# Weighted Urn Problem With Replacement Ending on Repeated Ball

Consider an urn with n different colored balls each with differing odds of being drawn. Balls are drawn with replacement. One draws from the urn and holds them in his hand until two of the same colored ball are drawn. Then upon finishing drawing, from the set of drawn balls one ball of the duplicate color is discarded.

This means the balls in hand have no duplicate colors and may be anywhere from 1 to n colors. The order they were drawn (besides what stopped the drawing) is irrelevant. It is straightforward to find the probability of holding any set of balls at the end.

But how would one calculate the odds of drawing a specific color independent of the number of balls drawn? Is there a simpler way than ($p_i$ is probability of drawing $i^{th}$ ball for one draw):

$p_i + \sum_{1=j\neq i}^n p_j*p_i + 2!*\sum_{1=j,k\neq i;j\neq k}^n p_j*p_k*p_i + ...$ until the case of n draws. This seems to resemble a weighted hypergeometric distribution.

• Your description is a little unclear: it permits many different plausible interpretations. Could you edit this post to (a) state exactly what gets recorded as an observation in this experiment; (b) explain what you mean by "combination"; and (c) explain what event is described by "drawing a particular ball"? – whuber Aug 19 at 18:01
• I just changed it. Let me know if this makes sense...thanks. – Sam Brett Aug 19 at 19:34