Question: Note: Your answer must be a computed number rounded to 4 digits after the decimal point.
What is the probability of drawing a 3-card hand (from a 52-card deck) containing cards of decreasing consecutive ranks, where all cards are of the same suit? For example, (5♥, 4♥, 3♥), (J♦, 10♦, 9♦), but not (3♠, 2♠, A♠). Use bijection to compute a number of favorable outcomes.
My Solution:
Since there are 52 cards in the deck, then there are S = 52C3 = 22,100 possible combinations of three-card hands possible. Drawing a 3-card hand (from a 52-card deck) containing cards of decreasing consecutive ranks, where all cards are of the same suit. There are 11 possible sequences: E = {(AKQ), (KQJ), (QJ10), (J109), (1098), (987), (876), (765), (654), (543), (432)} ***(32A) is not counted. There are 3! ways to draw for each subset of E. For example: (AKQ) = AKQ, AQK, KAQ, KQA, QAK, QKA Since there are 4 suits, then the number of 3-card straight flushes possible is just 11 * 6 * 4 = 246. ⇒ Pr[E] = |E| / |S| = 246/22100 = 0.0119.
Close, but not quite. In the denominator, you have $\binom{52}{3}$, which means that you are not taking the order in which the cards are dealt into account, but just considering the combinations of $3$ cards; the number of distinct $3$-card hands. So you shouldn't take order into account in the nuerator either. You should not have multiplied by $6$. The answer is $$\frac {4\cdot11}{\binom{52}3}$$