For Major League Baseball the expected win percentage is given by
For the 2021 season the Cincinnati Reds scored 786 runs and allowed 760 runs.
Based on the Win Ratio equation what is the expected win ratio to the nearest thousandth?
(Note that the thousandth is the 3rd digit after the decimal so a number such as 0.4279 would be rounded to 0.428.)
In a 162-game season how many wins would this have given the Reds (their real record was 83-79)? Round to the nearest whole number.
Show all work!
To calculate the expected win ratio, we can use the following equation:
Win Ratio = (Runs Scored^2) / (Runs Scored^2 + Runs Allowed^2)
Let's plug in the numbers given:
Runs Scored = 786
Runs Allowed = 760
Win Ratio = (786^2) / (786^2 + 760^2)
Now, let's calculate it:
Win Ratio = (616996) / (616996 + 577600)
Dividing the numerator by the denominator:
Win Ratio = 0.516
To round to the nearest thousandth, we look at the fourth decimal place:
The fourth decimal place for 0.516 is 6, which is less than 5, so we don't round up.
Therefore, the expected win ratio to the nearest thousandth is 0.516.
Next, let's calculate the number of wins for a 162-game season:
Expected Wins = Win Ratio * Number of Games
Using the expected win ratio of 0.516 and the number of games (162):
Expected Wins = 0.516 * 162
Calculating the multiplication:
Expected Wins = 83.832
Rounding to the nearest whole number, we get:
Expected Wins = 84 wins
Therefore, based on the win ratio equation, the Reds would be expected to have 84 wins in a 162-game season. This differs from their real record of 83-79.