There is a standard deck of cards and two of the cards are chosen at random. Compute the probability of choosing a third card at random whose number is in between the numbers of initially chosen cards.
Edit: A standard deck of cards has 52 cards, from A to K with spades, hearts, diamonds and clubs.
If the value of the last card has to be between the values of the first two, the three cards must have distinct numbers. There are ${13 \choose 3} = 286$ possible card combinations. There are two ways to draw each of these combinations: either pick the lowest number first, or pick the highest number first. Furthermore, each card can have one of four suits, so there are $4^3 = 64$ valid suit combinations. The probability of choosing a third card whose number is in between the numbers of initially chosen cards, thus equals:
$$\frac{286 \cdot 2 \cdot 64}{52 \cdot 51 \cdot 50} = \frac{36608}{132600} \approx 0.276$$
Alternatively, we can immediately calculate the probability $p$ of drawing three distinct values. We find:
$$p = \frac{52}{52} \frac{48}{51} \frac{44}{50} \approx 0.828$$
Since there are three possible numbers to be drawn last, we simply have to divide by three. The probability of choosing a third card whose number is in between the numbers of initially chosen cards, thus equals:
$$\frac{p}{3} \approx \frac{0.828}{3} = 0.276$$