Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, one chip has the number 3, and the other chip has the number 5. Miguel must choose two chips, and if both chips have the same number, he wins $2. If the two chips he chooses have different numbers, he loses $1 (–$1).
Let X = the amount of money Miguel will receive or owe. Fill out the missing values in the table. (Hint: The total possible outcomes are six because there are four chips and you are choosing two of them.)
What is Miguel’s expected value from playing the game?
Based on the expected value in the previous step, how much money should Miguel expect to win or lose each time he plays?
What value should be assigned to choosing two chips with the number 1 to make the game fair? Explain your answer using a complete sentence and/or an equation.
To find Miguel's expected value from playing the game, we need to calculate the weighted average of the possible outcomes.
In this case, there are six possible outcomes:
1. Choosing both chips with the number 1: Miguel wins $2
2. Choosing one chip with the number 1 and one chip with the number 3: Miguel loses $1
3. Choosing one chip with the number 1 and one chip with the number 5: Miguel loses $1
4. Choosing both chips with the number 3: Miguel wins $2
5. Choosing one chip with the number 3 and one chip with the number 5: Miguel loses $1
6. Choosing both chips with the number 5: Miguel wins $2
To calculate the weighted average, we multiply each outcome by its probability and sum them up.
Probability of outcome 1: (2/4) * (1/3) = 1/6
Probability of outcomes 2 and 3: (2/4) * (2/3) = 4/9
Probability of outcome 4: (1/4) * (1/3) = 1/12
Probability of outcome 5: (1/4) * (2/3) = 1/6
Probability of outcome 6: (1/4) * (1/3) = 1/12
This gives us the following expected value calculation:
Expected value = (1/6 * $2) + (4/9 * -$1) + (1/12 * $2) + (1/6 * -$1) + (1/12 * $2)
= $0.33 - $0.44 + $0.16 - $0.17 + $0.16
= -$0.04
Therefore, Miguel's expected value from playing the game is -$0.04.
Based on the expected value, Miguel should expect to lose an average of $0.04 each time he plays the game.
To fill out the missing values in the table, let's first list out all the possible outcomes:
1) Two chips with the number 1: Miguel wins $2
2) Two chips with the number 3: Miguel loses $1
3) Two chips with the number 5: Miguel wins $2
4) One chip with the number 1 and one chip with the number 3: Miguel loses $1
5) One chip with the number 1 and one chip with the number 5: Miguel wins $2
6) One chip with the number 3 and one chip with the number 5: Miguel loses $1
Now, let's fill out the missing values in the table:
Outcome | Probability | Money (X)
-------------------------------------------------
Two 1's | ? | +$2
Two 3's | ? | -$1
Two 5's | ? | +$2
One 1 and one 3 | ? | -$1
One 1 and one 5 | ? | +$2
One 3 and one 5 | ? | -$1
The next step is to calculate the probabilities for each outcome. Since Miguel is choosing two chips out of four, the total number of possible outcomes is 4C2 = 6. Therefore, each outcome has a probability of 1/6.
Outcome | Probability | Money (X)
-------------------------------------------------
Two 1's | 1/6 | +$2
Two 3's | 1/6 | -$1
Two 5's | 1/6 | +$2
One 1 and one 3 | 1/6 | -$1
One 1 and one 5 | 1/6 | +$2
One 3 and one 5 | 1/6 | -$1
Now, let's calculate Miguel's expected value. The expected value (E) is calculated by summing the products of each outcome's probability and the corresponding money.
E = (1/6 * $2) + (1/6 * -$1) + (1/6 * $2) + (1/6 * -$1) + (1/6 * $2) + (1/6 * -$1)
E = ($2/6) + (-$1/6) + ($2/6) + (-$1/6) + ($2/6) + (-$1/6)
E = $1/6
Therefore, Miguel's expected value from playing the game is $1/6.
Based on the expected value in the previous step, Miguel should expect to win or lose $1/6 each time he plays.
To make the game fair, the value assigned to choosing two chips with the number 1 should result in an expected value of $0. Since the current expected value is $1/6, we need to subtract the money won for choosing two 1's to make it equal to zero.
E - (1/6 * $2) = 0
$1/6 - (1/6 * $2) = 0
$1/6 - $2/6 = 0
-$1/6 = 0
Since this equation is not true, there is no value that can be assigned to choosing two chips with the number 1 to make the game fair.
To fill out the missing values in the table, let's first list all the possible outcomes when choosing two chips from the box:
1. Choosing two chips with the number 1: There are two ways to do this: chip 1 and chip 2, or chip 1 and chip 3. Therefore, the outcome is 2.
2. Choosing one chip with the number 1 and one chip with the number 3: There are two ways to do this: chip 1 and chip 4, or chip 2 and chip 3. Therefore, the outcome is -2.
3. Choosing one chip with the number 1 and one chip with the number 5: There is one way to do this: chip 1 and chip 5. Therefore, the outcome is -2.
4. Choosing two chips with the number 3: There is one way to do this: chip 3 and chip 4. Therefore, the outcome is 2.
5. Choosing one chip with the number 3 and one chip with the number 5: There is one way to do this: chip 3 and chip 5. Therefore, the outcome is -2.
6. Choosing two chips with the number 5: There are two ways to do this: chip 4 and chip 5, or chip 5 and chip 6. Therefore, the outcome is 2.
Now, let's fill out the table:
```
| Chip 1 | Chip 3 | Chip 5
--------|-------------|------------|-----------
Chip 1 |-2 (loss) | -2 (loss) | 2 (win)
Chip 3 | -2 (loss) | 2 (win) | -2 (loss)
Chip 5 | 2 (win) | -2 (loss) | 2 (win)
```
To calculate Miguel's expected value from playing the game, we need to find the average outcome (X) by summing up all the possible outcomes and dividing by the number of outcomes:
X = (-2 - 2 + 2 - 2 + 2 + 2) / 6 = 0 / 6 = 0
Therefore, Miguel's expected value from playing the game is $0.
Based on the expected value, Miguel can expect to neither win nor lose money on average each time he plays. This means that over multiple plays of the game, his winnings will even out to zero.
To make the game fair, the value assigned to choosing two chips with the number 1 should be adjusted so that the expected value is $0. Since we already have two chips with the number 1, the value of winning should be assigned as follows:
2 (winning) * P(choosing two chips with the number 1) = -2 (losing) * P(choosing one chip with the number 1 and one chip with another number)
Let's assume "x" is the value assigned to winning, then:
2 * x = -2 * (2 / 4)
2x = -1
x = -1/2
Therefore, the value assigned to choosing two chips with the number 1 to make the game fair would be -$0.50 or -$1/2.