<p><a href="https://www.prepswift.com/quizzes/quiz/prepswift-expected-value" target="_blank">Expected Value Exercise</a></p><p><strong><span style="color:#8e44ad;">Expected Value</span></strong> is, in many ways, exactly what it sounds like. If you roll a fair, $6$-sided die enough times, what is the average value you can expect? In this way, expected value is much like weighted average (see previous mountain entry). Each outcome has a certain probability and value, and if we multiply those numbers together and add all the outcomes up, we get the expected value, or the weighted average.</p>
<p><strong><span style="color:#27ae60;">The Expected Value of a Roll of the Die</span></strong></p>
<p>$$\left(\frac{1}{6}\right)1 + \left(\frac{1}{6}\right)2 + \left(\frac{1}{6}\right)3 + \left(\frac{1}{6}\right)4 + \left(\frac{1}{6}\right)5 + \left(\frac{1}{6}\right)6 = 3.5$$</p>
<p style="margin-left: 40px;">So on average, we can "expect" a value of $3.5$ for each roll of a fair die.</p>
<p><strong>But what if the die is <span style="color:#e74c3c;"><u>NOT</u></span> fair?</strong></p>
<p><em>What would be the expected value of one roll of a die where each even number was three times as likely as each odd number? </em></p>
<p style="margin-left: 40px;"><strong><u>Step 1</u></strong>: Assume the probability of rolling an odd number is $x$. Knowing that all outcomes must sum to $1$, we can write the following equation:</p>
<p style="margin-left: 40px;">$$P_1 + P_2 + P_3 + P_4 + P_5 + P_6 = 1$$</p>
<p>$$x + 3x + x + 3x + x + 3x = 1$$</p>
<p>$$12x = 1$$</p>
<p>$$x = \left(\frac{1}{12}\right)$$</p>
<p style="margin-left: 40px;">This lets us know that the probability of rolling an odd number is $\frac{1}{12}$, and the probability of rolling an even number is three times that, $\frac{3}{12} = \frac{1}{4}$.</p>
<p style="margin-left: 40px;"><strong><u>Step 2</u></strong>: Do the weighted average calculation.</p>
<p style="margin-left: 40px;">$$\left(\frac{1}{12}\right)1 + \left(\frac{1}{4}\right)2 + \left(\frac{1}{12}\right)3 + \left(\frac{1}{4}\right)4 + \left(\frac{1}{12}\right)5 + \left(\frac{1}{4}\right)6$$</p>
<p style="margin-left: 40px;">$$=3.75$$</p>
<p><strong><span style="color:#27ae60;">Expected Value Also Deals with Money A Lot</span></strong></p>
<p>You'll often encounter expected value questions related to money. For example, imagine a gameshow host says the following to you:</p>
<p style="margin-left: 40px;"><em>If you correctly guess a randomly chosen positive integer from $1$ to $100$, inclusive, you will win $\$25$,$000$. However, to play this game you must pay $\$300$.</em> </p>
<p>What should you do? Should you play? This is where expected value becomes useful. Correctly guessing a randomly chosen integer from $1$ to $100$ has a probability of $\frac{1}{100}$, or $0.01$. So we can expect to win this game about $1\%$ of the time. The expected value is this value multiplied by the award or the prize money.</p>
<p style="text-align: center;">$0.01 \times \$25$,$000 =$ $\$250$</p>
<p>Given that we can expect to win $\$250$ on average every time we play this game, but have to pay $\$300$ to play the game one time, this is <u><strong>NOT</strong></u> a good deal. In fact, we can say the <strong><u>TRUE</u></strong> expected value of playing this game is...</p>
<p style="text-align: center;">$\$300 - \$250 =$ <span style="color:#e74c3c;">$-\$50$</span></p>
<p>It <strong><span style="color:#e74c3c;">COSTS</span></strong> you money to play this crappy game. So expected value can be very helpful in cases such as these.</p>