Is there a combinatorical proof for this probability?

Question

Introduction

Assume you have a deck of $d$ cards and initially draw $v$ of them (without putting them back). $x$ cards have a certain color and you want to draw at least $y$ of them (order unimportant). But there is also $1$ special card that allows you to draw $z$ more cards. We want to know the probability of drawing at least $y$ out of $x$ in this scenario.

My approach

Then the event $A = \{$at least $y$ cards of $x$ will be drawn$\}$ can be partitioned into $\tilde A=\{$at least $y$ cards of $x$ will be drawn with $v$ draws, but not the special card$\}$ and $A_k=\{$exactly $k$ cards of $x$ and the special card will be drawn with $v$ draws, and after that at least $y-k$ cards of $x-k$ with $z$ draws$\}$.

$\tilde A$ can be calculated with the cumulative function of the hypergeometric distribution as such $$P(\tilde A) = \sum_{j=y}^x \binom{x}{j}\binom{1}{0}\binom{d-x-1}{v-j}\binom{d}{v}^{-1}$$

$A_k$ would be calculated like this (I'm not 100% sure right now) $$P(A_k) = \binom{x}{k}\binom{1}{1}\binom{d-x-1}{v-k-1}\binom{d}{v}^{-1}\sum_{\ell=y-k}^x \binom{x-k}{\ell}\binom{d-v-x+k}{z-\ell}\binom{d-v}{z}^{-1}$$

In summary (summing $P(A_k)$ over all $k$ and writing $j=k$), I would get

$$P(A) = \sum_{j=y}^x {\textstyle \binom{x}{j}\binom{d-x}{v-j}\binom{d}{v}^{-1}} + \sum_{j=0}^{y-1} {\textstyle \binom{x}{j}\binom{1}{1}\binom{d-x-1}{v-j-1}\binom{d}{v}^{-1}}\sum_{\ell=y-j}^x {\textstyle\binom{x-j}{\ell}\binom{d-v-x+j}{z-\ell}\binom{d-v}{z}^{-1}}$$

This is by the way another description of $A$: "Draw at least $y$ out of $x$ or less, but in the latter case also draw the special card and then enough out of $x$ to make up for the difference." But when I started writing the question I couldn't remember this outcome :)

Question

Is there a more combinatorical proof of the probability above? I used to use probability trees and the falling factorial to calculate $P(\tilde A)$ when I was a kid until a math professor I met by chance pointed out that the combinatorical way of calculating $P(\tilde A)$ like above is much faster. Now with the constraint of the special card suddenly I'm introducing conditional probability again in $P(A_k)$ and an implicit probability tree, so I want to know if I can get rid of this again?

Answer 1

To analyze this problem effectively, eliminate the special card. There's only two relevant scenarios.

1. The special card is in the first $v$ cards. In this scenario you draw $v + z - 1$ colored cards.

2. The special card is not in the first $v$ cards. In this scenario you draw simply $v$ colored cards.

Scenario 1 happens $v/d$ of the time, scenario 2 happens $1 - v/d$ of the time. In both scenarios you can now ignore the existence of the special card and combine their probabilities afterwards.