Reformatted so that it's easier to read.

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.

No problem at all! Let's go through each part of the problem step by step and check your work.

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

You correctly used the combination formula, nCr, to calculate this. The formula is given by n! / (r! * (n-r)!), where n is the total number of items and r is the number of items chosen. Plugging in the values, we have:

nCr(52, 5) = 52! / (5! * (52-5)!)
= 52! / (5! * 47!)
= 2598960

Your answer of 2598960 is correct.

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.

You correctly identified that you need to calculate the number of ways of being dealt two cards of the same value. You used the combination formula, nCr, correctly again. The value of n is 4, since there are 4 cards of each value, and r is 2, since you want to choose 2 cards. Plugging in the values, we have:

nCr(4, 2) = 4! / (2! * (4-2)!)
= 4! / (2! * 2!)
= 6

Your answer of 6 is correct for this part.

ii) This happens twice, for the other set of two cards.

You are correct that this happens twice for the other set of two cards. You don't need to account for the other two cards you drew, as you are calculating the number of ways to choose only two cards of the same value. Therefore, the answer is still 6.

iii) Determine how many ways these two group values can be dealt to you.

This part is a bit different. You are correct that there are 26 possible cards for each of the two pairs (spades + hearts and clubs + diamonds). However, you want to determine the number of ways these two group values can be dealt to you. This means you need to take into account the order in which the two pairs are dealt.

To calculate this, you need to use the concept of permutations. The formula for permutations is nPr = n! / (n - r)!, where n is the total number of items and r is the number of items chosen in order. In this case, the total number of items is 26, since there are 26 possible cards for each pair, and you want to choose both pairs, so r is 2. Plugging in the values, we have:

nPr(26, 2) = 26! / (26-2)!
= 26! / 24!
= 26 * 25
= 650

So the number of ways these two group values can be dealt to you is 650. You don't need to multiply by 2, as you already accounted for both pairs when calculating the number of ways to choose two cards of the same value.

iv) Determine the number of ways the single card can be dealt.

You correctly subtracted the 4 already dealt cards from the total of 52 to get 48. Then you used the combination formula, nCr, to calculate the number of ways to choose 1 card from 48. The value of n is 48 and r is 1. Plugging in the values, we have:

nCr(48, 1) = 48! / (1! * (48-1)!)
= 48! / 47!
= 48

So the number of ways the single card can be dealt is 48.

v) Determine the different ways altogether these cards can be dealt using the Fundamental Counting Principle.

The Fundamental Counting Principle states that if there are p ways to do one thing and q ways to do another thing, then there are p * q ways to do both things together.

In this case, you have determined the number of ways of the two pairs being dealt (650) and the number of ways the single card can be dealt (48). To find the different ways altogether these cards can be dealt, you multiply these two numbers together:

Different ways altogether = 650 * 48

vi) Determine the probability of this happening.

To determine the probability of this happening, you need to divide the number of favorable outcomes (found in part v) by the total number of possible outcomes, which you correctly calculated in part a.

Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
= (650 * 48) / 2598960

If you evaluate this expression, you will get the probability of being dealt a hand of two pairs in 5-card stud poker.

I hope this helps you check your math and complete your project! Let me know if you have any further questions.