Question

Determine the poker odds of drawing the following hand from a standard card deck. (4 suits,...

Determine the poker odds of drawing the following hand from a standard card deck. (4 suits, 13 ranks in each suit.) What are the odds of drawing a royal flush (AKQJ10 all of the same suit)? Drawing cards is WITHOUT replacement. (NOTE: If necessary, lay out 52 cards on a table and do a dry run before computing the probabilities!) In order to get a royal flush in spades, you must pick the following 5 cards:

What is the probability that you will pick the A of spades for your first card?

What is the probability that you will pick the K of spades for your second card?

What is the probability that you will pick the Q of spades for your third card?

What is the probability that you will pick the J of spades for your fourth card?

What is the probability that you will pick the 10 of spades for your fifth card?

Now we picked the cards in the order A K Q J 10 We could have also picked the cards in many other orders such as A Q K J 10, 10 J Q K A, etc. Therefore, you have to multiply your answer by a factor F to get the correct probability of getting a royal flush in spades. This factor is F = 5! What is the value of F? Now there are 4 suits in a deck, not just spades. So we need another factor G = 4 to compute the probability of getting a royal flush of ANY suit (spades, hearts, diamonds, or clubs).

Multiply the above numbers (a) to (g) to get the final probability. (Note: The probability of getting a pair is 1/649,740. If you did not get close to this number, then you did something wrong.)

Homework Answers

Answer #1

P(A of spades is the first card)=1/52 (since there are total of 52 cards in the day and a A of spade can occur in only 1 way thus the probability comes to 1/52)

P(K of spades occur as the second card)=1/51 (as in the previous example, K of spades can occur in only one way, however, the total number of cards now present in the deck is 51, as one card has already been selected in the first draw and the procedure is without replacement)

similarly, proceeding in this way, we have,

P(Q of spades occur as the second card)=1/50

P(J of spades occur as the second card)=1/49

P(10 of spades occur as the second card)=1/48

here, there are 5 cards AKQJ10 which can occur in any particular sequence

and 4 suits so another 4 is to be multiplied.

P(a royal flush occurs)= 1/52*1/51*1/50*1/49*1/48*5!*4

=1/649740

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