Question

Five cards are dealt from a standard 52-card deck. We define a "straight hand as one that consists of five cards where all the five cards are consecutive (regardless of their suit). Note that the order of the cards are Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King and that a "straight" is not cyclic (for example 9, 10, J, K, Q is a straight but J,Q,K, A, 2 is not). Find the probability of the following events: i. A straight consisting of 1 Ace, 1 two, 1 three, I four, 1 five. ii. A hand consisting of 1 Ace, 1 three, 1 five, 1 seven, 1 nine. iii. Any straight as defined in the problem statement.

Answer 1

The probability of drawing a straight hand consisting of 1 Ace, 1 two, 1 three, 1 four, and 1 five is approximately 0.039%.

Step-by-step explanation:

To find the probability of drawing a straight hand consisting of 1 Ace, 1 two, 1 three, 1 four, and 1 five, we need to determine the number of ways we can arrange these five cards. There are 4 Aces, 4 twos, 4 threes, 4 fours, and 4 fives in a standard deck of 52 cards. So, the total number of ways to choose one of each card is 4x4x4x4x4 = 1024. Now, we need to find the total number of five-card hands that can be dealt from a standard deck of 52 cards. This can be calculated using the formula for combinations: C(52, 5) = 52! / (5!(52-5)!) = 2,598,960. Therefore, the probability of drawing a straight hand consisting of 1 Ace, 1 two, 1 three, 1 four, and 1 five is 1024 / 2,598,960, which simplifies to approximately 0.00039 or 0.039%.