A game has the following rules. A fair die is tossed. The tosser wins $7 if a
1 or 2 shows and wins $11 if a 6 shows. The tosser pays $14 if a 3 shows and $6 if a 4 or 5
shows. The tosser's expected winnings for one toss is closest to?
Each side has a 1/6 chance of occurring. Multiply each value by the chance for each side. The either-or probability of the 6 choices is found by adding these products.
I hope this helps. Thanks for asking.
To find the tosser's expected winnings for one toss, we need to calculate the probability of each possible outcome and multiply it by the corresponding amount won or lost.
Let's calculate the probabilities first:
- The probability of getting a 1 or 2 is 2 out of 6, since there are two favorable outcomes out of six possible outcomes. So the probability is 2/6 = 1/3.
- The probability of getting a 6 is also 1/6, since there is one favorable outcome out of six possible outcomes.
- The probability of getting a 3 is 1/6, same as the probability of getting a 6.
- The probability of getting a 4 or 5 is 2 out of 6, so the probability is 2/6 = 1/3.
Now, let's calculate the expected winnings:
- For getting a 1 or 2, the tosser wins $7. So the expected winnings for this outcome is (1/3) * $7 = $7/3.
- For getting a 6, the tosser wins $11. So the expected winnings for this outcome is (1/6) * $11 = $11/6.
- For getting a 3, the tosser pays $14. So the expected winnings for this outcome is (1/6) * (-$14) = -$14/6.
- For getting a 4 or 5, the tosser pays $6. So the expected winnings for this outcome is (1/3) * (-$6) = -$6/3.
Now, we can add up the expected winnings for each outcome:
Expected winnings = ($7/3) + ($11/6) + (-$14/6) + (-$6/3)
= ($14/6) + ($11/6) + (-$14/6) + (-$6/6)
= ($14 + $11 - $14 - $6)/6
= $5/6
So, the tosser's expected winnings for one toss are $5/6.