1)/2 = \frac{9 - 3 \sqrt{5}}{4}$$

  • $\begingroup$ two questions: firstly, I think you used $-X_n$ instead of you $X_n$ in your recursions. Secondly, I don't understand how do you get the recursion relation on $u(x)$. How do you express $u(x)$ with the $X_n$, what means "starting point" here? ($n,X_0$??) $\endgroup$
    – MiKiDe
    Jun 22, 2019 at 6:33
  • $\begingroup$ and another point that I don't understand: how to show that the probability that we want is $1/2 (u(-1)+u(2))? $\endgroup$
    – MiKiDe
    Jun 22, 2019 at 8:07
  • $\begingroup$ I'm pretty sure I didn't make that mistake. $u(x)$ is the probability, for a walk starting at $X_0 = x$ and making each step $+1$ or $-2$ with equal probabilities, that $X_n$ is ever $0$. The recursion is obtained by "first step analysis": starting at $x \ne 0$, with probabilty $1/2$ the first step is $+1$ and you are at $x+1$, and with probability $1/2$ the first step is $-2$ and you are at $x-2$, so $u(x) = (1/2) u(x+1) + (1/2) u(x-2)$. $\endgroup$ Jun 24, 2019 at 20:29
  • $\begingroup$ Our actual process starts at $0$, but we don't want to count that as "twice as many heads as tails", so we look again at the result of the first step: $(1/2) u(1) + (1/2) u(-2)$. $\endgroup$ Jun 24, 2019 at 20:31

I have an idea about this which might work.
So imagine the number of tosses is $3n$. Now you have to pick $2n$ of them and place Head over there and for every place, the probability of it being head is $1/2$. So the probability of having $2n$ heads and n tails is $${3n \choose 2n}\frac{1}{2}^{2n}\frac{1}{2}^n$$

For n = 6,
$P(H = 4) = {6 \choose 4}\frac{1}{2}^{4}\frac{1}{2}^2$

Comment if I did something wrong or intended question was different.

  • $\begingroup$ That's true, but not enough because we want at least the probability of the reunion of such events. $\endgroup$
    – MiKiDe
    Jun 21, 2019 at 20:01
  • $\begingroup$ Probability of getting more than or equal to 2n Heads? $\endgroup$
    – wild_fox
    Jun 21, 2019 at 20:03

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