A deck of cards is dealt out. What is the probability that the first ace occurs on the 14th card?

A deck of cards is dealt out. What is the probability that the first ace occurs on the 14th card?

Solution 1.

$$4\cdot\frac{48\cdot...\cdot36}{52\cdot...\cdot39}= 4\cdot\frac{48!/35!}{52!/38!}=4\cdot\frac{48!\cdot38!}{52!\cdot35!}\approx0.03116$$

Solution 2.

$$\frac{\binom 41 \binom {48}{13} 13!}{\binom {52}{14}14!}\approx0.03116$$

In both solutions only 14 draws are considered. If one continues to draw cards, until all 52 cards have been drawn, would the probability that the first ace is drawn on the 14th turn be the same. If so, why?

The short answer to your question is that the cards after the fourteenth don't matter. The $15$th through $52$nd cards don't affect the outcome at all.

You draw thirteen non-ace cards ($_{48}P_{13}$) then draw an ace ($4$). Divide by the number of fourteen-card permutations ($_{52}P_{14}$).

Now, if you consider the remaining $38$ cards, you just get $38!$ on top and bottom again.

• Thank you very much, I understand now. – David Nov 25 '16 at 20:27

Let's instead suppose that all the cards are dealt, so that there are $52!$ different orders in which to deal the cards.

Now, for success, we want the first $13$ cards to be chosen from the $48$ non-ace cards--there are $\frac{48!}{(48-13)!}$ ways to do this. Given any such ordering of the first $13$ cards, we require the next card to be one of the $4$ aces, and there are $4$ ways to do this. After that, the order in which the remaining $38$ cards appear will not affect the success, so any of the $38!$ orderings of these cards will work. Consequently, there are $$\frac{48!}{(48-13)!}\cdot 4\cdot 38!=\frac{48!}{35!}\cdot 4\cdot 38!=4\cdot48!\cdot\frac{38!}{35!}$$ successful ways to deal out all the cards. Thus, the probability of success is again $4\cdot\frac{48!}{52!}\cdot\frac{38!}{35!}.$

• Thank you very much for the calculations, I get it now. – David Nov 25 '16 at 20:26

$\underline {Simple\; way,\; considering\; only\; aces\; in\; 52\; slots}$

The favourable cases necessarily have an ace in slot $14$, and three in the subsequent $38$ slots,

thus $Pr = \dfrac{\binom{38}3}{\binom{52}4}$