Probability Question (Cards)

May 2010
Not sure if this is a university level question, so please move if needed. I'm working on designing and understanding card games and had a lot of questions relating to probability.

Ran into this example on the internet.

1) What is the probability of receiving four of a kind when dealt 13 cards from a regular 52 card deck?

Internet answer given:
\(\displaystyle [ C(13,1)*C(48,9) - C(13,2)*C(44,5) + C(13,3)*C(40,1) ] / C(52,13)\)

The subtraction and addition confuse me on their purpose. The explanation given was to compensate for the fact that some hands were being counted twice since you could have two or three four-of-a-kind sets with 13 cards.

It was my understanding that r-combinations were unordered selections with no repetition using the C(n,r) formula. So why would you have to compensate for double counting? And is that addition of C(13,3)*C(40,1) supposed to be a subtraction?

2.) Given that I have a 60 card deck that has 5 ORANGE cards and 55 other cards. If I drew 3 cards, what is the probability that it has at least one of the ORANGE cards. Emphasis on at least one.

My initial approach: \(\displaystyle C(5,1) * C(55,2) / C(60,3) \)
1 card to choose from the 5 desired ORANGE ones and need to choose 2 more.

But, it looks to me that solution is for the probability of drawing specifically one of the orange cards. Should it be C(59,2) rather than C(55,2) to take into consideration the other 4 still being an option.

3.) If I were to continue the card game started in question 2, with a second player drawing his hand of 3 cards from the diminished deck. How do I calculate the probability of obtaining an ORANGE card when the cards the first player drew is unknown to the second and the second draw is dependent on the first. (game theory?) Or do I have to just have specific cases that designate what the first player drew?

Suggestions on reading material to help with these sort of questions would be appreciated as well! Thanks.
May 2010
you can can reading material for this in any basic probability textbook, good ones that i know of with plenty of examples are elementary probability (stirzaker) there is usually a copy floating about on the internet somewhere, and a first course in probability (sheldon ross)


MHF Hall of Honor
May 2006
Lexington, MA (USA)
Hello, Darkammo!

Here's a little help . . .

2) Given a 60-card deck that has 5 Orange cards and 55 other cards.
If I drew 3 cards, what is the probability that I haveat least one Orange card?

The opposite of "at least one Orange" is no Oranges (3 Others).

. . \(\displaystyle P(\text{3 Others}) \;=\;\frac{_{55}C_3}{_{60}C_3} \;=\;\frac{5247}{6844}\)

Therefore: .\(\displaystyle P(\text{at least one Orange}) \;=\;1 - \frac{5247}{6844} \;=\;\frac{1597}{6844}\)