Paul flips a coin twice. If two heads come up, he wins $3. If two tails come up, he wins $2. Otherwise, he loses $2.
What is the expected value for this game of chance ?
The sample space for this event is:
\left\{ HH, HT, TH, TT \right\}
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | Value (Random Variable X ) | Probability of X, P\left(X\right) |
| Two Heads Come Up (HH) | $3 | \dfrac{1}{4} |
| Two Tails Come Up (TT) | $2 | \dfrac{1}{4} |
| Head First, Tail Second (HT) | -$2 | \dfrac{1}{4} |
| Tail First, Head Second (TH) | -$2 | \dfrac{1}{4} |
The expected value can be calculated as:
E\left(X\right)=\dfrac{1}{4}\cdot3+\dfrac{1}{4}\cdot2-\dfrac{1}{4}\cdot2-\dfrac{1}{4}\cdot2=\dfrac{1}{4}=0.25
The expected value is 0.25.
There are 3 red marbles and 4 blue marbles in a bag. Richard is playing a game with his friend. If he picks a red marble out of the bag, then he gets $14. If he picks a blue marble, then he loses $7.
What is the expected value of this game?
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | Value (Random Variable X ) | Probability of X, P\left(X\right) |
| Richard picks a Red marble | $14 | \dfrac{3}{7} |
| Richard picks a Blue marble | -$7 | \dfrac{4}{7} |
The expected value can be calculated as:
E\left(X\right)=\dfrac{3}{7}\cdot14-\dfrac{4}{7}\cdot7=2
The expected value is 2.
Emma rolls a die. If she gets 6, then she wins $10. If she gets 1, then she loses $12. If she gets 2, 3, 4, or 5, then she wins $2, $3, $4 or $5 respectively.
What is the expected value for this game of chance ?
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | Value (Random Variable X ) | Probability of X, P\left(X\right) |
| 1 | -$12 | \dfrac{1}{6} |
| 2 | $2 | \dfrac{1}{6} |
| 3 | $3 | \dfrac{1}{6} |
| 4 | $4 | \dfrac{1}{6} |
| 5 | $5 | \dfrac{1}{6} |
| 6 | $10 | \dfrac{1}{6} |
The expected value can be calculated as:
E\left(X\right)=-\dfrac{1}{6}\cdot12+\dfrac{1}{6}\cdot2+\dfrac{1}{6}\cdot3+\dfrac{1}{6}\cdot4+\dfrac{1}{6}\cdot5+\dfrac{1}{6}\cdot10
E\left(X\right)=-\dfrac{12}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}+\dfrac{10}{6}=2
The expected value is 2.
In a standard 52 deck, we randomly pick a card. If a face is drawn, then we win $3. If an ace is drawn, then we win $13. Otherwise, we lose $1.
What is the expected value for this game of chance?
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | Value (Random Variable X ) | Probability of X, P\left(X\right) |
| We pick a face | $3 | \dfrac{12}{52} |
| We pick an ace | $13 | \dfrac{4}{52} |
| No aces and no faces | -$1 | \dfrac{36}{52} |
The expected value can be calculated as:
E\left(X\right)=3\cdot \dfrac{12}{52}+13\cdot \dfrac{4}{52}-1\cdot \dfrac{36}{52}=1
The expected value is 1.
In a game of chance, we roll two dice simultaneously. Let n be the sum of the numbers we get.
- If n \ge 10, then we win $n.
- If n \le 4, then we lose $n.
- If 5 \le n \le 9, we don't win or lose anything.
What is the expected value for this game of chance ?
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | Value (Random Variable ) | Probability |
| n=10 | $10 | \dfrac{3}{36} (4,6) (5,5) (6,4) |
| n=11 | $11 | \dfrac{2}{36} (5,6) (6,5) |
| n=12 | $12 | \dfrac{1}{36} (6,6) |
| n=4 | -$4 | \dfrac{3}{36} (1,3) (2,2) (3,1) |
| n=3 | -$3 | \dfrac{2}{36} (1,2) (2,1) |
| n=2 | -$2 | \dfrac{1}{36} (1,1) |
The expected value can be calculated as:
E\left(X\right)=\dfrac{3}{36}\cdot10+\dfrac{2}{36}\cdot11+\dfrac{1}{36}\cdot12-\dfrac{3}{36}\cdot4-\dfrac{2}{36}\cdot3-\dfrac{1}{36}\cdot2=\dfrac{11}{9}
The expected value is \dfrac{11}{9}.
Louise and Paul are playing a game. Paul writes a number n from 0 to 9 on a piece of paper and asks Louise to guess the number. If she can guess the number correctly, then Paul will give her $10. Otherwise, she must give $2 to Paul.
What is the value of this game?
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | Value (Random Variable X ) | Probability of X, P\left(X\right) |
| Correct number | $10 | \dfrac{1}{10} |
| Wrong number | -$2 | \dfrac{9}{10} |
The expected value can be calculated as:
E\left(X\right)=\dfrac{1}{10}\cdot10-\dfrac{9}{10}\cdot2=-0.8
The expected value is -0.8.
We flip a coin 3 times. For each heads up we win $2, and for each tail we lose $1.
What is the value of this game of chance?
The expected value is the sum of each value, called X, times its probability P\left(X\right) :
E\left(X\right)=\sum_{}^{}X\cdot P\left(X\right)
| Event | value | Probability |
| (H,H,H) | 6 | \dfrac{1}{8} |
| (H,H,T) | 3 | \dfrac{1}{8} |
| (H,T,H) | 3 | \dfrac{1}{8} |
| (H,T,T) | 0 | \dfrac{1}{8} |
| (T,H,H) | 3 | \dfrac{1}{8} |
| (T,H,T) | 0 | \dfrac{1}{8} |
| (T,T,H) | 0 | \dfrac{1}{8} |
| (T,T,T) | -3 | \dfrac{1}{8} |
The expected value can be calculated as:
E\left(X\right)=\dfrac{1}{8}\cdot6+\dfrac{1}{8}\cdot3+\dfrac{1}{8}\cdot3+\dfrac{1}{8}\cdot0+\dfrac{1}{8}\cdot3+\dfrac{1}{8}\cdot0+\dfrac{1}{8}\cdot0-\dfrac{1}{8}\cdot3
E\left(X\right)=\dfrac{1}{8}{\cdot\left(6+3+3+3-3\right)}=\dfrac{3}{2}
The expected value is \dfrac{3}{2}