[Math] 14th card is ace clarificarion.

probability

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.

Best Answer

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

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

NOTE:

  • 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

Related Question