How does revealing information affect a discrete distribution? Assume that there are a total of five different cards, $c_1, c_2, c_3, c_4, c_5$. Three cards out of these five get picked and we know that the probability for each card getting picked is $p_i$ for $i=1,2,3,4,5$ with $0\leq p_i \leq 1$ and $\sum_i p_i = 3$. For example, lets say the probabilities are $[p_1,p_2,p_3,p_4,p_5]=[\frac{1}{2}, 1, \frac{5}{6}, \frac{4}{6}, 0]$. Now, one of the three picked cards gets revealed at random, e.g. $c_1$. How does this affect the other probabilities?
This would lead to some new probabilities $p_i', i=2,3,4,5$ with $\sum_i p_i'=2$. I am looking for a general formula - all my approaches would also change probabilities that were $0$ or $1$ in the beginning, which cannot be right (e.g. normalising the remaining probabilities: $p_i'= 2\cdot \frac{p_i}{\sum_{2\leq i \leq5} p_i}$).

In my specific setting I have 20 cards of which 5 get chosen and this task has to be done over and over. So I'd be looking for an efficient method to (approximately) update the probabilities. Is there some more intuitive way than normalising the remaining probabilities? Thanks for all your quick answers!
 A: \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.
