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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.
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.

8 comments

[–] [Deleted] 2 points (+2|-0)

aren't the odds of rolling 1 in 6 on an even dye 1x2x3x4x5x6? apply that...
[–] TheRedArmy [OP] 2 points (+2|-0)

You'll have to clarify for me.

I know the odds of rolling X on a die is one in six - six possible outcomes, one of which is good. The bigger issue here comes from multiple drawings. It would be the same as rolling the die 4 times, and seeing the odds of rolling any fours. Also, the denominations in the various fractions are different, so having them interact with each other is hard.

So in this case, the odds of a drawing a red card are 2/20 (1/10) for the first draw, then 2/19 for the second draw, then 2/18 (1/9) for the third, and 2/17 for the last draw.

So the first draw isn't hard, it's a 10% chance. The second draw is obviously better than that, it's about 10 and a half percent chance. The third one is 11%, and the fourth is almost 12% (11.7). What I don't know how to do is...

  • combine these percentages into the single chance of drawing any red card
  • easily calculate these on the fly for different variables

And the second one in particular is what I need, although both bits are helpful.
[–] smallpond 4 points (+4|-0)

After my other reply, it's obviously easier to work out the chance of drawing all blue cards:

18/20 x 17/19 x 16/18 x 15/17 = (16 x 15)/(20 x 19) = 240/380 = 63.1%

So, 36.9% chance of drawing red.
[–] smallpond 3 points (+3|-0) Edited

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...
[–] TheRedArmy [OP] 1 points (+1|-0)

OK, I understand all this, and I do get how to generalize it, thanks a ton for your time and effort.
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