Assuming that we do not care about labels (i.e. -1 DeAza(outfield), -2 Par (d.h.), -3 Young (outfield),... is considered to be the same outcome as -1 DeAza(d.h.), -2 Par (outfield), -3 Young (outfield),...), as Andre points out this is a perfect time to use the Addition Principle.
Our lineup will consist of 1 catcher, 3 outfielders, 4 infielders, and one additional person (the designated hitter) from one of those categories or from the pitcher category. Split it up into the four cases:
- A) The designated hitter is a catcher
- B) The designated hitter is an infielder
- C) The designated hitter is an outfielder
- D) The designated hitter is a pitcher
Notice, that there is absolutely no way that when we count scenarios from each of these cases, that we accidentally count the same scenario a second time in a different case. I.e., these events are mutually exclusive and therefore we can apply the addition principle and add the amount of possibilities from each individual case to get the number of possibilities total.
Let us examine what happens in one of the cases in closer detail.
- Case A) The designated hitter is a catcher.
This means that among our nine players in the batting lineup, two of them are catchers, 3 of them are outfielders, and 4 of them are infielders. We wish to count how many different lineups satisfy this statement. Here we will use the Multiplication Principle. Set up a sequence of questions/choices to determine which lineup we have.
- Pick which two of the catchers are in the lineup ($\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup)
- Pick which three of the outfielders are in the lineup ($\binom{6}{3} = 20$ choices since there are six outfielders available and we want to choose three of them to appear in our lineup)
- Pick which four of the infielders are in the lineup ($\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup)
- Pick which zero of the pitchers are in the lineup ($\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the empty choice)
- Pick which order the nine people chosen appear in the lineup ($9! = 362880$ choices since every way of ordering the players counts as "different")
Thus, by the multiplication principle there are $\binom{2}{2}\binom{6}{3}\binom{5}{4}\binom{12}{0}9!=36288000$ different outcomes for the first case (case A).
Calculate the number in case B, case C, and case D similarly (in your comment on the other copy you only used a $\binom{12}{1}$ for case D, however that only counts how many ways to pick one pitcher to be the designated hitter; you still need to pick the rest of the team and where they are in the lineup).
For part B, we can approach this two ways. The easier way I think is this: first ask the question "What is the probability that Jones is not picked to be in the starting lineup?" By answering this question, we are able to then easily answer the follow up question "What is the probability that Jones is in the starting lineup?".
We will answer the question of the probability that he is not exactly like how we answered for the first part of the question with the following adjustment: since we want to count the number of cases Jones is not in the lineup, redo the calculations for the first part as though Jones weren't on the team at all (I.e. remove Jones from the list of available outfielders, making it so that there are only $5$ outfielders available).
We again break it into the same cases as before: A) d.h. is a catcher, B) d.h. is an infielder, C) d.h. is an outfielder, D) d.h. is a pitcher.
Let us examine the first case in detail again:
- Pick which two of the catchers are in the lineup ($\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup)
- Pick which three of the outfielders are in the lineup ($\binom{5}{3} = 10$ choices since there are $\color{red}{\star}$five$\color{red}{\star}$ outfielders available and we want to choose three of them to appear in our lineup) (since we don't want Jones)
- Pick which four of the infielders are in the lineup ($\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup)
- Pick which zero of the pitchers are in the lineup ($\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the empty choice)
- Pick which order the nine people chosen appear in the lineup ($9! = 362880$ choices since every way of ordering the players counts as "different")
For a total of $\binom{2}{2}\binom{5}{3}\binom{5}{4}\binom{12}{0}9!$ different outcomes for the first case. Calculate the other cases similarly.
Now that we have a tally of how many lineups do not have Jones on the roster, take the answer from the first half of the problem and subtract the number just calculated here to get the number of different ways that Jones IS on the starting lineup (as per a simple application of inclusion-exclusion ).
