Combinatorics : Which side is heavier? n coins are given, among which exactly 3 are bad and heavier than the good ones. A balance is used to identify the bad coins. Assume k coins are picked in both sides of the balance at a time. What is the probability of 


*

*left side being heavier

*right side being heavier 

*both sides being equal in weight


Thanks.
 A: I'm not great at statistics (which is why I try answering these questions sometimes), but here it goes and I'll count on the community to tell me if I'm wrong.
Because I'm not great, I have to break these problems down into small quite elementary chunks.
First, I started with the probability that each side is equal.  I believe that that could be expressed as
$$\mathbb{P}(\text{Both Sides Equal}) = \mathbb{P}(\text{0 Heavy}) + \mathbb{P}(\text{2 Heavy}) \cdot \mathbb{P}( \text{Even Split})$$
I decided that that's the probability that you don't choose any heavy coins (thus they must be equal) or that you choose 2 heavies and one is on the left.
I started out by saying that: $\mathbb{P}(\text{0 Heavy})  = \frac{\binom{n - 3}{k}}{\binom{n}{k}}$
After trying to figure out $\mathbb{P}(\text{2 Heavy}) $ I think that I stumbled on this: $\mathbb{P}(i\text{ Heavy})  = \frac{\binom{3}{i}\binom{n - 3}{k - i}}{\binom{n}{k}}$
I also decided that most of the time I wanted to look at selecting both sides at once and then splitting them evenly so I started thinking of $k$ as $2k$.
OK, so that covered $\mathbb{P}(\text{0 Heavy})$ and $\mathbb{P}(\text{2 Heavy}) $, but I had to figure out the probability that given 2 heavies 
Thanks to Gerry Myerson, I'm convinced this section was incorrect.

I'd randomly split them evenly between the sides of the scale.  I
  decided I'd flip the heavy coins and if they were heads I'd put them
  on the left and tails I'd put on the right.  That's metaphorically of
  course.  But, basically, there are $2^2$ outcomes of flipping 2 coins
  and 2 of them result in an even split so $\mathbb{P}(\text{Even Split}) = 2 / 2^2 = 2 / 4 = .5$.

My new thought is that there will always be only 2 ways to have the 2 heavy coins together: both on the left or both on the right.  There are $\binom{k}{\frac{k}{2}}$ ways to split the coins.  Thus, I think this is correct:
$$\mathbb{P}(\text{Even Split}) = 1 - \frac{2}{\binom{k}{\frac{k}{2}}}$$
If that's correct, I think, the probability of both sides weighing the same is:
$$\mathbb{P}(\text{Both Sides Equal}) =\frac{\binom{n - 3}{k}}{\binom{n}{k}}+\mathbb{P}(\text{Even Split})\cdot\frac{\binom{3}{2}\binom{n - 3}{k - 2}}{\binom{n}{k}}$$
OK.  So, the conditions for the left side being heavier than the right side are the same as the conditions for the right side being heaver than the left so $\mathbb{P}(\text{Left Heavier}) = \mathbb{P}(\text{Right Heavier})$.
Thus, I only tried to calculate $\mathbb{P}(\text{1 Side Heavier})$.  I spent more time than I'd like to admit before I realized that $\mathbb{P}(\text{1 Side Heavier}) = \mathbb{P}(\neg\text{Both Sides Equal})$.
Nervously, I submit this answer as I run some tests to make sure I'm at least on the right track.
A: Hint:  one side is heavier if it has  more bad ones than the other.  That can happen if one side has one bad coin and the other doesn't have any, or if one side has two bad ones and the other doesn't have any or ... There aren't too many possibilities.  Can you see a relation between part 1 and part 2?
The number of unordered ways to pick left side coins that include one bad one are $3\binom {n-3} {k}$  You might see Wikipedia on combination if this is the question.
A: We first assume that $n \geqslant 2k$, and $k \geqslant 3$
Let's first sample $2k$ coins from the heap on $n$. The number of bad coins $X$ in the sample follows the hypergeometric distribution $\operatorname{Hyp}(2k,3,n)$. Then we sample $k$ coins to put on the left side from these $2k$ coins. The number of bad coins in this sample $Y$, conditioned on $X=x$ also follows the hypergeometric distribution, $Y|X=x \sim \operatorname{Hyp}(k,x,2k)$. The left side is heavier if $Y > X-Y$, i.e. $2Y > X$.
$$ \begin{eqnarray}
  \mathbb{P}(2Y > X) &=& \sum_{x=0}^3 \mathbb{P}(2(Y|X=x) > x) \mathbb{P}(X=x) =
      \sum_{y=0}^3 \sum_{x=0}^3 I(2y > x) \mathbb{P}(X=x) \mathbb{P}(Y|(X=x)=y) \\
   &=& \frac{k \left(10 k^2-9 k n+6 k+3 n^2-6 n+2\right)}{(n-2) (n-1) n}
\end{eqnarray}
$$
Similarly, the probability of right side being heavier is 
$$\begin{eqnarray}
   \mathbb{P}(2Y < X) &=& \sum_{y=0}^3 \sum_{x=0}^3 I(2y < x) \mathbb{P}(X=x) \mathbb{P}(Y|(X=x)=y) \\
   &=& \frac{k \left(10 k^2-9 k n+6 k+3 n^2-6 n+2\right)}{(n-2) (n-1) n}
\end{eqnarray}
$$
Then, the probability of both sides being equal is
$$
   \mathbb{P}(Y=X-Y) = 1 - \mathbb{P}(2Y>X) -\mathbb{P}(2Y<X) = \frac{(n-2 k) \left(-4 k n+2 k (5 k+3)+n^2-3 n+2\right)}{(n-2) (n-1) n}
$$
A: Probability of either left or right side of balance being heavier is:

Probability of left/right side getting 2 heavy coins + probability of
  left / right side getting one heavy coin * probability of other side
  getting 0 heavy coins.
(k)C2 + {K(n-2k)/(n)C2 } / (n)C(2)

