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) |
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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