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I'm finding difficulties in cracking this probability problem. Let's say that we have $n$ players, who are supposed to be part of two teams, red and blue. They are divided with the following procedure. A number $Y$ is chosen randomly from the set $ \left \{1,2,...,n-1 \right\}$, with all values equally likely. Then $Y$ of the $n$ players are chosen to form the red team, with all possible sets of size $Y$ equally likely. The remaining $n-Y$ players form the blue team.

Let's consider a particular player, called player A. I have to find the probability that the team containing player A has size $k$ for $k=1,2,...,n-1$. Moreover, I need to find the mean of the size of the team that contains player A.

After the teams have been chosen, each team selects a captain, who is equally likely to be any member of that team. I have to find the conditional distribution of the size of the team containing Player A, given that Player A is the captain.

The results I get aren't consistent with the data given: maybe I'm setting up the problem badly...

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    $\begingroup$ How are you setting up the problem and what results are you getting? $\endgroup$ – whuber Sep 23 '15 at 16:16
  • $\begingroup$ I have tried to set up the first step of the problem like this: $P(A \in R | Y=y) = \frac{P(A \in R \cap Y=y)}{P(Y=y)} = \frac{n}{2}$, but I think that I've made a mistake... For the next two steps I don't really know how to proceed... $\endgroup$ – james42 Sep 23 '15 at 16:30
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    $\begingroup$ It might help to look at the situations with $n=2,3,4$ in detail. There are few cases to examine, so you can work out all probabilities by making an exhaustive list of all the possible team choices. If you approach your enumeration in an orderly fashion, it will reveal helpful symmetries and relationships. $\endgroup$ – whuber Sep 23 '15 at 16:46
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ale42, in your comment above you are trying to determine the conditional probability $P(A \in R| Y = y)$ by looking at a quotient. However in this problem the joint distribution in the numerator is one of the harder things to get at, whereas the conditional probability you were trying to find is much simpler. So let's focus for now on:

$$ P(A \in R | Y=y). $$

If $Y=y$, then the Red team will have $y$ out of the $n$ people on it. There are $\binom{n-1}{y-1}$ options that have player $A$ on the Red team. Meanwhile, there are $\binom{n}{y}$ total options. That ratio comes out to

$$ \frac{\binom{n-1}{y-1}}{\binom{n}{y}} = \frac{(n-1)!}{(y-1)!(n-y)!} \cdot \frac{(y)!(n-y)!}{(n)!} = \frac{y}{n} $$

The much more intuitive way of seeing this is that with each choice $\frac{k}{n}$ of the players end up on the Red team. So because no player is more likely to end up on the Red team than anyone else, everyone must have probability $\frac{k}{n}$.

Now because of the way this problem (and many problems like this) is set up, the conditionals are easier to work with than the joint distribution. But once we have the joint distribution we can get whatever we want by summing. So the questions are, what is the joint distribution? And what do we want?

I'll help get you started on that. Ultimately what we care about is the size of teams, and which one player $A$ is on. Let us call A's team $T_A$. We could answer the question if we knew:

$$ P(T_A = x, Y=y) = P(T_A=x|Y=y) \cdot P(Y=y). $$

I think you can work that out. My question to you is, what is $P(A$ is on a team of size $k)$ in terms of $P(T_A = x, Y=y)$?

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  • $\begingroup$ Whe should have $P(Y=y) = \frac{1}{n-1}$ whereas for $P(T_A = x | Y = y) = \frac{Y}{n}$... Correct me if I'm wrong $\endgroup$ – james42 Sep 23 '15 at 22:00
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    $\begingroup$ Lower case $y$ in the numerator at the end but that's totally correct. Now can you answer the last question in my post? $\endgroup$ – jlimahaverford Sep 24 '15 at 0:21
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    $\begingroup$ Ah to be precise. $P(T_A = R| Y=y)=\frac{y}{n}$. But $P(T_A = B| Y=y)=\frac{n-y}{n}$. $\endgroup$ – jlimahaverford Sep 24 '15 at 0:28
  • $\begingroup$ Thus we should have $P(T_A = R, Y= k) = \frac{k}{n(n-1)}$ for the red team, whereas for the blue we should have $P(T_A = B, Y= k) = \frac{n-k}{n(n-1)}$... Anyway the two step forward are much more messing... $\endgroup$ – james42 Sep 24 '15 at 7:26
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    $\begingroup$ @ale42 good now can you use the joint distribution to get the probability that A is on a team with $k$ people on if? $\endgroup$ – jlimahaverford Sep 24 '15 at 12:38

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