# Combinatorics and Probability

Cosmossos

## Homework Statement

in a poker game we want to calculate the Probability to get differnet combinations. in a card deck there are 52 cards from 4 different series. each series has 13 cards. we assume that each card get 5 random cards.

What is the number of combinations to get the cards? (The order isn't important)

what is the number of combinations if we don't distinguish?
My answer: 52C5-13*12*4 (the number of combinations to choose 5 cards - the number of combinations which repeat themselves)

What is the probability to get K,J,Q,A,Flush and all the cards are from the same series?

what is the probability to get a one pair (two card of the same kind)?

We play N times. What's the probabilty that until the Nth game we won't get one pair and in the last game we will get one pair?

Is it correct?
thanks

Homework Helper
Gold Member

## Homework Statement

in a poker game we want to calculate the Probability to get differnet combinations. in a card deck there are 52 cards from 4 different series. each series has 13 cards. we assume that each card get 5 random cards.

What is the number of combinations to get the cards? (The order isn't important)

The "series" are usually called "suits". Do you mean to ask how many possible different 5 card hands can be dealt? If so, your answer is wrong. You have given the number of permutations of 52 things 5 at a time, P(52,5), which counts different orders as different hands. You want combinations, C(52,5), which doesn't distinguish different orders of the same 5 card hands.
what is the number of combinations if we don't distinguish?
My answer: 52C5-13*12*4 (the number of combinations to choose 5 cards - the number of combinations which repeat themselves)

Don't distinguish what? This is a confused paragraph; see above.

What is the probability to get K,J,Q,A,Flush and all the cards are from the same series?

No. How many royal flushes are there? How many possible 5 card hands are there? Use those figures.

what is the probability to get a one pair (two card of the same kind)?

It's more complicated than that. Normally, when you say you have one pair it means you don't have a better hand. So you can't have a full house, which includes a pair or 3 or 4 of a kind.

Cosmossos
Don't distinguish what? This is a confused paragraph; see above.
If we don't distinguish between the suits...

No. How many royal flushes are there? How many possible 5 card hands are there? Use those figures.
I don't understand what you meant. I need to find the probability to get a shape/color and all the cards are from the same suit.

It's more complicated than that. Normally, when you say you have one pair it means you don't have a better hand. So you can't have a full house, which includes a pair or 3 or 4 of a kind.
So what do I do?

The other answers are correct?

Homework Helper
Gold Member
If we don't distinguish between the suits...

I don't understand what you meant. I need to find the probability to get a shape/color and all the cards are from the same suit.

So what do I do?

The other answers are correct?

What in the world does "shape/color and all the cards from the same suit" mean?

I doubt your answers are correct, but it's hard to tell given that I can't figure out what you are really intending to ask. If these questions are from a text, please copy them verbatim so I know what they really say.

Cosmossos
O.k , Forget about it, I need help only with the following qeustions:
what is the number of combinations if we don't distinguish between the suits?
And:
We play N times. What's the probabilty that until the Nth game we won't get only one pair and in the last game we will get one pair?
Trying to answer: Is like (1-get one pair)^(n-1) * (get one pair)?