Probability questions involving chance of drawing particular poker hands.

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jazwidz

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I am trying to design a program to calculate the probability of getting a set of poker hands, but I've been having trouble finding where to start. It would really help me out if someone could assist me with a couple of simple probability questions, and then I should be able to figure out the rest on my own.
Here are the questions I would like help with:


Please note that the cards "in hand" must be used for the "final hand":

1.) With a 51 card deck, if you already have the 10 of clubs in hand, what is the probability that you will draw at least one more 10 from the deck after drawing 20 more cards?
2.) With a 51 card deck, if you already have the 10 of clubs in hand, what is the probability that you will get a royal flush (same suit, 10-J-Q-K-A) after drawing 25 more cards?
3.) If you already have a 3 of hearts in hand, what are the odds that you will be able to get a straight (any 5 card sequence with 2-3-4-5-6-7) if the deck only has two 5s and three 6s, after drawing 20 more cards (deck only has 48 cards left in this case)?


Any suggestions/guidance is very much appreciated!
 
jazwidz said:
Please note that the cards "in hand" must be used for the "final hand":
1.) With a 51 card deck, if you already have the 10 of clubs in hand, what is the probability that you will draw at least one more 10 from the deck after drawing 20 more cards?
2.) With a 51 card deck, if you already have the 10 of clubs in hand, what is the probability that you will get a royal flush (same suit, 10-J-Q-K-A) after drawing 25 more cards?
3.) If you already have a 3 of hearts in hand, what are the odds that you will be able to get a straight (any 5 card sequence with 2-3-4-5-6-7) if the deck only has two 5s and three 6s, after drawing 20 more cards (deck only has 48 cards left in this case)?
Click to expand...
NOTATION: \(\displaystyle \dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}\)

1) At least one is the complement of none. So \(\displaystyle 1-\dfrac{\binom{48}{20}}{\binom{51}{20}}\).

2) \(\displaystyle \dfrac{\binom{47}{21}}{\binom{51}{25}}\).
 
Last edited:
pka said:
NOTATION: \(\displaystyle \dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}\)

1) At least one is the complement of none. So \(\displaystyle 1-\dfrac{\binom{48}{20}}{\binom{51}{20}}\).

2) \(\displaystyle \dfrac{\binom{47}{16}}{\binom{51}{20}}\).
Click to expand...

Thanks! I'm assuming the bottom 20 should be a 25, and you just didn't notice the change in amount of cards drawn. Would you mind elaborating about the process involved to reach these answers?
 
jazwidz said:
Thanks! I'm assuming the bottom 20 should be a 25, and you just didn't notice the change in amount of cards drawn. Would you mind elaborating about the process involved to reach these answers?
Click to expand...

Yes, I edited the post.
 
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