# What is the probability that exactly two jacks, two queens, and two kings have been picked from the deck when the second ace turns up

Suppose that cards are picked at random from a full deck of 52 cards

What is the probability that exactly two jacks, two queens, and two kings have been picked from the deck when the second ace turns up?

NO. of ways to select 2 jack:$4\choose 2$

NO. of ways to select 2 queen: $4\choose 2$

NO. of ways to select 2 king : $4\choose 2$

NO. of ways to select 2 ace : ${4\choose 1} * {3\choose 1}$

$\therefore \text{ required probability is} \frac{{4\choose 2}*{4\choose 2}*{4\choose 1}*{3\choose 1}}{52\choose 8}$

• Have you any reason to believe that it is correct? Please explain. Further describe what's happening. Are cards randomly drawn until the second ace turns up? – drhab Apr 3 '18 at 9:46
• As mentioned in the question, i assumed cards are randomly drawn without replaccement. – DRPR Apr 3 '18 at 9:52
• So the description of the process in my former comment is correct? Btw, you haven't answered my first question yet. – drhab Apr 3 '18 at 10:00
• Yes you are right . And as for your first question i worked it out by myself , so i thinks the answer may be correct ,but i am not sure since there are no answers given. – DRPR Apr 3 '18 at 10:07
• Well, surely this depends on when the second Ace turns up. To see this, suppose the second Ace turned up on the 50th draw from the deck... – Theoretical Economist Apr 3 '18 at 11:27

We only have to look at jacks, queens, kings and aces. The other cards may be neglected.

Also no distinction is relevant between e.g. the jack of aces and the jack of spades.

So actually we are dealing with equiprobable strings of length $16$ with the letters letters J,K,Q,A each $4$ times.

There are: $$\frac{16!}{4!4!4!4!}=63063000$$ strings in total.

It remains to find the number of strings with A on spot $8$, and next to that exactly twice a J, twice a Q, twice a K and once an A on the spots $1,2,3,4,5,6,7$.

That number is: $$\frac{7!}{2!2!2!1!}\times\frac{1!}{1!}\times\frac{8!}{2!2!2!2!}=1587600$$

So we end up with probability:$$\frac{1587600}{63063000}=\frac{15876}{630630}\approx0,0252$$

Actually it is the probability that among the first $8$ cards there are exactly $2$ of each sort divided by $4$ (because the ace, jack, queen and king have equal probability to be the card on spot $8$).

So you could write the probability also as:$$\frac14\times\frac{\left(\frac{8!}{2!2!2!2!}\right)^2}{\frac{16!}{4!4!4!4!}}=\frac14\times\frac{\binom42^4}{\binom{16}8}$$

In its looks that comes closer to the (wrong) answer that you gave in your question.