In a deck of cards that are turned up one at a time until the first A appears. Is the next card more likely to be the A of spades or the 2 of clubs?

Question

This is Example 5j, from Sheldon Ross's First Course in Probability 8th ed, page 38. I don't understand why the following is true.

Solution.To determine the probability that the card following the first ace is the aceof spades, we need to calculate how many of the (52)! possible orderings of the cardshave the ace of spades immediately following the first ace. To begin, note that each ordering of the 52 cards can be obtained by first ordering the 51 cards different from the ace of spades and then inserting the ace of spades into that ordering.

I don't see how this (sentence in italics) can be true. For example if we have $S=\{ 1, 2, 3\} $ the number of orderings that can be obtained are $3!=6$. Following the solution's reasoning we could calculate the orderings for $S$ by ordering the cards different form $3$ and then inserting in into that ordering, that is $ 2!$ . What am I missing? perhaps the sentence in italics does not mean what I think it does?

Also, the solution given is a probability of $ \frac{1}{52} $ for both, I understand why but I have a different solution that also seems valid:

My solution

• Ordering in which the card following the first ace is the ace of spades; We have 3 other aces so we put $A_i A_s $, with $i = c, d, h $, together as one unit and count the number of permutations $ = 51! $. As we have three of these such pairs $$ P(N_a) = \dfrac{3\cdot 51!}{52!} $$

• Ordering in which the card following the first ace is the two of clubs By a similar argument we put $A_i A_s $, with $i = c, d, h, s$, so $$ P(N_c) = \dfrac{4\cdot 51!}{52!} $$

Can someone tell me what is the error in this reasoning?

Answer 1

If you order the numbers $1$ and $2$ and then insert $3$, you can get all $6$ permutations of $1,2$, and $3$. From the initial ordering $1,2$ you get $3,1,2$, $1,3,2$, and $1,2,3$, and from the initial ordering $2,1$ you get $3,2,1$, $2,3,1$, and $2,1,3$, depending in both cases on where you insert the $3$. The insertion can be done anywhere in the shorter sequence.

I’m afraid that I can’t follow the reasoning in your solution: the factors of $3$ and $4$ really don’t make sense, because treating the spade ace together with another ace or together with a two doesn’t make sense when you’ve already counted $51!$ permutations of the cards other than the space ace.

Answer 2

In your analysis of set $S$, you didn't take it far enough. As you indicated, there are $3!$ total orderings.

Further, as you indicated, there are $2!$ total orderings of the elements besides the $3$. Each of these $2!$ orderings in effect have gaps before the 1st element, between the two elements, and after the 3rd element. In order for the $3$ to immediately follow the first element in any of the $2!$ orderings, the $3$ must go in the 2nd gap. There is only 1 way that this can occur.

Thus, with respect to your set $S$, you have a fraction where the denominator is $3!$, and the numerator is $2!$.

For critiquing your solution to the 52 card deck:

In your

"Ordering in which the card following the first ace is the ace of spades"

your enumeration of the numerator is wrong, because you are overcounting.

Suppose that you couple the Ah with As. This only pertains if the Ah happens to be the first Ace, among the Ah, Ad, Ac.

That is, when Ah-As are coupled, your enumeration incorrectly counts Ac,Ah-As,Ad as "favorable".

"Ordering in which the card following the first ace is the two of clubs"

I am unable to critique this, because I can not re-construct (i.e. reverse engineer) what you mean re "by a similar argument".

Answer 3

I don't see how this (sentence in italics) can be true. For example if we have S={1,2,3} the number of orderings that can be obtained are 3!=6. Following the solution's reasoning we could calculate the orderings for S by ordering the cards different form 3 and then inserting in into that ordering, that is 2! . What am I missing? perhaps the sentence in italics does not mean what I think it does?

You are missing that there are $3$ positions to put the number $3$ is so there are $2! \times 3$ results.

Your example:

Shuffle the three cards and there are $3!=6$ options. They are $1,2,3|1,3,2|2,1,3|2,3,1|3,1,2|3,2,1$.

Now do it the book's way.

Shuff the $1,2$. There are $2!$ ways to do it. $1,2$ or $2,1$.

Now put the $3$ into the deck. You seem to think there is only one way to do that but there are $3$ positions it can be placed in. If you shuffled $1,2$ as $a,b$ you can put the $3$ if the first position; $3,a,b$ or the second; $a,3,b$; or the third $a,b,3$.

So the total number of ways is $2!\times 3=6$.

They are: if $1,2$ is shuffled as $1,2$ then $3,1,2|1,3,2|1,2,3$. And if $1,2$ is shuffled as $2,1$ then $3,2,1|2,3,1|2,1,3$.

We have 3 other aces so we put AiAs, with i=c,d,h, together as one unit

This tells allows that the $A_s$ con procede any other ace. BUt you did not specify that that ace is the first $A$. If we have $A_cA_s$ we have only a $\frac 13$ chance that $A_c$ is the first ace. If $A_c$ is the second or third it doesn't count.

So the result your way would be $\frac {\frac {3\cdot 51!}3}{52!}$.

And for the $2$ of clubs you'd have $\frac {\frac {4\cdot 51!}4}{52!}$.

Answer 4

Your argument (the second one) seems to miss the FIRST ace aspect of the situation. The $3*51!$ orderings in the numerator have the spade ace immediately following another ace but some have yet a different ace occurring before both of them, so they shouldn't be counted.

Answer 5

Given that the first ace is the 20th card to appear: This suggests no Ace was chosen in the first 19 card flips. No other information was given, so every other card is subject to random choice in the first 19 card flips. On the 20th flip we are told an Ace is flipped. This gives the equation for P(A)

$P(A) = \binom{48}{19} * \binom{4}{1}$

On the 21st flip we are told an Ace of Spades is flipped. This means there are only 3 Aces to be randomly selected on the 20th flip. On the 21st flip there are 32 cards to chose from and we need the probability that the Ace of Spades is chosen.

$P(BA) = \binom{48}{19} * \binom{3}{1} * \dfrac1{32}$

So the final equation is

$P(B \mid A) = \dfrac{\binom{48}{19} * \binom{3}{1} * \dfrac1{32}}{\binom{48}{19} * \binom{4}{1}}$

On the 21st flip we are told the Two of Clubs is flipped. This means in addition to the 4 Aces, the Two of Clubs is not chosen in the first 19 flips. This leaves 47 cards to be chosen at random in the first 19 flips. On the 20th flip one of four Aces are chosen. On the 21st flip the Two of Clubs is chosen.

$P(CA) = \binom{47}{19} * \binom{4}{1} * \dfrac1{32}$

So

$P(C \mid A) = \dfrac{\binom{47}{19} * \binom{4}{1} * \dfrac1{32}}{\binom{48}{19} * \binom{4}{1}}$