Take this newest number and divide by the answer found in the first half to get the probability that Jones is in the starting lineup as per the definition of probability in an equiprobable sample space:
$$Pr(A) := \frac{|A|}{|S|}$$
Complete the problem by noting that, given that Jones is in the starting lineup, he is equally likely to be in each of the spots and apply the multiplication rule (probability).
$$Pr(\text{Jones bats}~4^{th}) = Pr(\text{Jones in lineup})\cdot Pr(\text{Jones bats}~4^{th}~|~\text{Jones in lineup})$$
There is another equally acceptable approach to solving the second part. Directly count the number of ways that Jones bats fourth. Do so by breaking it into cases:
- A) The designated hitter is a catcher
- B) The designated hitter is an infielder
- C) The designated hitter is a pitcher
- D) The designated hitter is an outfielder
Looking at one of these cases more closely, looking at A again: Break it up via multiplication principle as before. Since the D.H. is a catcher, that means our lineup will consist of 2 catchers, Jones, 2 other outfielders, and 4 infielders.
- Pick which two of the catchers are in the lineup ($\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup)
- Pick which two of the other outfielders are in the lineup ($\binom{5}{2} = 10$ choices since there are $\color{red}{\star}$five$\color{red}{\star}$ outfielders available and we want to choose two of them to appear in our lineup) (since we already know Jones is in the lineup)
- Pick which four of the infielders are in the lineup ($\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup)
- Pick which zero of the pitchers are in the lineup ($\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the empty choice)
- Pick which order the nine people chosen appear in the lineup ($\color{red}{\star}8! = 40320\color{red}{\star}$ choices since Jones is already occupying the fourth slot, and we want to arrange the other eight players around him)
Do so similarly for the other cases to find the number of ways that Jones appears in the fourth batting position. Complete by applying the definition of probability (num outcomes for event / num outcomes regardless).
Let $E_i$ denote the expected number of additional hitters who will reach base given that there have been $i$ outs thus far. The desired quantity is $E_0$ since we want to know how many hitters will reach base over one inning (which starts with $0$ outs).
If $x$ is the (common) $\mathsf{OBP}$ of each batter, then the following equations hold:
$$\begin{align}
E_0 &= x (1 + E_0) + (1 - x) E_1 \\
E_1 &= x (1 + E_1) + (1 - x) E_2 \\
E_2 &= x (1 + E_2) + (1 - x) \cdot 0.
\end{align}
$$
The reason for the $x(1 + E_i)$ term is that with probability $x$ a hitter will reach base, and that adds $1$ to the count and puts the pitcher in the same situation of only having $i$ outs that they were in before that batter hit (hence $E_i$).
The reason for the $(1 - x)E_{i+1}$ term is that with probability $1 - x$ the count remains the same, and the pitcher is now in the situation that they have one more out to their advantage. Of course, the last equation has a $0$ there because if we already have two outs, and the batter is put out, then the pitcher’s inning is over, and there is nothing to add.
Working backwards, we find that $E_2 = x/(1 - x)$, $E_1 = 2 x/(1 - x)$, and $E_0 = 3x/(1 - x)$. Since you have already done the work to find that $x = 1 - 2^{-1/3}$, we have
$$E_0 \approx 0.779763.$$
Best Answer
Basically you can't, unless you know a lot more information. For example, suppose we have three batters $a,b,c$ with averages $0.75,0.5,0.25$ and three pitchers $A,B,C$ with BAAs $0.75, 0.5, 0.25$. If we multiply by $600$ at bats by each, $200$ for each pitcher/batter combination you are filling in a matrix $$\begin {array} {c|c|c|c|}&A&B&C&Total\\ \hline \\a&&&&450\\ \hline \\b&&&&300\\ \hline \\c&&&&150\\ \hline \\&450&300&150 \end{array}$$ You have six equations in nine unknowns, so lots of freedom. Each cell can contain any number from $0$ through $200$
Added: compare these two sets of hits: in each case, each pitcher pitches to each batter 200 times. The batting averages and BAAs are the same.
$$\begin {array} {c|c|c|c|}&A&B&C&Total\\ \hline \\a&200&150&100&450\\ \hline \\b&150&100&50&300\\ \hline \\c&100&50&0&150\\ \hline \\&450&300&150 \end{array}$$
$$\begin {array} {c|c|c|c|}&A&B&C&Total\\ \hline \\a&200&200&50&450\\ \hline \\b&200&100&0&300\\ \hline \\c&50&0&100&150\\ \hline \\&450&300&150 \end{array}$$
The first approximates the intuitive view you seem to have. The second supports the same data, and the worst batter is hitting $0.500$ against the best pitcher.