2
$\begingroup$

What's the probability of a 5 card hand is dealt 4 Kings, given that the hand has the King in Spades and Hearts?

Here's my attempt:

E= KKKK_ F= KK_ _ _

so $P(E) = 48/(\phantom{}_{52}C_5)$ and $P(F) = (\phantom{}_4C_2)(\phantom{}_{50}C_3)/(\phantom{}_{52}C_5)$.

The intersection is $P(E)$, so $P(E)/P(F)$.

I got the same answer as if the question didn't specify the Hearts and Spades.

$\endgroup$
2
$\begingroup$

We can use the formal machinery of conditional probability. Let $A$ be the event the hand has $4$ Kings (and of course a fifth card). Let $B$ be the event the hand has the $\spadesuit$ K, and the $\heartsuit$ K, together with $3$ other cards, some of which may be Kings.

We want $\Pr(A|B)$, the probability of $A$ given $B$. By the definition of conditional probability, we have $\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$. We compute the probabilities on the right.

We have $\Pr(B)=\frac{\binom{2}{2}\binom{50}{3}}{\binom{52}{5}}$ and $\Pr(A\cap B)=\frac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$. Divide.

Remark: Note that the solution of msh210 is more intuitive, and therefore better.

$\endgroup$
  • $\begingroup$ How did you come up with the intersection? $\endgroup$ – Probably Lost Feb 18 '14 at 3:24
  • $\begingroup$ The formula $\Pr(A|B)=\dots$ is the usual (for $\Pr(B)\ne 0$) definition of conditional probability. Or maybe you are asking how the expression for $\Pr(A\cap B)$ was found in this case. We have $A$ and $B$ if there are $4$ Kings including the $\spadesuit$ and $\heartsuit$ K. In this case, $A\cap B)=A$. The $4$ spades can be chosen from the $4$ available in $\binom{4}{4}$ ways. Of course this is $1$, so could have been left out. Then the fifth card is any card chosen from the $48$ non-Kings. $\endgroup$ – André Nicolas Feb 18 '14 at 3:32
  • $\begingroup$ Okay yeah that makes sense. For some reason I thought you did some strange math to get that. Thanks! $\endgroup$ – Probably Lost Feb 18 '14 at 3:56
  • $\begingroup$ You are welcome. I tried to use absolutely standard methods, to make the answer fit in as much as possible with what you are learning. $\endgroup$ – André Nicolas Feb 18 '14 at 3:58
3
$\begingroup$

You're essentially dealing a three-card hand from a deck of 50 (missing the spade and heart kings), wanting to know the probability of getting the remaining two kings among those three. The number of ways of dealing three cards is $_{50}C_3=19600$; the number of hands with two kings is $48$ (one for each of the remaining cards in the hand); so the probability is $48/19600=3/1225$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.