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Poker - 52 cards
4 suits of 13 cards each
A deck will have 4 aces

Let say 5 player - 10 cards pre flop

Poker probability

What is the chance exactly 2 aces are out (in the 10 cards dealt)?
Is this correct?

$$\binom{4}{2} \binom{12}{8} \binom{4}{1}^8 \over \binom{52}{10}$$
(aces)(other)(suits)/(possible)

What are the chances 2 or more aces are out (dealt)?
Another way to say that is at least 2.
Or 2+. Is this even close?
$$\binom{4}{2} \binom{50}{8} \over \binom{52}{10}$$
(aces)(number of random hands left) / (possible)

If this should be two separate questions then fine.

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  • $\begingroup$ Could you explain the reasoning that led you to each formula? And please tell us what "2 aces out" means. Does that refer to two aces appearing in ten randomly dealt cards, or to two aces remaining in the 42 undealt cards, or to two aces appearing in one of the five hands, or maybe to something else? $\endgroup$
    – whuber
    Jan 31, 2016 at 19:48
  • $\begingroup$ @whuber Two aces out means two dealt - common poker term. In the 10 cards dealt there are 2 aces. My best guess looking at the probability in that link. I provided () reason on the first. If I knew the answer I would not be asking. $\endgroup$
    – paparazzo
    Jan 31, 2016 at 19:50
  • $\begingroup$ We're not a poker site, so you shouldn't expect your readers to understand poker terminology: you need to explain it. And, since we are a stats site and not a poker site, the appearance of your question here indicates you are interested in any statistically related issues in this question--not just the numerical answer. Please consult our help center for more information about how this site works and what is on topic. $\endgroup$
    – whuber
    Jan 31, 2016 at 20:01
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    $\begingroup$ @whuber And I updated the the question. Are you upset with me? I (think) I am asking about statistics that happens to be a poker related question. I say poker as most people will know that is a deck of 52 cards with 4 suits (each card repeats 4 times). $\endgroup$
    – paparazzo
    Jan 31, 2016 at 20:04
  • $\begingroup$ @whuber Two in one hand is a possible subset but I know that number. This would be more like if I am holding JJ and and ace comes up on the flop (next three common cards) what are that chances that hit a hand. $\endgroup$
    – paparazzo
    Jan 31, 2016 at 20:08

4 Answers 4

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Let's look at the first question. You want exactly 2 aces and 8 other cards. The probability is given by $${{4 \choose{2} }{48 \choose {8}}} \over {{52} \choose {10}}$$

Do you see now how to get the answer to your second question?

As JohnK noted, question 2 requires finding the probability of getting 3 and 4 aces and then adding those probabilities to the answer to question 1.

So, for 3 aces, $${{4 \choose{3} }{48 \choose {7}}} \over {{52} \choose {10}}$$

Following this approach and continuing on by finding the probability for 4 aces and summing up over 2, 3, and 4, we find the probability of having 2 or more aces out in 10 cards to be 0.1625

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  • $\begingroup$ Clearly not, because then I would interpret my second as correct. For the first I copied how the did a pair in 5 in the link in my answer. $\endgroup$
    – paparazzo
    Jan 31, 2016 at 23:47
  • $\begingroup$ @Frisbee The problem with your calculation in the first question is that you have counted the ways you can obtain 8 different cards other the aces. But surely, as long as two aces are out, there is no problem if say there are also two fives among the remaining cards. $\endgroup$
    – JohnK
    Feb 1, 2016 at 0:11
  • $\begingroup$ OK, I think I follow that but then again I would assume my second is correct if this is the answer to the first. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 0:16
  • $\begingroup$ @Frisbee The reason that it is not is that you have double-counted the aces. To see the correct result more clearly, not that the event that at least two aces turn up is the event that you get two or three or four aces. Soakley's answer deals with the case of two so you have that but now, based on his answer can you figure out in how many ways you can get three aces and then four of them? You can then add these results together to get your answer. By the way, this is a hypergeometric problem, en.wikipedia.org/wiki/Hypergeometric_distribution in case you want to read more. $\endgroup$
    – JohnK
    Feb 1, 2016 at 0:28
  • $\begingroup$ @JohnK I get you are trying to help but you are introducing way to much stuff for me to direst. "The reason that it is not is that you have double-counted the aces." I don't even know what "it" is. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 1:10
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@Frisbee, the probability of having at least 2 is the probability of having 2, 3, or 4. By the exact same logic as in my previous answer, the probability of having three aces is

$$ \frac{\binom{4}{3} \binom{48}{7}}{\binom{52}{10}} $$

and the probability of having four aces is

$$ \frac{\binom{4}{4} \binom{48}{6}}{\binom{52}{10}} $$

These are disjoint events, so add them all up to get the total probability....

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  • $\begingroup$ Please visit stats.stackexchange.com/help/merging-accounts to merge your accounts and then you can edit your previous answer to include this extra information. $\endgroup$
    – whuber
    Feb 1, 2016 at 16:03
  • $\begingroup$ Thanks, I follow. Question 2 is not 2,3, or 4 as in 3 numbers. It is "two or more" in ONE number. Or you could say "at least 2". What confuses me is based on the answer to question 1 then my answer to question 2 is correct but soakley said you should be able to figure it out which to me implies it is wrong. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 16:04
  • $\begingroup$ @Frisbee, "2 or more" when the maximum is 4 is the same as "2 or 3 or 4". This is no longer a probability point, but an english language comprehension issue. $\endgroup$
    – linksys
    Feb 1, 2016 at 16:11
  • $\begingroup$ @linksys Thanks, and agree there is a English language comprehension issue. No where in my current question am I asking for probability of exactly 3 (or 3 or more). Exactly two is not the same as at least two. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 16:19
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    $\begingroup$ I did add "at least" to the question already. And it that not the correct answer in my question. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 16:22
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After you question, I tried reproducing the frequency figures that can be calculated with combinations, as posted in Wikipedia by simply computer-simulating each hand in R, and repeating $5$-card draws millions of times.

I don't know if, strictly speaking, this is an "answer" to your question, but I think it is nice to confirm that combinatorics does indeed reflect perfectly what you would encounter if you were to just tally the outcomes of millions of random draws.

Here is the code, and the results so that you don't even have to run it.

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  • $\begingroup$ In full disclosure, I got stuck at the beginning of writing the code I linked in what turned out to be a silly oversight in an R loop: for (i in iterations) instead of for (i in 1:iterations), and I did ask for help on SE. Although there were a number of suggestions made besides fixing this mistake, I can honestly say that the code I'm sharing is entirely my work, and mostly the result of trial and error on my own solutions. $\endgroup$ Feb 1, 2016 at 16:19
  • $\begingroup$ Thank, if I would going to try and write a full poker odds calculator (e.g. all the was to river) I would look at r. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 16:32
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@Frisbee, the "it" you keep asking about is obviously referring to your proposed answer. As the other commenter said, your answer is wrong because you double counted two aces in the group of cards you're choosing from (and no idea where the $\binom{4}{1}^8$ part came from).

Once you choose exactly 2 out of the 4 aces, of which there are $\binom{4}{2}$ (read "4 choose 2") possible ways to do, there are 48 cards left and you choose 8 of them, and there are $\binom{48}{8}$ ways to do that. So the number of possible ways both can happen is $\binom{4}{2} \binom{48}{8}$

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  • $\begingroup$ OK, thank but there are 3 questions - guess that is my problem. If that is the correct first answer then again I fails to understand how my guess and the second is not correct. $\endgroup$
    – paparazzo
    Feb 1, 2016 at 3:17

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