# Conditional Probability in a card game

1. Jul 2, 2008

### SiegeX

Although this problem may look like homework, I assure you it is not. It is a question that arose from a trading card game that I am stuck on. The problem is as follows (with simplified cards)

You have a deck of 53 cards, and 11 of those cards are red and 42 are black. If you were to randomly draw 7 cards from the deck, how often would your hand of 7 have *less* red cards in it if you had removed 1 red card from the deck prior to drawing your cards compared to leaving it in?

I'm fairly certain this is a hypergeometric distribution and I have calculated the probabilities of drawing 0-7 red cards in a hand of 7 with 11 red cards in a 53 card deck as well as the probability of drawing 0-7 red cards in a hand of 7 with 10 red cards in a 52 card deck. These numbers are listed below. Where to go from here I am not sure. This seems to be a conditional probability but there shouldn't be any dependence since the 7 drawn cards are replaced on each trial.

Code (Text):

0   1   2   3   4   5   6   7
53  17.50%  37.44%  30.35%  11.98%  2.46%   0.26%   0.01%   0.00%
52  20.17%  39.21%  28.61%  10.04%  1.80%   0.16%   0.01%   0.00%

Last edited: Jul 2, 2008
2. Jul 2, 2008

### CRGreathouse

First of all, I get a different chart from yours, starting
Code (Text):
19.63%  35.98%  28.27%
I'll need more context before I can understand you. Let's say your question was "how often would your hand of 7 have *less* red cards in it if you had removed 0 red cards from the deck prior to drawing your cards compared to leaving it in?" (bold part changed). Would the answer to this question be 0 (since the probabilities would of course be the same) or positive (since some hands would improve)?