The Problem If a player rolls a pair of dice and gets a sum of 2 or 12, he wins \$20. If the person gets a 7, he wins \$5. The cost to play the game is \$3. Find the expection (or expected value) of this game.

In order to find the expected value which is the mean of the probability distribution, we must first construct the probability distribution. We know that there are

6 x 6 = 36 possible outcomes when a pair of dice is rolled.

We know that the event, gets a sum of 2 or 12, contains 2 outcomes, namely (1,1) and (6,6). Further, when one of these outcomes occurs, then the person wins \$20 - \$3 = \$17. The probability of this event thus is = .

We also know that the event, gets a 7, has 6 possible outcomes namely (3,4), (4,3), (2,5), (5,2), (6,1) or (1,6). The winning amount for these outcomes is \$5 - \$3 = \$2. The probability of this even is = .

There are 28 remaining ways to loose \$3. The probability of this even is = .

Thus the probability distribution is as follows:

 x (amount of win) \$17 \$2 - \$3 P(x) (probability of win)

Using the formula E(x) = , we enter the x values in and enter the P(x) values in as shown below.

We can calculate the required sum of the x times p(x) values by pressing from the home screen 2nd STAT[LIST], pressing right arrow twice, and then pressing 5 to select 5:Sum(. Now press 2nd 1 to type an , press , and then press 2nd 2 to type an . Press ) followed by ENTER. The results are shown below.

Thus, we expect a LOSS of about \$1.06.

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