7.2 Expected Value and Standard Deviation for a Discrete Probability Distribution

A men’s soccer team plays soccer zero one or two days a week.  The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3.  Find the long-term average or expected value of the number of days per week the men’s soccer team plays soccer.

Solution:

First, let the random variable [latex]X[/latex] be the number of days the men’s soccer team plays soccer per week. [latex]X[/latex]  takes on the values 0, 1, 2. The table below shows the probability distribution for [latex]X[/latex], and includes an additional column [latex]x \times P(x)[/latex] that we will use to calculate the expected value.  In this new column, we will multiply each [latex]x[/latex] value by its corresponding probability.

Add the last column to find the long-term average or expected value:

[latex]\begin{eqnarray*}E(x) & = & (0 \times 0.2) +(1 \times 0.5)+ (2 \times 0.3) \\ & = & 0+ 0.5+ 0.6 \\ & = & 1.1 \\ \end{eqnarray*}[/latex]

The expected value is 1.1. The men’s soccer team would, on average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men’s soccer team plays soccer week after week after week.

Let [latex]X[/latex] be the number of times per week a newborn baby’s crying wakes its mother after midnight.  The probability distribution for [latex]X[/latex] is:

Find the expected value and standard deviation of the number of times a newborn baby’s crying wakes its mother after midnight.

Solution: 

For the expected value:

[latex]\begin{eqnarray*} \\ \mu & = & 0 +0.22+ 0.92 +0.54 +0.32+ 0.1 \\ & = & 2.1 \\ \\ \end{eqnarray*}[/latex]

On average, a newborn wakes its mother after midnight 2.1 times per week.

For the standard deviation:  For each value [latex]x[/latex], multiply the square of its deviation by its probability (each deviation has the format [latex]x-\mu[/latex]).

Add the values in the third column of the table and then take the square root of this sum:

[latex]\begin{eqnarray*}\sigma&=&\sqrt{1.05}\\&=&1.024...\end{eqnarray*}[/latex]

Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.  A computer randomly selects five numbers from zero to nine with replacement.  You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000).  Over the long term, what is your expected profit of playing the game?

Solution:

To do this problem, set up an expected value table for the amount of money you can profit.  Let [latex]X[/latex] be the amount of money you profit.  The values of [latex]x[/latex] are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.  Because you are interested in your profit (or loss), the values of [latex]x[/latex] are $100,000 and −$2 dollars.

To win, you must get all five numbers correct, in order.  The probability of choosing one correct number is
[latex]\displaystyle{\frac{1}{10}=0.1}[/latex] because there are ten numbers.  You may choose a number more than once.  The probability of choosing all five numbers correctly and in order is

[latex]\displaystyle{\frac{1}{10} \times\frac{1}{10} \times\frac{1}{10} \times\frac{1}{10} \times\frac{1}{10}=0.00001}[/latex]

Therefore, the probability of winning is 0.00001 and the probability of losing is [latex]\displaystyle{1-0.00001=0.99999}[/latex].

The expected value is as follows:

[latex]\begin{eqnarray*} \\ E(x) & = & -1.99998 +1 \\ & = & -0.99998 \\ \\ \end{eqnarray*}[/latex]

Because –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play.  However, each time you play, you either lose $2 or profit $100,000. The $1 is the average (or expected) LOSS per game after playing this game over and over.

Suppose you play a game with a biased coin where the probability of heads is [latex]\displaystyle{\frac{2}{3}}[/latex].  You play each game by tossing the coin once.  If you toss a head, you pay $6. If you toss a tail, you win $10.  If you play this game many times, will you come out ahead?

1. Define a random variable [latex]X[/latex].
2. Construct the probability distribution for [latex]X[/latex].
3. What is the expected value? Do you come out ahead?

Solution:

1. Let [latex]X[/latex] be the amount of profit per game.  The values of [latex]x[/latex] are –$6 (for a loss) and $10 (for a win).

2. [latex]x[/latex]	[latex]P(x)[/latex]
   [latex]10[/latex]	[latex]\displaystyle{\frac{1}{3}}[/latex]
   [latex]-6[/latex]	[latex]\displaystyle{\frac{2}{3}}[/latex]

3. [latex]x[/latex]	[latex]P(x)[/latex]	[latex]x \times P(x)[/latex]
   [latex]10[/latex]	[latex]\displaystyle{\frac{1}{3}}[/latex]	[latex]\displaystyle{\frac{10}{3}}[/latex]
   [latex]-6[/latex]	[latex]\displaystyle{\frac{2}{3}}[/latex]	[latex]-4[/latex]

   [latex]\begin{eqnarray*} \\ E(x) & = & \frac{10}{3}+ (-4) \\ & = & -0.67 \\ \\ \end{eqnarray*}[/latex]

     On average, you lose $0.67 each time you play the game, so you do not come out ahead.