### Instruction: A very bizarre weighted coin comes up heads with probability $\frac12$, tails with probability $\frac13$, and rests on its edge with probability $\frac16$.  If it comes up heads, I win 1 dollar.  If it comes up tails, I win 3 dollars.  But if it lands on its edge, I lose 5 dollars.  What is the expected winnings from flipping this coin?  Express your answer as a dollar value, rounded to the nearest cent.### Response: This is a problem about expected value, which is the average outcome of a random event over many trials.
To find the expected value, I need to multiply each possible outcome by its probability, and then add them up.
The possible outcomes are 1, 3, and -5 dollars, and their probabilities are $\frac12$, $\frac13$, and $\frac16$ respectively.
So the expected value is $(1)(\frac12) + (3)(\frac13) + (-5)(\frac16)$.
To simplify this expression, I need to find a common denominator for the fractions.
The lowest common denominator is 6, so I can rewrite the fractions as $\frac36$, $\frac26$, and $\frac16$.
Now the expected value is $(1)(\frac36) + (3)(\frac26) + (-5)(\frac16)$.
Distributing the numbers, I get $\frac36 + \frac66 - \frac56$.
Adding the fractions, I get $\frac{3+6-5}{6} = \frac46$.
Simplifying the fraction, I get $\frac23$.
This means that on average, I expect to win $\frac23$ of a dollar, or about 67 cents, from flipping this coin.