Probability?

Question

Suppose a pitcher faces a batter who never swings and furthermore the pitcher throws balls ans strikes at random. For how many ball/strike sequences will the batter be called out on strikes on the 5th pitch?

please explain your answer so I can understand how this is done...thanks

Answer 1

Since the batter never swings, every pitch is either a ball or a strike (never a foul ball).

Since the batter is called out of the 5th pitch, the 5th pitch has to be a strike, and the batter has to have 2 strikes PRIOR to the 5th pitch (i.e., after 4 balls, the count is "2 and 2").

Therefore, the question is:
In how many ways can the first four pitches include exactly 2 strikes?

This is the same thing as asking:
Given 4 objects (i.e., pitches 1, 2, 3, and 4),
in how many ways can we select 2 objects (i.e., the two pitches that will be strikes)?

Here's the mathematical approach to the question:
There are 4 ways to choose one of the first 4 pitches to be a strike (i.e., it could be 1, 2, 3, or 4).
After choosing that one pitch, there are 3 ways to choose a second pitch to be a strike (i.e., it could be any of the other 3 pitches).
So there are 12 ways to select one pitch, then a different pitch (4 ways to make the first choice, times 3 ways to make the second choice, equals 12).
But those 12 sets of choices include each pair of pitches twice, because the 2 pitches could be selected in either order. (For example, one of the 12 pairs would be the 2nd and 4th pitches, and another would be the 4th and 2nd pitches. But that's the same pair of pitches, so we've counted it twice.)
Since each pair of pitches is counted twice, we have to divide by 2, and the final answer is 12 / 2 = 6. There are 6 possible pitch sequences that produce a strikeout on the 5th pitch.

Here's the "enumeration" (or "listing") approach to finding out how many ways there are:

The first two strikes could occur on pitches 1 & 2 (call that "1-2") and in 5 other ways for a total of 6 ways. In each case, the third strike occurs on the fifth pitch.
Here are the 6 possible combinations:
1-2
1-3
1-4
2-3
2-4
3-4

Answer 2

The 5th pitch would need to be the 3rd strike.

This means that you have to arrange exactly 2 strikes in the first 4 positions:

If you place a strike 1st, then you can place the second strike 2nd, 3rd, or 4th:

SSBBS
SBSBS
SBBSS

If you place a ball 1st, instead, then there's only one ball between 2nd, 3rd, and 4th, and it can also be in any of those three positions:

BBSSS
BSBSS
BSSBS

That's a total of six combinations.

This was a simple enough problem to work out all the possibilities by hand. You could also do it mathematically:

With 2 strikes in 4 positions, there are:
* 4 slots to choose from for one strike,
times
* 3 remaining slots for the 2nd strike,
divided by
* 2 permutations for the two positions

(The reason for the division by two: if you pick a strike 1st then a strike 3rd, that's the same as picking a strike 3rd and then a strike 1st.)

This calculation is called "4C2" ("four choose two") in combinatorics. aCb is equal to a!/(b!(a-b)!), which in this case is 4!/(2!(4-2)!) = 4*3*2*1/(2*1 * 2*1) = 4*3/2 = 6.

If this were a more difficult problem (like throwing 10 strikes in the first 15 pitches), it would not be practical to list all the combinations. But the calculation is easy: 15!/(10!(15-10)!) = 15*14*13*12*11 / 5*4*3*2 = 3,003 combinations