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A box contains 4 normal coins (probability of getting "head" equals $\frac{1}{2}$) and 3 fake coins (probability of getting "head" equals $\frac{1}{3}$).

One coin is randomly chosen from the box and being tossed again and again until the first time it shows "head". What is the expectation of the number of required tosses?

I was thinking this - I will define two variables ($Y$ and $X$) and where $Y = 1$ means we tossed "head" and then calculate using $E(Y = 1 | X)$ , but I'm really not sure if this it the right direction at all.

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  • $\begingroup$ What would $X$ be? $\endgroup$ – Henrik - stop hurting Monica Dec 26 '16 at 12:54
  • $\begingroup$ Maybe $X = 1$ is "a normal coin" and $X = 2$ is "a fake coin" ? $\endgroup$ – UFC Insider Dec 26 '16 at 12:55
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Let $E_1$ be the expectation for the fair coin and $E_2$ the expectation for the biased coin. Letting $E$ denote the answer you seek, we have $$E=E_1\times \frac 47 +E_2\times \frac 37$$

To compute $E_1$: toss once to see that $$E_1=1\times \frac 12+\left(E_1+1\right)\times \frac 12\implies E_1=2$$

To compute $E_2$: toss once to see that $$E_2=1\times \frac 13+\left(E_2+1\right)\times \frac 23\implies E_2=3$$

Thus $$E=2\times \frac 47+3\times \frac 37=\frac {17}7$$

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Let $X$ denote the number of tosses until the first head, then:

$P(X=n)=\frac47\cdot\left(1-\frac12\right)^{n-1}\cdot\frac12+\frac37\cdot\left(1-\frac13\right)^{n-1}\cdot\frac13$

Therefore:

$E(X)=$

$\sum\limits_{n=1}^{\infty}n\cdot P(X=n)=$

$\sum\limits_{n=1}^{\infty}n\cdot\left(\frac47\cdot\left(1-\frac12\right)^{n-1}\cdot\frac12+\frac37\cdot\left(1-\frac13\right)^{n-1}\cdot\frac13\right)=$

$\sum\limits_{n=1}^{\infty}n\cdot\frac47\cdot\left(1-\frac12\right)^{n-1}\cdot\frac12+n\cdot\frac37\cdot\left(1-\frac13\right)^{n-1}\cdot\frac13=$

$\left(\sum\limits_{n=1}^{\infty}n\cdot\frac47\cdot\left(1-\frac12\right)^{n-1}\cdot\frac12\right)+\left(\sum\limits_{n=1}^{\infty}n\cdot\frac37\cdot\left(1-\frac13\right)^{n-1}\cdot\frac13\right)=$

$\left(\frac47\cdot\color\red{\sum\limits_{n=1}^{\infty}n\cdot\left(1-\frac12\right)^{n-1}\cdot\frac12}\right)+\left(\frac37\cdot\color\green{\sum\limits_{n=1}^{\infty}n\cdot\left(1-\frac13\right)^{n-1}\cdot\frac13}\right)=$

$\frac47\cdot\color\red2+\frac37\cdot\color\green3=$

$\frac{17}{7}$

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Let $Z$ denote the number of "failures", i.e. the number of tosses before the first head appears.

$$E[Z]=\frac47E[Z\mid \text{Normal coin is chosen}]+\frac37E[Z\mid \text{Fake coin is chosen}].$$

Now, based on the second kind of formulas given for the geometric distribution in Wiki, we have

$$E[Z\mid \text{Normal coin is chosen}]=\frac{1-\frac12}{\frac12}=1$$

and

$$E[Z\mid \text{Fake coin is chosen}]=\frac{1-\frac13}{\frac13}=2.$$

So

$$E[Z]=\frac47+2\frac37=\frac{10}7.$$


If $Z$ denotes the total number of trials (failure +1 success) then the conditional expectations are:

$$E[Z\mid \text{Normal coin is chosen}]=\frac1{\frac12}=2$$

and

$$E[Z\mid \text{Fake coin is chosen}]=\frac1{\frac13}=3.$$

Ad then

$$E[Z]=2\frac47+3\frac37=\frac{17}7=\frac{10}7+1.$$

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