experiment repeating the same experiment n times, probability of event
A occurring r times is
(^) nCr×Pr×( 1 −P)n−r
A hitter with a batting average of three hundred is considered a decent
hitter. Hitting 3 times out of 10, which means 7 failures(outs) are
allowed. More failures than successes sounds very paradoxical, but the
complicated combinations of them supports that it is not that easy.
Now, I calculated the probability using the same batting average for each
time at bat. However, if you watched baseball, you will know that the
batting averages constantly update each time at bat throughout the
game.
For example, let's say the batter A had hit 3 out of 9 to get an average of
⅓. , and we calculate the same probability: making a single hit three times
at bat. There were three combinations, HOO, OHO, and OOH, of an
outcome.
In outcome HOO, probability of making a hit is 3/9. Remember batting
average = H/AB. After making a hit in the first at bat, the batter’s average
becomes 4/10. Now on second at bat, the calculation is based on the new
batting average. Because the batter failed to make a hit on second at bat,
the number of hits does not increase, so the new average is 4/11.
The variation of a batting average before each time at bat are
3/9, 4/10, 4/11. In combination HOO, a batter is out on second and
third at bat. Our probability of this batter hitting a hit out of three times
at bat is 3/9x4/10x4/11=7/55. This probability is the same for
combination of OHO and OOH, 7/55x3= 21/55. 21/55 is slightly lower