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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!

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  • $\begingroup$ Have you had any thoughts yourself? $\endgroup$ – Henry Mar 16 '16 at 22:48
  • $\begingroup$ @Henry I have. I took a course on statistics a couple of years ago, but I can't find any of my notes. I believe it is quite easily calculated. The only problem is that it's a little too specific to find relevant examples online. $\endgroup$ – user3495690 Mar 16 '16 at 22:57
  • $\begingroup$ What do you mean by "be able to get"? Is this just a roundabout way of saying "get", or something more involved? $\endgroup$ – joriki Mar 16 '16 at 23:10
  • $\begingroup$ @joriki Yes, sorry. I fell out of the "math question" persona there for a second xD $\endgroup$ – user3495690 Mar 16 '16 at 23:12
  • $\begingroup$ Use the search box in this web to find questions about cards. Im not sure but I think there is a tag for it, for cards. Just put in the search box "poker", press ENTER and enjoy. $\endgroup$ – Masacroso Mar 16 '16 at 23:42
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For 1, to fail you have to draw your $20$ cards out of the $48$ non-$10$s. The chance of that is $\frac {48 \choose 20}{51 \choose 20}$, so the chance of success is $1-\frac {48 \choose 20}{51 \choose 20}$ You are correct, the others are similar.

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  • $\begingroup$ Is the 20 in the top "combination" also due to the amount of cards you draw? If so, are there any circumstances where this number would differ from the bottom "combination"'s? $\endgroup$ – user3495690 Mar 17 '16 at 15:02
  • $\begingroup$ Yes it is. You have to select $20$ cards out of the $48$ acceptable ones. "Any circumstance?", yes. Say I want the chance of drawing exactly $2$ kings and $3$ tens out of $13$ cards. It is $\frac {{4 \choose 2}{4 \choose 3}{44 \choose 8}}{52 \choose 13}$ You have to read and think about the problem. $\endgroup$ – Ross Millikan Mar 17 '16 at 15:12
  • $\begingroup$ I see. Thanks! One problem, though.. I've been told that $\frac {47 \choose 16}{51 \choose 25}$ is the answer to the second question, but I can't seem to find the reasoning behind this. Do you understand it? $\endgroup$ – user3495690 Mar 17 '16 at 18:13
  • $\begingroup$ I don't see where that comes from. Do you want the chance of a club royal flush or any royal flush? The second is rather harder. $\endgroup$ – Ross Millikan Mar 17 '16 at 20:16
  • $\begingroup$ I'm looking for a club royal flush, I guess. It seems to make sense to me that it would be $\frac {{1 \choose 1}{1 \choose 1}{1 \choose 1}{1 \choose 1}{47 \choose 21}}{51 \choose 25}$ which equals $\frac {47 \choose 21}{51 \choose 25}$. Does that make sense to you? $\endgroup$ – user3495690 Mar 18 '16 at 18:48

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