Probability that of $5$ cards drawn from a shuffled deck the 1st, 3rd and 5th will be the same suit.

Question

Given a well shuffled deck, what is the probability that the first, third and fifth card will be of the same suit.

Initially I was inclined to quickly say $\frac{4 \times {13 \choose 3} \times 49 \times 48}{{52 \choose 5}}$ ($4$ ways to choose $3$ of the same suit and the rest can be whatever) but clearly the order matters here.

So it seems that the simple, brute force way to do it would be a lot of conditioning:

i.e, the first card can be whatever.

Now condition on whether or not the second card is of the same suit or not.

And then each of these above splits into two more conditionals - whether the fourth card is the same suit or not.

So you'll have a sum of $4$ probabilities:

$\frac{52 \times 39 \times 12 \times 38 \times 11}{52 \times 51 \times 50 \times 49 \times 48} + \frac{52 \times 12 \times 11 \times 39 \times 10}{52 \times 51 \times 50 \times 49 \times 48} + \frac{52 \times 39 \times 12 \times 11 \times 10}{52 \times 51 \times 50 \times 49 \times 48} + \frac{52 \times 12 \times 11 \times 10 \times 9}{52 \times 51 \times 50 \times 49 \times 48}$

This looks about right to me, but I'm guessing there is a sleeker, more compact way to express this?

$\frac{4 \times {13 \choose 5}}{{52 \choose 5}} + \frac{4 \times {13 \choose 4} \times {39 \choose 1}}{{52 \choose 5}} \times 2 + \frac{4 \times {13 \choose 3}\times {39 \choose 2}}{{52 \choose 5}}$

This is clearly the same thing as the above, but I guess it makes me uncomfortable - the question very clearly enforces a kind of order, which isn't really being paid attention to in the combinations solution. Perhaps I'm overthinking it. Why can we ignore considerations of order in the expression above, or is it just the wrong solution entirely?

Answer 1

The probability is the same as the probability that the first three cards are of the same suit, or equivalently that the second and third card are of the same suit as the first card, which is:$$\frac{12}{51}\frac{11}{50}=\frac{22}{425}$$

This is also the probability that e.g. the $9$-th, the $17$-th and the $48$-th card are of the same suit.

The numbering of the cards does not affect this probability (why should it?).

Answer 2

There are $52!$ possible deck configurations. If you fix the first, the third and the fifth card there are just $49!$ possible configurations, so the answer will be: $$\frac{49!\times \#\{\text{ways we can fix the first,third and fifth card so that they have the same suit}\}}{52!}$$ Clearly for each suit we have $13\times 12 \times 11$ ways to fix those three cards so: $$\frac{49!\times 4 \times 13 \times 12 \times 11}{52!}=\frac{22}{425}$$