Question

A game consists of tossing 2 coins. If they are both heads, you win $4. If they are both tails, you win $6. If one of each comes up, you lose $8. What are your expected winnings?

Question 10 options:

$2.00


-$1.50


-$2.00


$1.50

Answer 1

To calculate the expected winnings from the game, we first need to determine the probabilities for each possible outcome when tossing 2 coins.

When tossing 2 coins, the possible outcomes are:
1. Heads & Heads (HH)
2. Tails & Tails (TT)
3. Heads & Tails (HT)
4. Tails & Heads (TH)

Each outcome has a probability of \( \frac{1}{4} \) because there are 4 equally likely outcomes.

Now we can calculate the winnings associated with each outcome:
- For HH (both heads), you win $4.
- For TT (both tails), you win $6.
- For HT (one head, one tail), you lose $8.
- For TH (one tail, one head), you lose $8.

Next, we can calculate the expected value (EV) of the winnings using the formula:

\[
EV = (P(HH) \times W(HH)) + (P(TT) \times W(TT)) + (P(HT) \times W(HT)) + (P(TH) \times W(TH))
\]

Where:
- \( P(HH) = \frac{1}{4}, W(HH) = 4 \)
- \( P(TT) = \frac{1}{4}, W(TT) = 6 \)
- \( P(HT) = \frac{1}{4}, W(HT) = -8 \)
- \( P(TH) = \frac{1}{4}, W(TH) = -8 \)

Now substituting in the values:

\[
EV = \left(\frac{1}{4} \times 4\right) + \left(\frac{1}{4} \times 6\right) + \left(\frac{1}{4} \times -8\right) + \left(\frac{1}{4} \times -8\right)
\]

Calculating each term:

\[
EV = 1 + 1.5 - 2 - 2 = 1 + 1.5 - 4 = 2.5 - 4 = -1.5
\]

Thus, the expected winnings are:

\[
\text{Expected Winnings} = -1.5
\]

The answer is \(-1.50\).