Friday

April 18, 2014

April 18, 2014

Posted by **Matt** on Sunday, October 30, 2011 at 3:57pm.

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.

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

I wrote:

nCr(52, 5)

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

=52!/(5!47!)

=311875200/120

=2598960

[B]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).[/B]

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

I wrote:

nCr(4, 2)

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

=4!/(2!2!)

[B]=6[/B]

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

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!)

[B]=6[/B]

[B][I][U]I get confused from part iii onwards:[/U][/I][/B]

[B]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:[/B]

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)= [B]650[/B]

[B]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.[/B]

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!

[B]=48[/B]

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

[/B]

I have no idea what this is even asking, so I don't even know where to start. :frustrating::bored::confused::unsure:

[B]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.[/B]

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. :laughing::blushed:

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