Examples

Gambling game - tokens in bins

Consider a game like this: a coin is flipped; if $H$ then draw a token from Bin 1, if $T$ then from Bin 2.

• Bin 1 contents: 1 token $1,000, and 9 tokens $1
• Bin 2 contents: 5 tokens $50, and 5 tokens $1

It costs $50 to enter the game. Should you play it? (A lot of times?) How much would you pay to play?

Solution

(1) Setup:

Let $X$ be a random variable measuring your winnings in the game.

The possible values of $X$ are 1, 50, and 1000.

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(2) Find the PDF of $X$:

For $k=1$ have $P_X(1)\gg \gg \frac{1}{2}\cdot \frac{9}{10}+\frac{1}{2}\cdot \frac{5}{10}\gg \gg \frac{7}{10}$

For $k=50$ have $P_X(50)\gg \gg \frac{1}{2}\cdot \frac{5}{10}\gg \gg \frac{1}{4}$

For $k=1000$ have $P_X(1000)\gg \gg \frac{1}{2}\cdot \frac{1}{10}\gg \gg \frac{1}{20}$

These add to 1, and $P_X(x)=0$ for all other $x$.

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(3) Find $E[X]$ using the discrete formula:

$$\begin{array}{c}E[X]=\sum _{k}k\cdot P_X(k)\gg \gg 1\cdot P_X(1)+50\cdot P_X(50)+1000\cdot P_X(1000)\\ \\ \gg \gg 1\cdot \frac{7}{10}+50\cdot \frac{1}{4}+1000\cdot \frac{1}{20}\gg \gg \approx 63.2\end{array}$$

Since $63.2>50$, if you play it a lot at $50 you will generally make money.

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Challenge Q: If you start with $200 and keep playing to infinity, how likely is it that you go broke?

Expected value: rolling dice

Let $X$ be a random variable counting the number of dots given by rolling a single die.

Then:

$$E[X]\gg \gg 1\cdot \frac{1}{6}+2\cdot \frac{1}{6}+\cdots +6\cdot \frac{1}{6}\gg \gg \frac{7}{2}$$

Let $S$ be an RV that counts the dots on a roll of two dice.

The PMF of $S$:

Then:

$$E[S]\gg \gg 2\cdot \frac{1}{36}+3\cdot \frac{2}{36}+4\cdot \frac{3}{36}+\cdots +12\cdot \frac{1}{36}\gg \gg 7$$

Notice that $\frac{7}{2}+\frac{7}{2}=7$.

In general, $E[X+Y]=E[X]+E[Y]$.

Let $X$ be a green die and $Y$ a red die.

From the earlier calculation, $E[X]=\frac{7}{2}$ and $E[Y]=\frac{7}{2}$.

Since $S=X+Y$, we derive $E[S]=7$ by simple addition!

Expected value by finding new PMF

Let $X$ have distribution given by this PMF:

Find $E[|X-2|]$.

Solution

(1) Compute the PMF of $|X-2|$:

PMF arranged by possible value:

$$\begin{array}{cc}P[|X-2|=0]& \gg \gg P[X=2]=\frac{1}{14}\\ P[|X-2|=1]& \gg \gg P[X=1]+P[X=3]=\frac{1}{7}+\frac{3}{14}=\frac{5}{14}\\ P[|X-2|=2]& \gg \gg P[X=4]=\frac{2}{7}\\ P[|X-2|=3]& \gg \gg P[X=5]=\frac{2}{7}\\ P[|X-2|=k]& \gg \gg 0\text{for}k\ne \mathrm{0,1,2,3.}\end{array}$$

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(2) Calculate the expectation:

Using formula for discrete PMF:

$$E[|X-2|]\gg \gg 0\cdot \frac{1}{14}+1\cdot \frac{5}{14}+2\cdot \frac{2}{7}+3\cdot \frac{2}{7}\gg \gg \frac{25}{14}$$

Variance for composite using PMF and simpler formula

Suppose $X$ has this PMF:

$k:$	1	2	3
$P_X(k):$	$1/7$	$2/7$	$4/7$

Find $\mathrm{Var}\left[\frac{1}{1+X}\right]$ using the formula $\mathrm{Var}[Y]=E[Y^2]-E{[Y]}^2$ with $Y=\frac{1}{1+X}$.

(Hint: you should find $E[Y]=\frac{13}{42}$ and $E[Y^2]=\frac{13}{126}$ along the way.)