Find the probability that the thirteenth spade will appear before the thirteenth diamond?

Question

The $52$ cards of an ordinary deck of cards are placed successively one after the other from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond?

This appears to be a hard problem since counting the favorable cases are hard to count. Clearly, the position of the $13$th diamond must be between $26$ and $N$. Now, I tried to count for specific values, but I find this difficult, perhaps there is a better way?

Note: I checked the answer and the favorable cases are supposed to be:

$$\sum_{k=26}^{52}(n-1)!(52-n)!\binom{26}{n-26}\cdot 13.$$

So I guess this ugly sum divided by $(52)!$ is equal to $1/2,$ right?

PS:

Question:

Answer:

Answer 1

Use symmetry.

Let us write $P_1$ as the probability that the thirteenth spade comes before the thirteenth diamond, and let $P_2$ be the probability that the thirteenth diamond comes before the thirteenth spade.

Clearly, one of these must occur, and they can't both occur. Thus, we have that $$P_1+P_2=1$$ Furthermore, since suits are arbitrary, we may exploit symmetry to say that $P_1=P_2$.

Thus, $P_1=P_2=1/2$.

Equivalently, the number of deck arrangements in which the last spade comes before the last diamond is $52!/2$.

Answer 2

By symmetry, the probability is $\frac 12$. Think of shuffling a deck, then swapping the spades and diamonds of matching rank, and asking the question. This bijects the cases where you get the thirteenth spade first with the ones where you get the thirteenth diamond first.

Answer 3

By symmetry, the probability is $\frac12$.

Formally, we may biject the deck orderings where the last spade comes out first, with the deck orderings where the last diamond comes out first, by the involution that switches $A\spadesuit$ with $A\diamondsuit$, $2\spadesuit$ with $2\diamondsuit$, ..., $K\spadesuit$ with $K\diamondsuit$. So there must be $\frac12 \cdot 52!$ orderings of each kind, which makes the probability $\frac12$.

Answer 4

In case if you are curious how to simplify the summation directly to $\frac12$: \begin{align} \sum_{n=26}^{52}(n-1)!(52-n)!\binom{26}{n-26}\cdot 13 =&{13\cdot 26!}\cdot 25!\cdot \sum_{n=26}^{52}\binom{n-1}{25} \\\stackrel{\text{H.S.}}=&13\cdot 26!\cdot 25!\cdot \binom{52}{26} \\=&52!\cdot \frac12 \end{align} In $\stackrel{\text{H.S.}}=$ we use the Hockey Stick identty.

Answer 5

As an alternative to the symmetry argument, reduce the problem size. Use a deck of 4 cards with just the aces. (Insert handwaving here why this does not change anything). It's easy to write down the 4! cases and count the number of times when the A♠ appears before A♢.

But why stop there? All the hearts and clubs don't make any difference. So we can reduce the deck to just the two relevant cards, yielding the obvious probability of 1/2.

Now back to the handwaving: With 13 cards of each suit, you can also remove all hearts and clubs. This reduces the original problem to "what is the probability for the last card to be diamonds?" which is obviously 1/2.

This is a wonderful example how we can transform a seemingly vast problem to a triviality!