Hi all I recently ran into this problem: I have two bins. Each bin contains N numbered balls, from 1 to N. For both the bins, the probability of the ball numbered k to be selected equals to P(ball-k-selected)=k/SUM(1:N) (in other versions this can be any given probability distribution) Simple case: Having selected 1 ball from the first bin, and 1 ball from the second bin, i want to find the probability of the ball having the same number. If i am correct, the probability for this is SUM(k=1:N) (P(ball-k-selected)^2). Complex case: Having selected m balls from the first bin, and m balls from the second bin, i need the probability of holding at least one pair of balls with the same number at the end of the experiment. Assumption: The selection is without replacement. However, for simplicity we can assume that the probability of a ball to be selected remains stable during the experiment, and is given by P(ball-k-selected)=k/SUM(1:N) If something is not clear, please let me know. Thanks in advance for any contributions!