1. **Problem statement:** Suppose you pick three cards randomly from a deck of 52 cards, and they happen to be aces. What is the probability that the next card drawn from the remaining deck (including the aces) is a red ace? 2. **Understanding the problem:** There are 4 aces in a deck: 2 red (hearts and diamonds) and 2 black (clubs and spades). You have already drawn 3 aces. We want the probability that the next card drawn is the remaining red ace. 3. **Step 1: Identify which aces are drawn.** Since you have 3 aces, there are several cases: - Case 1: You have drawn both red aces and one black ace. Then no red ace remains. - Case 2: You have drawn one red ace and two black aces. Then one red ace remains. - Case 3: You have drawn zero red aces and three black aces (impossible since only 2 black aces exist). 4. **Step 2: Calculate the probability that the remaining ace is red given 3 aces drawn.** The total number of ways to choose 3 aces out of 4 is $\binom{4}{3} = 4$. Number of ways to choose 3 aces including exactly one red ace: - Choose 1 red ace out of 2: $\binom{2}{1} = 2$ - Choose 2 black aces out of 2: $\binom{2}{2} = 1$ - Total: $2 \times 1 = 2$ Number of ways to choose 3 aces including both red aces: - Choose 2 red aces out of 2: $\binom{2}{2} = 1$ - Choose 1 black ace out of 2: $\binom{2}{1} = 2$ - Total: $1 \times 2 = 2$ 5. **Step 3: Probability that the remaining ace is red:** Only in the case where exactly one red ace is drawn (2 ways) does the remaining ace include one red ace. So probability = $\frac{2}{4} = \frac{1}{2}$. 6. **Step 4: Probability that the next card drawn is the remaining red ace:** After drawing 3 aces, there are $52 - 3 = 49$ cards left. If the remaining ace is red (probability $\frac{1}{2}$), then there is exactly 1 red ace left in the 49 cards. So the probability that the next card is the red ace is $\frac{1}{2} \times \frac{1}{49} = \frac{1}{98}$. If the remaining ace is not red (probability $\frac{1}{2}$), then the probability is 0. 7. **Final answer:** $$\boxed{\frac{1}{98}}$$