There are 2 questions.

I need to find probablity that 14th card is ace and that first ace occurs on 14th card.

For the first part, I though it like this.

Choose any of 1 ace in ways = ${4 \choose 1}$

And choose 13 cards from remaining 51 cards= ${51 \choose 13}$

Choose 14 cards from 52 = ${52 \choose 14}$

Probability = $\frac{{4 \choose 1}{51 \choose 13}} {{52 \choose 14}}$

For Second Part

I removed all four aces, cards remaining = 48. Choose 13 from these = ${48 \choose 13}$ And any one ace = ${4 \choose 1}$ Probability = $\frac{{4 \choose 1} {48 \choose 13}} {{52 \choose 14}} $

Both answers do not match.


First part: $$\frac{\binom 41 \binom {51}{13} 13!}{\binom {52}{14}14!}$$

Second part: $$\frac{\binom 41 \binom {48}{13} 13!}{\binom {52}{14}14!} $$


  • Order matters.($13!$ and $14!$)

Your answer consider these two(and many more) cases the same: $$[2,3,\color{green}{4,5},6,7,8,9,10,J,Q,K,2,A]\leftrightarrow[2,3,\color{red}{5,4},6,7,8,9,10,J,Q,K,2,A]$$ because you only select those $13$ cards, not arrange them before the $14$th ace card

  • $\begingroup$ can you explain the concept of order. Both answers are suddenly correct. :D $\endgroup$ – Kartik Sharma Aug 31 '14 at 10:18
  • $\begingroup$ @KartikSharma does that explain it? $\endgroup$ – RE60K Aug 31 '14 at 13:50

For the first question I would consider four cases:

1) First 13 cards no aces, then an ace.

2) First 13 cards one ace, then an ace.

3) First 13 cards two aces, then an ace.

4) First 13 cards three aces, then an ace.

1) $\frac{{4 \choose 0}{48 \choose 13}} {{52 \choose 13}}\cdot \frac{4}{39}$

2) $\frac{{4 \choose 1}{48 \choose 12}} {{52 \choose 13}}\cdot \frac{3}{39}$

And so on. Then you have to sum these probabilities.


For all, who do not believe that the answer is correct, here is the result.


For (1), your denominator is wrong, because of the $14$ cards chosen order is important (for at least one of the cards). (1) would be easier if you imagined picking the $14$th card first (or if you prefer, laying out the first $13$ cards face down, and then turning up the $14$th card.

For (2) your denominator is again wrong. You might try doing this one with order being important throughout the problem. (I.e., choose a non-ace for slot 1, choose a non-ace from the remaining cards for slot 2, etc.)


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.