# In a 52 card deck, with 2 white, 2 black, and the rest blue, what is the probability of picking 4 cards at once with 2 white and 2 black?

Cards: 52

Blue: 48

Black: 2

White: 2

What is the probability of picking a 4 card bundle (all cards at once) in which you have 2 black cards, and 2 white, in any order?

I've already tried to solve this, but the fact that we take all cards at once confuses me as I'm not sure whether we subtract cards from the 'total probabilities' side after picking one. Perhaps this is a hypergeometric problem? I'd appreciate any help. I wanted to give this problem to my friend and realized I could not solve it myself.

• Probability is $$P = \displaystyle\frac{\text{ number of successful outcomes }}{\text{ number of total outcomes }}$$ Here, the number of total outcomes is the number of ways to choose $4$ cards from a set of $52$, which is called "$52$ choose $4$", it's a binomial coefficient. $$\,$$ The number of successful outcomes is the number of ways to choose $2$ white cards and $2$ black cards, which is just $1$, because there are only $2$ white cards and $2$ black cards. Nov 3 '20 at 15:35
• "I've already tried to solve this, but the fact that we take all cards at once confuses me" There's no difference between taking four cards at once, taking four cards one at a time but only looking at them in the end or taking them one at a time looking as you do so but being paralyzed and not allowed to do anything. Or to take one card in your hand, removing 12 from the deck, and repeating four times. You get 4 cards. Period. As long as it's random nothing changes about how you get them. Nov 3 '20 at 15:48
• "Perhaps this is a hypergeometric problem?" Um.... what's a hypergeometric problem??? Nov 3 '20 at 15:49
• @fleablood a hypergeometric series can be used to determine the probability of an event occurring when we take a sampling from a larger data set. For example, if in a 1000 marble population we pick 100 marbles every time, the probability that we get a combination of 50 of type 1 marble and 50 type 2 marble is clearly not the same probability as 1 and 99, assuming that in our initial population we have it split 500/500. Hypergeometric distributions should help determine the probability (8%). Nov 3 '20 at 16:49
• Don't overthink it. You pick 4 cards. That's all. Nov 3 '20 at 17:00

$$\frac{1}{\binom{52}{4}}=\frac{1}{270,725}$$
So you have $$4$$ good cards and $$48$$ bad cards and $$52$$ total cards.
You choose $$4$$ cards. There are literally "$$52$$ choose $$4$$" $$= {52 \choose 4}$$ ways of doing that. And only one of them are your $$4$$ good cards.
So the probability $$\frac 1{52 \choose 4}$$.
$$\frac 1{52\choose 4} = \frac 1{\frac{52!}{48!4!}}= \frac {4!48!}{52!}=\frac {4!}{49*50*51*52}= \frac {1*2*3*4}{49*50*51*52}=\frac 1{49}*\frac 1{25}*\frac 1{17}*\frac 1{13}=\frac 1{270725}$$