# Probability of the card following first ace being ace of spades or two of clubs

I am learning probability from Scheldon Ross' book.

A deck of 52 playing cards is shuffled, and the cards are turned up one at a time until the first ace appears. Is the next card – that is, the card following the first ace – more likely to be the ace of spades or the two of clubs?

Solution given is:

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. Furthermore, for each of the 51! orderings of the other cards, there is only one place where the ace of spades can be placed so that it follows the first ace. Thus,

$$P{(\text{the ace of spades follows the first ace})} = \frac{51!}{52!}=\frac{1}{52}$$ Similarly, the probability that the two of clubs (or any other card) follows the first ace is also $$\frac{1}{52}$$.

which sounds reasonable. However the author carefully notes the confusion usually made while solving this problem. I got his point, but somehow the explanation given to eliminate the confusion is still confusing me. The author notes below:

Many people’s common reaction is to suppose initially that it is more likely that the two of clubs follows the first ace, since that first ace might itself be the ace of spades. This reaction is often followed by the realization that the two of clubs might itself appear before the first ace, thus negating its chance of immediately following the first ace.

Upto this I got it, but did not get how the probability of first ace itself is ace of spade and probability that the two of clubs comes before the first ace are same, as author explains:

However, as there is one chance in four that the ace of spades will be the first ace (because all 4 aces are equally likely to be first) and only one chance in five that the two of clubs will appear before the first ace (because each of the set of 5 cards consisting of the two of clubs and the 4 aces is equally likely to be the first of this set to appear), it again appears that the two of clubs is more likely. However, this is not the case as explained above.

1. I did not get this: only one chance in five that the two of clubs will appear before the first ace (because each of the set of 5 cards consisting of the two of clubs and the 4 aces is equally likely to be the first of this set to appear)

2. What I think is we have to prove: $$P(\text{first ace itself is an ace of spade}) = P(\text{two of clubs appear before first ace})$$ How authors' explanation proves this. And if it does not, then how it can be proven?

• What you mention under 2) is a statement that is not true. This because $\frac14\neq\frac15$. Aug 1 '15 at 8:54
• ok so I understood that the fact of ace of spade itself may be the first ace is balanced by the fact that the two of clubs may appear before first ace. But their probabilities are different: $\frac{1}{4}$ and $\frac{1}{5}$. Then how are both events equally likely? No further explanation is given elaborating how the two events are equally likely despite of these unequal probabilities. I am not able to figure out what is missing.
– Maha
Aug 4 '15 at 10:56

You are telling contradictory things. Firstly you say "did not get how the probability of first ace itself is ace of spade and probability that the two of clubs comes before the first ace are same" [ they are not! ], then go on to say

".. did not get [how ] only one chance in five that the two of clubs will appear before the first ace (because each of the set of 5 cards consisting of the two of clubs and the 4 aces is equally likely to be the first of this set to appear)"

The second one is the correct statement of probabilities, and the simple way to understand it is to realize that when considering the relative sequence in which a subset of the deck appears, you only need to consider the subset, and ignore the rest. So when you consider the relative sequence of the subset comprising of the two of clubs and 4 aces, only these 5 cards are material, and the two of clubs will be first among them $\frac 15$ of the time.

• ok so I understood that the fact of ace of spade itself may be the first ace is balanced by the fact that the two of clubs may appear before first ace. But their probabilities are different: $\frac{1}{4}$ and $\frac{1}{5}$. Then how are both events equally likely? No further explanation is given elaborating how the two events are equally likely despite of these unequal probabilities. I am not able to figure out what is missing.
– Maha
Aug 4 '15 at 10:57
• I think that you should delete (at least from your mind !) the entire portion after "Similarly, the probability that the two of clubs (or any other card) follows the first ace is also $\frac{1}{52}$." All that he is describing thereafter is various confusions people go through, and the wrong argument they give to prove that the ace of spades and the two of clubs are equally likely to follow the first ace. Aug 4 '15 at 11:31
• Yeah I can delete. But then for some similar but different problem, I will induce confusion myself. I think understanding how the confusion is confusing :p is important. There has to be some argument explaining how those two events are equally likely despite above explanation says their probabilities are different.
– Maha
Aug 4 '15 at 21:20

I am revisiting the question after two years. And now I feel the whole confusion is dues to misunderstanding of the problem. When the question asks "Is the next card – that is, the card following the first ace – more likely to be the ace of spades or the two of clubs?", it presumes following:

• first ace is not of spades and
• two of clubs dont come before first ace

Since, only after above assumptions only, the whole problem makes sense. Also, notice that the solution given first orders 51 other cards giving 51! ordering and then inserts spade-ace and club-two after first ace in the ordering. With this approach taken in the solution, its clear that above two points are indeed presumptions and hence should not be considered otherwise in order to avoid any confusion.