How does revealing information affect a discrete distribution?

\begin{align} p_1 = {} & \text{probability that $c_1$ is chosen} \\[8pt] = {} & \text{probability that the three-card} \\ & \text{hand is a member of the set} \\ & \left\{ \begin{array}{lll} c_1c_2c_3, & c_1c_2c_4, & c_1c_2c_5, \\ c_1c_3c_4, & c_1c_3c_5, & c_1c_4c_5 \end{array} \right\} \\[8pt] = {} & \text{the sum of the six probabilities} \\ & \text{of the six hands in this set.} \end{align} Similarly $p_2$ is the sum of six probabilities, as are $p_3,$ $p_4,$ and $p_5.$

That gives us five equations. A sixth equation says the sum of the probabilities of all of the hands is $1.$

But there are ten hands: $$ \left\{ \begin{array}{lll} c_1c_2c_3, & c_1c_2c_4, & c_1c_2c_5, \\ c_1c_3c_4, & c_1c_3c_5, & c_1c_4c_5, \\ c_2c_3c_4, & c_2c_3c_5, & c_2c_4c_5, \\ & c_3c_4c_5 \end{array} \right\} $$ Six linear equations in six unknown quantities is not enough to completely determine them. Thus there are infinitely many probability distributions on the set of ten outcomes that satisfy the constraints that say $p_1,p_2,p_3,p_4,p_5$ are the five specified numbers.

Besides these equations, there is a constraint that says all of the probabilities must be nonnegative. So we get a convex set of solutions with four degrees of freedom.

If we are given that $c_1$ is one of the chosen cards, the probabilities are then adjusted via the usual definition of conditional probability. Maybe I'll add more on this later.