To solve this, we first determine the expected value (EV) of a single game.

Step 1: Compute the probability of winning

You win if you get exactly two heads in three coin tosses. The probability of this happening can be found using the binomial formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n = 3$ (total tosses),
• $k = 2$ (successful heads),
• $p = 0.5$ (probability of heads).

$P(2H) = \binom{3}{2} (0.5)^2 (0.5)^1 = 3 \times 0.25 \times 0.5 = 0.375$

So, the probability of winning is 0.375, and the probability of losing is:

$P(\text{Loss}) = 1 - 0.375 = 0.625$

Step 2: Compute the Expected Value of One Game

The expected value formula is:

$EV = (\text{Win Payout} \times P(\text{Win})) + (\text{Loss Payout} \times P(\text{Loss}))$

Substituting the given values:

$EV = (55 \times 0.375) + (-28 \times 0.625)$

$EV = 20.625 - 17.5 = 3.125$

So, for each game, you expect to win $3.125 on average.

Step 3: Compute the Total Expected Value for 914 Games

Multiply the expected value per game by the number of games:

$\text{Total EV} = 3.125 \times 914$

$= 2856.25$

Final Answer:

If you played this game 914 times, you would expect to win $2856.25.

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Do you have any questions, or would you like more details?

Here are five related questions to explore further:

1. What is the probability of getting exactly one head in three tosses?
2. How does the expected value change if the payout for winning increases?
3. If you played only 100 games, what would your expected earnings be?
4. What is the variance and standard deviation of this game’s earnings?
5. How does the law of large numbers apply to this game?

Tip: Expected value does not guarantee an exact amount but gives an average outcome over many trials!