Counting Problem: Briefcases Suppose you and a friend are presented with nine identically-looking briefcases, three of which have $100 in them while the rest are empty. You both take turns picking briefcases (with the first person choosing 5 and the second choosing 4), and only at the end are the briefcases opened. I have done 1-4 parts of these HW questions but am unsure how to do the last:

*

*If you go first, what is the probability that you pick an empty briefcase on your first turn.

*Say you went first and picked an empty briefcase. What is the probability that your friend also picks an empty briefcase on their first turn, given that you also picked an empty one your first turn.

*Say you went first and picked a briefcase with 100 dollars. What is the probability that your friend picks an empty briefcase on their first turn, given that you also picked one with 100 dollars on your first turn.

*If the game were to end after each person has only one go, show that the probability that you choose a briefcase with 100 dollars is the same regardless of whether you start or not.

*If the game were to end once each briefcase was picked, would it be better to start or go second?

EDIT: Goal of the game is to make as little money as possible since 'gambling is bad'.

My attempt:

*

*The answer is the no. of ways you can pick an empty briefcase divided by the no. of ways you can choose a briefcase. This is 6C1 (since there are 6 empty briefcases) divided by 9C1 which is equal to 2/3.


*I believe the answer to this is 2/3 (from part 1) multiplied by 5C1 (since now there are 5 empty briefcases remaining and the friend picks one of these) divided by 8C1 (since now there are only 8 remaining cases to choose from). This is approx. 0.4167.


*I believe this is $\frac{3C1}{9C1}\cdot\frac{6C1}{8C1}$ since the first person must pick one of the three money briefcases out of a total of 9, while the second person must choose one of the 6 empty briefcases out of the 8 remaining. This is 0.25.


*The way I thought to approach this was by considering four different events, where:

*

*Event A: You go first and get 100 dollars, then friend chooses empty.

*Event B: You go first and get 100 dollars, then friend also gets 100 dollars.

*Event C: Your friend goes first and gets 100 dollars, then you also get 100 dollars.

*Event D: Your friend goes first and chooses empty, then you get 100 dollars.

Therefore, if you go first the probability of getting 100 dollars is the probability of event A plus the probability of event B. If you go second, it's the probability of event C plus the probability of event D. Both these expressions are $\frac{3C1\cdot 6C1}{9C1 \cdot 8C1}+\frac{3C1\cdot 2C1}{9C1 \cdot 8C1}$ which is 1/3.


*This part I am confused about how to approach, since there are many different sequences possibly of choosing 6 empty briefcase and 3 with 100 dollars.
Any advice would be greatly appreciated.
 A: If you go first, the number of "money" suitcases you get follows a hypergeometric distribution, since your five suitcases are drawn without replacement from a population of $3$ "money" suitcases and $6$ empty ones.
$$P(X=k) = \begin{cases}
\frac{\binom{3}{0} \binom{6}{5}}{\binom{9}{5}} = \frac{6}{\binom{9}{5}} & k=0
\\
\frac{\binom{3}{1} \binom{6}{4}}{\binom{9}{5}} = \frac{45}{\binom{9}{5}}
& k = 1
\\
\frac{\binom{3}{2} \binom{6}{3}}{\binom{9}{5}} = \frac{60}{\binom{9}{5}}
& k = 2
\\
\frac{\binom{3}{3} \binom{6}{2}}{\binom{9}{5}} = \frac{15}{\binom{9}{5}}
& k = 3
\end{cases}$$
If you go second, the corresponding probabilities will be in reverse order. You can see that the player that goes first tends to get more money suitcases.
A: Your answer in #2 is wrong.

Say you went first and picked an empty briefcase. What is the probability that your friend also picks an empty briefcase on their first turn, given that you also picked an empty one your first turn.

You are required to assume that you have already gone first and picked an
empty briefcase.  This means that there are 8 briefcases left of which only
5 are empty.
Answer is
$$\frac{5}{8}.$$
For the same reason as above, the answer to #3 is
$$\frac{6}{8}.$$
#4 is tricky and can be attacked in two different ways.
The pedestrian approach is that

*

*your chance of briefcase with money, if you go first is $\frac{3}{9} = \frac{1}{3}.$


*if you go second there are two possibilities - either your friend
picked an empty briefcase, and then you picked one with money, or your friend picked a briefcase
with money and then you also picked one with money.
The combined probabilities here are
$\left(\frac{6}{9} \times \frac{3}{8}\right) ~+~ 
\left(\frac{3}{9} \times \frac{2}{8}\right) ~=~ \frac{24}{72} = \frac{1}{3}.$
The more elegant analysis is to reason that picking a briefcase is mathematically
equivalent to being handed a briefcase at random.  If a moderator picks out
two briefcases at random, one for you, and one for your friend, the chance that
your briefcase contains money is unaffected by whether the moderator hands
you your briefcase before or after the moderator hands your friend his briefcase.
Edit
I misinterpreted

If the game were to end once each briefcase was picked, would it be better to start or go second?

If you go first, you are going to be given 5 briefcases and your friend will only be given 4 briefcases.  This is equivalent to pretending that there are 10 briefcases, instead of 9, and that the 10th briefcase is forced to be empty.
Therefore, if you go first, you will have 5 briefcases, any of which may have money.  If you go second, one of your briefcases (the imaginary 10th briefcase) is forced to be empty.
Therefore, the person who goes second can expect to get less money.
