Conditional probability on card drawing

Question

I take an ordinary shuffled deck of cards and keep flipping them over until the first ace appears, which is card #14. What's the probability the next card is the eight of hearts.

At first glance, one might think the answer to this is 1/51. There are 52 cards, one of which has a known position. Therefore, there are only 51 possibilities for the 15th card. However, this is not taking the other known information into consideration. We might not know exactly where the remaining aces are, but we can be certain they are lying somewhere after card #14. Thus, we have some condition to our probability.

To tackle this, let's work on two probabilities: first, what is the probability the eight of hearts is in the second split, and second, what is the probability the eight of hearts is the next card given that it is in the second split. (multiplying a conditional probability with the condition gives you the intersection of those probabilities.)

We can utilize compliments and rephrase the probability of the eight of hearts being in the second split as the probability it is not in the first split. There are 13 cards (excluding the known ace) in the first split. The total possible number of cards that can be in that split is 50, as all the aces come at index 14 and after. Thus, the probability the eight of hearts is in the second split is 1 - 13/50 = 37/50

Given that the eight of hearts is in the second split, what is the probability it is next? There are 38 cards in the remaining split (again not including the known ace) all of which are equally likely to be selected next resulting in a 1/38 conditional probability.

Multiplying these probabilities gives us 37/1900, the probability that the 15th card is the eight of hearts. Is this correct?

Answer 1

There are ${52-5 \choose 13}$ ways to choose first $13$ cards without $4$ Aces and eight of hearts.

There are ${4 \choose 1}$ ways to choose an Ace for $14$th card.

There is only one way to place eight of hearts at $15$th location.

There are $(52-15)!$ ways of placing rest of the cards starting from $16$ th position.

So overall there are ${52-5 \choose 13} \times {4 \choose 1} \times 1 \times (52-15)!$ arrangements with first ace at position $14$ followed by eight of hearts in position $15$.

And overall there are ${52-4 \choose 13} \times {4 \choose 1} \times (52-14)!$ arrangements with first ace at position $14$ in position $15$.

So conditional probability that first ACE is in position $14$ followed by eight of heats given that first ACE is in position $14$:

$$\frac{{52-5 \choose 13} \times {4 \choose 1} \times 1 \times (52-15)!}{{52-4 \choose 13} \times {4 \choose 1} \times (52-14)!} = \frac{35}{48 \times 38}$$

Answer 2

No, that's not right.

The total possible number of cards that can be in that split is 50, as all the aces come at index 14 and after.

You are forgetting the datum: the card at position 14 is the first ace.

Let $U$ be the event that our special card (eight of hearts) appears in the upper section, ie, after the position 14. And let $U^c$ be its complementary event.

Let $S$ be success event: the special card appears in position 15.

Let $E$ be the given event: "the first ace occurred in position 14".

We want $P(S|E)$. To introduce $U$ :

$$\begin{align} P(S \mid E) &= P(S, U \mid E) + P(S, U^c \mid E) \tag{1}\\ &= P(U \mid E)\, P(S\mid U,E) \tag{2}\\ \end{align} $$

Now, $P(U\mid E) = 1 - P(U^c \mid E) = 1 - \frac{13}{48}=\frac{35}{48}$

And, yes, $P(S\mid U,E) = \frac{1}{38}$

Answer 3

No. The method works, but you miscalculated a value.

Let us try another route ...

When the first ace is card #14, then there are $\binom{52-14-1}{3}$ ways to place the aces with the $8\heartsuit$ following the first ace out of the $\binom {52-14}{3}\binom{52-4}{1}$ ways to place the aces and $8\heartsuit$ in total.

$$\dfrac{37\cdot36\cdot35}{38\cdot37\cdot39\cdot48} =\left(1- \dfrac{13}{48}\right)\cdot\dfrac 1 {38}$$