# Baseball Combinations Problem

Two part question (My work below).

For both questions will use the orioles current roster:

-Current orioles roster:

12 pitchers, 2 catchers, 5 in-fielders, and 6 out-fielders: Similar to the list here but not identical http://www.foxsports.com/mlb/baltimore-orioles-team-roster?season=2015

In the American League, the pitcher in the game does not bat. Each of the other 9 players on the team are put into a “batting order”. (Somebody bats first, somebody bats second, etc.).

Below is a typical batting order:

-1. De Aza (outfielder)

-2. Paredes (designated hitter)

-3. Young (outfielder)

-4. Jones (outfielder)

-5. Davis (infielder)

-7. Joseph (catcher)

-8. Cabrera (infielder)

-9. Navarro (infielder)

Question A: How many different batting orders are possible for the Orioles?

Hints: -You have to have 1 catcher, three outfielders, four infielders, and a designated hitter (who could be anyone, even those players who are listed under the “pitchers” section of the roster).

My answer: I'm choosing 1 catcher, 3 out-fielders, 4 in-fielders, and 1 designated hitter of what is left of the roster. So: $${2 \choose 1}*{6 \choose 3}*{5 \choose 4}*{17 \choose 1} = 3400$$

Question B.: If the batting order is selected randomly (assume each viable batting order from the previous question is equally likely), then what is the probability that Jones will be batting in the fourth position? (Be careful: Jones could either be in the game as an outfielder or as designated hitter!)

Thinking about B a lot but have no idea on how to approach this one. Any hints and suggestions appreciated thank you.

• For the first, we are interested in batting orders. Any viable choice of people for the batting rotation gives rise to $9!$ distinct batting orders. For the second, I am not happy about the assumption that all viable batting orders are equally likely. There is a complication because of the pitcher. – André Nicolas May 2 '15 at 16:38
• Ah so I should approach this as a permutation. Would this approach work then (13 choose 1)(11 choose 3)(8 choose 4)(17 choose 1). My logic is I have 13 out of the 25 players to choose for a player that is not a designated player. So I choose 1 catcher, 3 out fielders, and 4 in fielders and at the end choose 1 designated hitter from remaining roster. – Retuxan May 2 '15 at 16:51
• We need to worry about the order of batting. I do not see from your comment how your calculation would go in detail. One way to do it is to break into cases, the DH is a catcher, an infielder, an outfielder, a pitcher. Calculate for each the number of ways to choose, add up, multiply by $9!$. – André Nicolas May 2 '15 at 17:28
• There are many ways. If we for example use an outfielder as DH, there are $\binom{6}{4}\binom{2}{1}\binom{5}{4}9!$ batting orders. Similar expressions for if we use a catcher, and so on. Add up. – André Nicolas May 2 '15 at 19:04
• possible duplicate of A Baseball Combinations Problem – apnorton May 3 '15 at 2:18

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).