you are being dealt a 7-card hand from a standard deck of cards. What is the probability of being dealt 3 number cards (A-10) and 4 face cards (J, Q, K)

Answer 1

First, we need to determine the total number of possible 7-card hands that can be dealt from a standard deck of cards. This can be found using the combination formula:

C(52,7) = 52! / (7! * (52-7)!) = 133,784,560

Next, we need to determine the number of ways to select 3 number cards and 4 face cards from the deck. We can do this by breaking it down into two steps:

Step 1: Selecting 3 number cards
There are 10 number cards in each suit (A, 2, 3, 4, 5, 6, 7, 8, 9, 10), and a total of 4 suits. Therefore, there are C(40,3) ways to select 3 number cards from the deck.

Step 2: Selecting 4 face cards
There are 3 face cards in each suit (J, Q, K), and a total of 4 suits. Therefore, there are C(12,4) ways to select 4 face cards from the deck.

The total number of ways to select 3 number cards and 4 face cards is the product of the two:

C(40,3) * C(12,4) = 91,914,240

Finally, we can calculate the probability of being dealt a 7-card hand with 3 number cards and 4 face cards:

P(3 number cards and 4 face cards) = 91,914,240 / 133,784,560 ≈ 0.6866

Therefore, the probability of being dealt 3 number cards and 4 face cards is approximately 0.6866 or ~68.66%.

Answer 2

The bot was correct in
C(40,3)*C(12,4) / C(52/7)

but then messed up the arithmetic again.

I got .036555.. , which is a more reasonable answer than the .68... by the bot

Answer 3

I apologize for any confusion caused. You are correct, the probability of being dealt 3 number cards and 4 face cards is:

C(40,3)*C(12,4) / C(52,7)

= 913,896,000 / 133,784,560

= 0.036555

This is approximately 3.66%, not 68.66%, as earlier indicated. Thank you for bringing this to my attention.