Back with a harder poker probability problem Okay, so last time I got help figuring out a simple binomial coefficient misunderstanding. Now I'm trying to figure out what happens if the following scenario occurs:
Player $1$ gets a $5$-hand of poker cards and tells everyone that he did not get an ace. (for the sake of the context, let's assume he's not lying)
How big is the chance of getting $2$ aces for the second player?
Tried solution
The chance percentage will be the quotient of:
combinations of $5$-hands with $2$ aces $\div$ maximum number of combinations of $5$-hands
The number of combinations of $5$-hands is $\binom{47}{5}$ considering player $1$ has been dealt $5$ cards.
Here's where I'm stuck. I would usually do the following to calculate combinations of, say $2$ aces:
$$P = \binom{1}{1}\cdot\binom{4}{2}\cdot\binom{a}{2}\cdot\binom{4}{1}^2 + \binom{1}{1}\cdot\binom{4}{2}\cdot\binom{a}{1}\cdot\binom{4}{2}$$
Where $a$ is the number of ranks we want to choose from. The issue is that the $a$ in our case would be totally dependent on what cards player $1$ got, right?
If player $1$ got 4 cards of the same rank, our $a$ would be $11$, and it could take on different values as well.
Is there any easy solution to this problem, or what would be the general method? I might add that I have not learned about Bayes' theorem, and so I won't be using it just yet.
Edit 1: I think the equation should actually be, not a $100\%$ sure though.
$$P = \binom{1}{1}\cdot\binom{4}{2}\cdot\left (\binom{a}{1}\cdot\binom{4}{1}\right )^3$$
 A: Let's assume we're trying to count the number of hands with exactly two aces. Since there are 47 remaining cards in the deck, which includes all 4 aces. Then, we pick the two suits for the two aces, for which there are $\binom{4}{2}$ ways of doing. Then, we pick the remaining three cards from the 43 non-ace cards. Hence, there are
$$\binom{4}{2}\binom{43}{3}$$
ways of the second player getting a hand with exactly two aces.
Also, if the problem was asking for a hand with at least two aces, then repeat the process for getting three aces and for getting all four aces.
Now, we divide by the total possible number of hands, $\binom{47}{5}$ to get the desired probability.
A: It depends what he means by "I did not get an ace".  What if he got 2, 3, or 4 aces would he be lying?  "An" ace implies only 1 ace.
If someone had 4 wives at the same time and you asked him "do you have a wife"?  Should he say yes or no?  Which is more accurate?  I would say the most accurate would be "no, I have 4 wives!" because if he said yes, he is stating he has one wife and implying he has no more than 1 wife.
