This problem compares a player who bats 300 against
all pitchers to a player that bats 200 against left handers and 400 against
right handers.
Question: Two
players have four plate appearances in every game. (I will assume all
plate appearances end in either a hit or an out.)
One player has a 300 batting average against all
pitchers. The other player hits 200
against left handers and 400 against right handers.
Over a 10game period, half the games are
against right handers and half the games are against left handers. (I am not allowing for substitution of right
handers for left handers or left handers for right handers in this problem.)
What is the likelihood that the 300 hitter has
at least one hit in all 10 games?
What is the likelihood that the hitter with a
200 average against left handers and a 400 average against right handers has at
least one hit in all ten games?
Answer: At least one hit
means 1, 2, 3, or 4 hits. It also means not 0 hits. Since 0 hits is
the complement of at least one hit we can calculate the probability of one hit
from:
P(at
least one hit) = 1  P(zero hit),
The
probability of zero hits is
P(zero
hits) = (1  BA)^{4}
The chart below has the probability of zero hits
for BA 200, 300, and 400.
Batting Average

P(Zero HIts)

P(at least one hit)

0.2

0.41

0.59

0.3

0.24

0.76

0.4

0.13

0.87

The likelihood of a hitter with a batting
average of 0.300 against both right and left handed pitching going 10 straight
games with at least one hit is 0.76^{10 } or 0.064.
The likelihood of a hitter with a batting
average of 0.200 against left handed hitters and 0.400 against right handed
hitters going 10 straight games with at least one hit 0.59^{5} x
0.87^{5 }is 0.036.
Both batters are 300 hitters. However, the
probability of a 10game streak is 1.77 times larger for the batter who can
high both lefthanded and righthanded pitchers at 300 compared to the hitter
who is 200 against left handers and 400 against right handers.
No comments:
Post a Comment