This is a problem that I occasionally run into when playing card games. The numbers and variables can change, but the basic idea is always the same. I would like both the answer to the problem below as well as how to solve similar problems in the future.
The problem: I have a deck of 20 cards. I will draw 4 cards from this deck. 2 of the cards are red, and the rest are blue. What are the odds of me drawing at least one red card?
If possible, an exact way to solve as well as a decent "estimate" way to solve it would be useful; the more exact way can be used during deckbuilding, when I can take as long as I need, where the quick way is useful in the middle of a game, and I don't (or can't) take too long to decide what to do.
Thanks in advance to everyone who helps me with this.
To do it exactly, you need to list all the possible ways of drawing at least one red card:
First draw:
r1) 2/20 chance of drawing red
b1) 18/20 chance of drawing blue
Second draw:
If r1 then r2) 1/19 chance of drawing red and b2) 18/19 chance of drawing blue.
If b1 then r2) 2/19 chance of drawing red and b2) 17/19 chance of drawing blue.
After two cards:
Chance of r1 and r2 = 2/20 x 1/19 = 2/380
Chance of r1 and b2 = 2/20 x 18/19 = 36/380
Chance of b1 and r2 = 18/20 x 2/19 = 36/380
Chance of b1 and b2 = 18/20 x 17/19 = 306/380
All add to probability 1. The chance of drawing one red card from two draws is 74/380, the sum of drawing red then red, red then blue and blue then red.
Hopefully you get the idea of how to generalize to three or more draws...