# probability to draw 2 cards in same hand

In the card game Star Realms one acquires cards and plays them. When played with another card from the same faction, they often have an ally bonus. Now it's been a while since I've used combinatorics and probabilities, so I could use some help.

Given deck_size= n (assume n as multiple of 5 for now) and hand_size= 5. There are two blue faction cards in the deck; what is the probability of drawing those 2 given cards in the same hand? (A deck consists of $\frac{deck\_size}{hand\_size}$ hands, non-overlapping)

I know that I have to calculate the chance of both cards appearing in one hand, then calculate the different permutations of the hand, then the different permutations that the hand can occur in the deck. But I'm at a loss how to combine these permutations, combinations and probabilities.

Any help or pointers would be appreciated.

• What do you mean by "the same faction" ? Can $n$ be anything ? You need to do some explaining ! – true blue anil Sep 25 '15 at 14:33
• Added a little more explanation. – Shondoit Sep 25 '15 at 14:52

Imagine the deck cut into hands. Choose some place for one of the two blue faction cards. Now there are $n-1$ places left for the other blue faction card, and of these $5-1=4$ are in the same hand. Since all $n-1$ places are equiprobable, the probability of the second blue faction card to be in the same hand as the first is $4/(n-1)$.

A more roundabout way to get the same result would be to say that there are $\binom n2$ ways of selecting two cards, $\binom52$ ways to select them from one particular hand, and $\frac n5$ hands to select them from, so the probability is

$$\frac n5\frac{\binom52}{\binom n2}=\frac4{n-1}\;.$$

Let me take $n = 20$, say

Also, I assume that you want the probability of getting the $2$ blue cards in your hand.

Of the $20$ slots, only $5$ are "yours", so $Pr = \dfrac{5}{20}\cdot\dfrac{4}{19} = \dfrac{1}{19}$

[The blue cards occupy any $2$ of your $5$ slots, and the rest of the cards are sure to be non-blue]

You can similarly compute for any other value of $n$

PS

If it is to be any of the hands, using the same approach $Pr = \dfrac{20}{20}\cdot\dfrac{4}{19} = \dfrac{4}{19}$