This question is a follow-up question on this one: Probability of drawing an ace or 2 after an ace

Background Out of a standard deck you draw cards (without replacement) one by one until you reach an ace. Then you continue to draw cards until you reach another ace or a "2" card. What is the probability that you reached a "2" card and not an ace?

The solution To shorten things up, let's assume the deck has only two aces and two "2" cards. Clearly, all that matters are the relative position of these two cards, not all the other cards. So we can think of the permutations of 4 items in a row.

The event "the card following the first ace is ace" occurs when the two aces are together. The sample space is ${4 \choose 2}$ (choosing 2 places for the aces) and there are 3 locations in which they are together (1-2,2-3,3-4) so the solution is $\tfrac{3}{{4\choose 2}}=0.5$.

Surprisingly, the answer $0.5$ remains correct even if there are 4 aces and 4 "2"s in the original deck and in fact, as @InterstellarProbe shows in the comments to the original question, if there are $n+1$ of each type, the answer is: $$\sum_{k=0}^{n+1} \frac{n\left(\begin{array}{c} n+k \\ k \end{array}\right)}{(n+k)\left(\begin{array}{c} 2 n+2 \\ n+1 \end{array}\right)}=\frac{1}{2}$$

The question The above result surprises me a bit. There is no obvious symmetry in the question and it is not obviously clear why there is the same number of arrangements in which after the first ace comes another one as arrangements in which after the first ace comes a "2". I spent a couple of months thinking, asked also fellow probability teachers, but no-one had a good idea.

So, returning the ball to this court. What is the intuitive explanation to this result? Why the permutations are symmetric and how can we get this $\tfrac{1}{2}$ without the algebra?

  • $\begingroup$ I would say that it looks asymmetric because you don't think about what happens before that first ace. You might have already discarded several 2's. $\endgroup$ – Marc Romaní Nov 24 '20 at 14:32

I'm not sure that the following solution is any more intuitive than others (in my experience, some probability problems are simply non-intuitive, at least with my intuition), but at least it is a different approach than those given so far.

To start with, we can ignore all the cards in the deck except for aces and twos. So suppose we have a deck of four aces and four twos, which can be arranged in $8!$ ways, all of which we assume are equally likely. We want to find the probability that the card following the first ace in the deck is also an ace.

As a further simplification, let's see if we can find the probability that the card immediately following the first ace is the ace of spades. We want to count the number of such arrangements. To do so, suppose we remove the ace of spades from the deck and deal out the remaining cards; this can be done in $7!$ ways. There is only one place where the ace of spades can now go so that it is immediately following the first ace, so the number of arrangements in which the ace of spades follows the first ace is $7!$, and the probability of this event is $$\frac{7!}{8!} = \frac{1}{8}$$

If we don't care which ace follows the first ace in the deck, there are $4$ possible choices, all of which have the same probability of $1/8$. So the probability that the card following the first ace is also an ace is $$4 \cdot \frac{1}{8} = \frac{1}{2}$$

  • $\begingroup$ It wasn't what I was looking for (searched for symmetry of the permutations) but this is an excellent answer. Following on that logic, from symmetry, the probability of any card to be after the first ace is equal so it must be $\tfrac{1}{8}$ and the probability that it would be an ace is $0.5$. The generalization to more cards and more types (ace, 2, and kings...) is straightforward. Great! $\endgroup$ – YJT Nov 24 '20 at 17:46

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