1
$\begingroup$

So let me try to explain my reasoning:

Since we know there are no Ace, there are 48 cards left. Now, we know that we have at least one Kings which mean we have (exactly one or two or three or four kings) while choosing the remaining cards from 44 leftover cards.

$$\frac{{4\choose1}{44\choose4}+{4\choose2}{44\choose3}+{4\choose3}{44\choose2}+{4\choose4}{44\choose1}}{52\choose5}$$

Now, there is another correct answer:

P(no Ace and at least one King) = P(no Ace) - P(no Ace and no King) $$\frac{{48\choose5}-{44\choose5}}{52\choose5}$$ if that's the case.. can I say for

P(no Ace and no Queen and at least one King):

$$\frac{{44\choose5}-{40\choose5}}{52\choose5}$$

EDIT: sorry, my question is: can someone explain why the other correct answer works too? I can't seem to grasp the logic of it. So let me explain further by saying that the logic behind it is somewhat similar to finding at least one king in a deck of 52 for a standard poker hand. $$\frac{{52\choose5}-{48\choose5}}{52\choose5}$$ But for this case P(no ace but at least one kings), we only start with 48cards.

$\endgroup$
  • $\begingroup$ What is the question? Probability of picking king before ace? $\endgroup$ – астон вілла олоф мэллбэрг Apr 13 '16 at 4:54
  • $\begingroup$ It sounds like he's talking about a hand of five cards, but this should have been specified in the question. $\endgroup$ – browngreen Apr 13 '16 at 4:55
  • 1
    $\begingroup$ In your first answer, you made a typo, the final term on the numerator should have been $\binom{4}{4}\binom{44}{1}$. Otherwise, yes. Both answers and methods are correct (but the first method is often seen as more tedious due to the case work. What if this was a million card deck with 100 kings of different suits? First method has a hundred things to add. Second method only needs two). Your follow up question is correct as well. $\endgroup$ – JMoravitz Apr 13 '16 at 5:01
  • $\begingroup$ In addition to the error in the first problem that @JMoravitz spotted, there is an error in your calculation of the probability that a five-card hand has neither an ace nor a queen but contains at least one king. You should have $\binom{44}{5} - \binom{\color{red}{40}}{5}$ in the numerator since a hand that contains no aces, kings, or queens is selected from the $52 - 3 \cdot 4 = 40$ other cards in the deck. $\endgroup$ – N. F. Taussig Apr 13 '16 at 9:39
  • $\begingroup$ @JMoravitz sorry for not posting the question! I've edited the post now. $\endgroup$ – misheekoh Apr 13 '16 at 17:22
3
$\begingroup$

Answer to the edited version of the question:

Find the probability that a five card hand contains no aces but at least one king.

As you know, the number of five card hands is $\binom{52}{5}$ since we are selecting five cards from a deck with fifty-two cards.

To calculate the number of hands that contain no aces but at least one king, we subtract the number of hands that contain neither aces nor kings from the number of hands that contain no aces. By taking the difference of these numbers, we find the number of five cards hands which contain no aces but do contain at least one king.

Since there are four aces in a deck of $52$ cards, the number of cards that are not aces is $52 - 4 = 48$. Therefore, the number of five card hands that contain no aces is $\binom{48}{5}$ since we must select five cards from the $48$ cards that are not aces.

Since there are four aces and four kings in a $52$ card deck, the number of cards that are neither aces nor kings is $52 - 2 \cdot 4 = 44$. Therefore, the number of ways of selecting five cards from the deck that are neither aces nor kings is $\binom{44}{5}$ since we must select five cards from the $44$ cards that are neither aces nor kings.

Hence, the number of five card hands that contain no aces but at least one king is $$\binom{48}{5} - \binom{44}{5}$$ from which we obtain the probability $$\frac{\dbinom{48}{5} - \dbinom{44}{5}}{\dbinom{52}{5}}$$ that a hand contains no aces but at least one king.

Answer to the original version of the question:

Your approach to both problems is correct, but there are some mistakes in the execution.

Find the probability that a five card hand contains no ace but at least one king.

In your first approach, as JMoravitz pointed out in the comments, the last term in the numerator should be $\binom{4}{4}\binom{44}{1}$ since you are selecting four kings and one additional card from the $52 - 2 \cdot 4 = 44$ cards that are neither aces nor kings, so the probability is $$\frac{\dbinom{4}{1}\dbinom{44}{4} + \dbinom{4}{2}\dbinom{44}{3} + \dbinom{44}{3}\dbinom{44}{2} + \dbinom{4}{4}\dbinom{44}{1}}{\dbinom{52}{5}}$$

Your second approach is correct. It also, as JMoravitz pointed out in the comments, is more efficient. It has the added virtue of containing fewer steps and, consequently, fewer opportunities to make an error than your first approach.

Find the probability that a five card hand contains no aces and no queens but at least one king.

Your approach of subtracting the number of hands that contain no aces, no queens, and no kings from the number of hands that contain no aces and no queens is correct. However, there are $52 - 3 \cdot 4 = 40$ cards that contain no aces, no kings, and no queens. Hence, the probability is $$\frac{\dbinom{44}{5} - \dbinom{40}{5}}{\dbinom{52}{5}}$$

$\endgroup$
  • $\begingroup$ I've edited the post to add the initial question I had in mind. Thank you. $\endgroup$ – misheekoh Apr 13 '16 at 17:23
  • $\begingroup$ I have edited my answer accordingly. $\endgroup$ – N. F. Taussig Apr 13 '16 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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