Question

Instructions for Question:

We are creating a new card game with a new deck. Unlike the normal deck that has 13 ranks (Ace through King) and 4 Suits (hearts, diamonds, spades, and clubs), our deck will be made up of the following.

Each card will have:

i) One rank from 1 to 11.
ii) One of 9 different suits.

Hence, there are 99 cards in the deck with 11 ranks for each of the 9 different suits, and none of the cards will be face cards! So, a card rank 11 would just have an 11 on it. Hence, there is no discussion of "royal" anything since there won't be any cards that are "royalty" like King or Queen, and no face cards!

The game is played by dealing each player 5 cards from the deck. Our goal is to determine which hands would beat other hands using probability. Obviously the hands that are harder to get (i.e. are more rare) should beat hands that are easier to get.

Question itself:

How many different ways are there to get exactly 1 pair (i.e. 2 cards with the same rank)?
The number of ways of getting exactly 1 pair is ________

Also, what is the probability of drawing one pair from the deck?? (Rounded to 7 decimal points)

My Attempted Answer:

11C1 * 10C 3 * 9C1 * 9C1 * 9C1 --> 11*120*9*9*9 = 962,280 different ways

962,280/71523144 = 0.0134541 probability

Can someone verify my answer?? thanks in advance!! Appreciate you guys.

Answer 1

In the calculations of selecting one pair of cards just the consideration of pair is not taken. That is 9C2 is not consider remaining is correct.

Thank you.