# 2008 iTest Problems/Problem 34

## Problem

While entertaining his younger sister Alexis, Michael drew two different cards from an ordinary deck of playing cards. Let a be the probability that the cards are of different ranks. Compute $\lfloor 1000a\rfloor$.

## Solutions

### Solution 1

Use complementary counting to count the number of ways one can draw two cards with the same rank. There are $13$ ranks, and each rank has $4$ cards. That means the probability of getting two cards with the same rank is $\tfrac{13 \cdot 4 \cdot 3}{52 \cdot 51} = \tfrac{3}{51}$, so the probability of getting two cards with different ranks is $\tfrac{48}{51}$. That means $\lfloor 1000a\rfloor = \boxed{941}$.

### Solution 2

The first card can be any card, so the probability is $1$. However, of the $51$ cards remaining, only $12 \cdot 4 = 48$ of them have a different rank. Thus, the probability of getting two cards with different ranks is $1 \cdot \tfrac{48}{51} = \tfrac{48}{51}$, so $\lfloor 1000a\rfloor = \boxed{941}$.