Sunday

April 20, 2014

April 20, 2014

Posted by **Matt** on Sunday, October 30, 2011 at 4:20pm.

I have a Statistics project due tomorrow on Combinatorics. Admittedly, while my math skills aren't too bad (I get 99th percentile on standardized tests and such), it has been a while since I last took a math class after a 2-year lapse in education, so my skills are really rusty. I would greatly appreciate some help checking to see if my math is correct.

I will show the work I currently have written down. My intention is not to simply ask for answers and be lazy; I am asking to see if I am on the right track.

The problem is:

"Suppose you and your friends are playing 5-card stud poker (standard deck of 52 playing cards). Using combinations, you want to determine what is the probability of being dealt a hand of "two-pair;" similar to the hand shown in the following figure (a hand of two-pair is two cards of one value; two cards of another value; and a single card of a third value).

The paper then shows 5 specific cards: an 8 of spades, and 8 of hearts, a 3 of clubs, a 3 of diamonds, and a King of diamonds.

a.) Determine the number of different ways of being dealt 5 cards in a game of poker.

I wrote:

nCr(52, 5)

=52!/[5!(52-5)!]

=52!/(5!47!)

=311875200/120

=2598960

b.) Determine the number of ways in a hand of 5 cards of being dealt two cards of one value; then two cards of another value; and then a single card of a third value (see figure on last page).

i) To do this, determine the number of ways of being dealt only two cards of the same value:

I wrote:

nCr(4, 2)

=4!/[2!(4-2)!]

=4!/(2!2!)

[B]=6[/B]

ii.) Of course this happens twice, for the other set of two cards:

I think I'm sure about this one, but do I have to account for the other two cards that I drew or does that not matter?

I wrote:

nCr(4, 2)

=4!/[2!(4-2)!]

=4!/(2!2!)

=6

I get confused from part iii onwards:

iii.) These outcomes are for these specific card values. But remember, there are 13 different card values in a standard deck of cards. Now determine how many ways these two group values can be dealt to you:

My reasoning here was that since I'm looking for an 8 of spades + 8 of hearts, and a 3 of clubs and a 3 of diamonds, and that there is 13 cards of each suit, that there are 26 possible cards for spades + hearts and 26 possible cards for clubs + diamonds, and that I'm choosing two from 26.

... so I wrote:

nCr(26, 2)

=26!/[2!(26-2)!]

=26!/[2!24!]

=325

325*2 (to account for both pairs)= 650

iv.) Finally, we determine the number of ways the single card can be dealt. Remember to keep in mind that other cards have already been dealt.

My reasoning here is that since the 4 other cards have been dealt, I subtract 4 cards from 52, giving me 48. Since I am choosing 1 card (King of Diamonds) from 48,

I write:

nCr(48, 1)

=48!/(1!(48-1)!)

=48!/47!

=48

v.) Finally, using the Fundamental Counting Principle we determine the different ways altogether these cards can be dealt:

I have no idea what this is even asking, so I don't even know where to start.

vi.) To determine the probability of this happening, divide the answer obtained from v) above by the total number of ways of being dealt a 5 card hand.

I have no idea how to approach this one, since it relies on an answer from v).

I hope all of you don't mind that I ask? Sorry for the long post! Don't make fun of me for failing at math.

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