Probability that two poker hands have no aces

Question

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 ${43 \choose 5}$.

Answer 1

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

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

Answer 2

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 $$\frac{{48 \choose 10}}{{52 \choose 10}} = \frac{246}{595} \approx 0.41$$

Answer 3

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 $$\frac{\binom{48}{5}}{\binom{52}{5}}\cdot\frac{\binom{43}{5}}{\binom{47}{5}}.$$