2
$\begingroup$

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...

$\endgroup$
  • 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 '19 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 '19 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 '19 at 19:30
  • $\begingroup$ And it needs to be exactly twice as many. $\endgroup$ – MiKiDe Jun 21 '19 at 19:40
  • $\begingroup$ By the law of large numbers the H/T ratio will $\to1$,so the probability of what you want $=0$. $\endgroup$ – herb steinberg Jun 21 '19 at 19:49
2
$\begingroup$

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 $$ \begin{align} \frac1{2\pi}\int_0^{2\pi}\frac1{1-e^{ix}}\,\mathrm{d}x &=\lim_{r\to1^-}\frac1{2\pi}\int_0^{2\pi}\frac1{1-re^{ix}}\,\mathrm{d}x\\[3pt] &=1\tag6 \end{align} $$

$\endgroup$
  • $\begingroup$ This is the same $57.2949\%$ that Robert Israel gets. $\endgroup$ – robjohn Jun 22 '19 at 11:49
  • $\begingroup$ Wow, I never thought like that. Thank a lot! $\endgroup$ – MiKiDe Jun 22 '19 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 '19 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 '19 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 '19 at 13:24
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '19 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 '19 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 $u(x) = (1/2) u(x+1) + (1/2) u(x-2)$. $\endgroup$ – Robert Israel Jun 24 '19 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: $(1/2) u(1) + (1/2) u(-2)$. $\endgroup$ – Robert Israel Jun 24 '19 at 20:31
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That's true, but not enough because we want at least the probability of the reunion of such events. $\endgroup$ – MiKiDe Jun 21 '19 at 20:01
  • $\begingroup$ Probability of getting more than or equal to 2n Heads? $\endgroup$ – wild_fox Jun 21 '19 at 20:03
  • $\begingroup$ It is the probabilty to get at least once during the infinite sequence, for some $n$, twice as many heads as tails during the first $n$ toss $\endgroup$ – MiKiDe Jun 21 '19 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.