Combinatorics and Probability

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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)
my answer: 52*51*50*49*48

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?
My answer: (52/52)(12/52)(11/52)(10/52)(9/52)

what is the probability to get a one pair (two card of the same kind)?
My answer: 3!*(52/52)(3/52)(48/52)(47/52)(46/52)

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?
My answer:[(1-3!*(52/52)(3/52)(48/52)(47/52)(46/52))^(n-1)]*(3!*(52/52)(3/52)(48/52)(47/52)(46/52))

Is it correct?
thanks
 

Answers and Replies

  • #2
LCKurtz
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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)
my answer: 52*51*50*49*48

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?
My answer: (52/52)(12/52)(11/52)(10/52)(9/52)

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)?
My answer: 3!*(52/52)(3/52)(48/52)(47/52)(46/52)

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.
 
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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?
 
  • #4
LCKurtz
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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.
 
  • #5
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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)?
 

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