# Probability that two poker hands have no aces

The probability of no two hands from the same poker deck having no aces is $$\frac{{48 \choose 5} + {48 \choose 5} - {48 \choose 10}}{{52 \choose 5}}$$ I am not sure why this is the answer, as the second draw is conditional on the first draw so it should be ${43 \choose 5}$.

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I think the first "no" is in excess? – joriki Aug 29 '11 at 15:37
Are you sure this is the right formula? In its current form, the numerator is negative. – Fixee Aug 29 '11 at 15:41
Where did you get that formula? The numerator is a negative number, and the expression evaluates to around - 2515. So that absolutely cannot be a probability... – Willie Wong♦ Aug 29 '11 at 15:47

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(BTW, what's the difference between "two poker hands" and "a draw of ten cards"?) – Willie Wong♦ Aug 29 '11 at 15:49

## 3 Answers

This is not a probability, since $\binom{48}{10}$ is a lot bigger than the other ones and would make this a large negative number.

It looks like someone was trying to work out the probability that at least one of the two hands has no aces using the inclusion-exclusion principle. The probability that one hand has no aces is

$$\frac{\binom{48}5}{\binom{52}5}\;,$$

and the probability that one or the other hand has no aces is twice that, except that would overcount the case where they both have no aces, which is being counted twice, so we have to subtract the probability of both hands having no aces, which is

$$\frac{\binom{48}{10}}{\binom{52}{10}}\;.$$

Note the $10$ in the denominator, which makes all the difference. So the probability of at least one hand having no aces is

$$\frac{{48 \choose 5} + {48 \choose 5}}{52 \choose 5} - \frac{{48 \choose 10}}{{52 \choose 10}}=\frac{576}{637}\approx0.90\;.$$

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Nice answer as usual, but it's called subtract, not substract. – Hans Lundmark Aug 29 '11 at 16:53

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@Hans: Thanks; fixed. – joriki Aug 29 '11 at 21:02

The formula in the question cannot be right (it has a negative numerator). Here's how I would find the probability:

The number of sets of 10 cards (two hands) with no aces is ${48 \choose 10}$. The number of set of 10 cards from the deck is ${52 \choose 10}$. So the overall answer is $$\frac{{48 \choose 10}}{{52 \choose 10}} = \frac{246}{595} \approx 0.41$$

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I agree that it is not clear whether this is the question being asked or the one joriki answered.  In any case, this is a good answer to this one. – Ross Millikan Aug 29 '11 at 19:25

I will assume, like Fixee did, that you want to find the probability that neither hand has an ace. Then Fixee's solution is the most efficient. A mild variant of it computes probabilities directly, bypassing binomial coefficients. The probability that the first card is not an ace is $48/52$. Given that the first card is not an ace, the probability that the second card is not an ace is $47/51$. So the probability that neither of the first two cards is an ace is $(48/52)(47/51)$. Continue in this way. We find that the required probability is $$\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{46}{50}\cdots\cdot\frac{40}{44}\cdot \frac{39}{43}.$$ There is a good deal of easy cancellation. After a while we arrive at $246/595$.

We can also find the answer by an argument that uses your idea.

Suppose that one hand is dealt, then the other. The probability that the first hand has no aces is $$\frac{\binom{48}{5}}{\binom{52}{5}}.$$ The probability that the second hand has no aces, given that the first hand has no aces, is $$\frac{\binom{43}{5}}{\binom{47}{5}}$$ (there are $47$ cards left, of which $43$ are non-aces). Thus the required probability is $$\frac{\binom{48}{5}}{\binom{52}{5}}\cdot\frac{\binom{43}{5}}{\binom{47}{5}}.$$

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So what is the point of using the inclusion/exclusion principle in this problem? Conditional probability makes much more sense in this problem. I still cannot see how you use the inclusion exclusion principle in this problem because you will have fewer possibilities to choose the second hand after you have chosen the first hand. Any ideas as to how inclusion/exclusion intuitively makes sense if this is a conditional probability problem? – lord12 Sep 2 '11 at 15:44
The people posting answers are uncertain about the exact question you are asking. Fixee and I interpret the question as asking for the probability that none of the cards (in either hand) is an Ace. The interpretation of joriki is that you are asking for the probability that at least one hand has no Ace.  These are different problems, with different answers.  It would be helpful if you clarified which of these two interpretations is the one you intended.  For the joriki interpretation, inclusion-exclusion is natural to use. (Continued) – André Nicolas Sep 2 '11 at 16:33
Finding the probability that hand $A$ has no Ace is easy, same for hand $B$. But if we add these two probabilities, we are double-counting the situations in which neither $A$ nor $B$ has an Ace, so we have to subtract. We could also solve the problem with joriki's interpretation by conditional probabilities, by finding first the probability that both hands have at least one Ace.  The problem is that this divides into cases: $A$ has exactly $1$, and $B$ has $1$. $2$, or $3$; $A$ has exactly $2$, and $B$ has $1$ or $2$; $A$ has $3$, and $B$ has $1$. Much more work! – André Nicolas Sep 2 '11 at 16:41
Even in the inclusion/exclusion case, why would the probability of no Ace in hand B be the same as no ace in A. Meaning, why isn't it {48\choose 10} as opposed to {48 \choose 3}. If A1 = {Hand 1 has no ace} and A2 = {Hand 2 has no ace}, the number of possibilities cannot be equal to the number of possibilities of hand 1. I may have a hard time understanding this. Could you give me a similar problem that is similar to this that uses the inclusion/exclusion principle? – lord12 Sep 2 '11 at 19:16
The cards don't "know" which hand is being dealt first. The probability that the second card is the Ace of Spades is the same as the probability that the ninth card is the Ace of Spades. Problem:   A bin has 30 identically wrapped candies, 10 chocolate and 20 strawberry. $A$ and $B$ each take $3$ candies. What is the probability at least one of them gets no chocolate candy? – André Nicolas Sep 2 '11 at 19:50
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