Conditional Probability 5 card hand

We have a 5 card hand from a standard deck. What is the probability that the hand is all Spades, given that it has at least two Spades?

I know the formula for conditional probability is: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

So in this case would it be:

$P(A|B) = \frac{13 \choose 5}{{13 \choose 5}+{13 \choose 4}{39 \choose 1}+{13 \choose 3}{39 \choose 2}+{13 \choose 2}{39 \choose 3}}$

• What does the $P(A|B)$ calculate to? I mean the final value? – John Smith Feb 25 '17 at 16:38
• P(A|B) comes out to 0.001349 – user1775500 Feb 25 '17 at 16:44
• Hold on: how did you define $A$ and $B$? – John Smith Feb 25 '17 at 16:48
• A = "hand is all spades" B = "at least two spades" I thought the intersection of A and B would, just be A since when a hand is all spades it contains at least two spades – user1775500 Feb 25 '17 at 16:51
• It looks right to me. – callculus Feb 25 '17 at 16:52

The number $X$ of spades can be modeled as a hypergeometric random variable. $P(X = k) = \frac{{13\choose k}{39 \choose 5-k}}{{52 \choose 5}},$ for $k = 0,1, \dots,5.$ Here is the PDF table from R statistical software:

k=0:5;  pdf = dhyper(k, 13, 39, 5)
cbind(k, pdf)
## k          pdf
## 0 0.2215336134
## 1 0.4114195678
## 2 0.2742797119
## 3 0.0815426170
## 4 0.0107292917
## 5 0.0004951981


I think your answer is OK, but I have not checked the numerical values directly from your formula. If you're using a calculator to get numerical values, it is easier to find $P(X \ge 2) = 1 - P(X \le 1).$

From R, I get your desired conditional probability to be 0.001349141 (as Commented by @user1775500).

[In R, dhyper is the PDF and phyper is the CDF.]

dhyper(5, 13, 39, 5)/(1 - phyper(1, 13, 39, 5))
## 0.001349141
dhyper(5, 13, 39, 5)/sum(dhyper(2:5, 13, 39, 5))
## 0.001349141


It is extremely rare to get five spades in a five-card hand (.0005). Knowing that there are at least two, you get more than double that probability (.0013), but still not a very large value.

Any hand with 5 spades has 2 spades so it is not a really and intersection problem.

First take out 2 spades. Now you have 50 cards left and 11 spades left. You need 3 more spades.

(11,3) / (50,3) = 0.84%

$\frac{11 \choose 3}{{50 \choose 3}} = 0.0084$