Help Me Help Me

Question

Jason picks two cards from a deck of 52 cards. What is the probability that the sum of these 2 cards are greater than 17.

 

Given that A=1, J,Q,K=10

Answer 1

Actually, the answer is slightly uglier than it initially seems, so I understand your worry, Richard.

The good news is that since the smallest card you can draw and still make more than 17 with the 2nd is 8, which means we need not worry about Ace through 7. Combinations that make 17 are:

8 with: 10, J, Q, K

9 with: 9, 10, J, Q, K

10 with: 10, J, Q, K (because we already paired it with 8 or 9)

J with: J, Q, K (again, it has already paired with smaller number)

Q with: Q, K (see above)

K with K

For the first row (the row for 8), we have a probability of (4/52) * (16/51) = 

For the "9" row: (4/52) * (19/51)

For the "10" row: (4/52) * (15/51)

For the Jack row: (4/52) * (11/521

For the queen row: (4/52) * (7/51)

For the king row: (4/52) * (3/51)

If you add up all of the probabilities, you can simplify the procedure  by multiplying the (4/52) = 1/13 by the sum of all of those second factors, which totaled: 71 / 51

In the end, you get 71 / 663. 

Please note that on most lines, we had to lower the probability whenever we picked out a 2nd card of the same kind.

Answer 2

As per stated word problem, any combination of 2 face cards J,Q or K will satisfy the condition of sum been greater than 17. There are 12 face cards in the deck. The probability to get the first face card is 12/52, to get a second card (without replacement) is 11/51. Finally the probability is P = (12/52) * (11/51) = 0.049773 (rounded)