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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?

user21820
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Vercingetorix
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2 Answers2

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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?).

KReiser
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drhab
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  • "The numbering of the cards does not affect this probability (why should it?)." Well my thinking was that the later the card occurs the more constrained its possibilities are (since other cards and suits have been exhausted by prior cards) but I can see the flaw now – Vercingetorix Jan 08 '23 at 15:34
  • @Vercingetorix This effect is balanced out by the fact that as you go along the number of cards in the deck decreases. This is why I like to think at these kind of problem combinatorially: it removes this kind of "chronological bias". – Kandinskij Jan 08 '23 at 15:36
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    @Kandinskij You could see it like that (balanced). But I would rather say that the that every triple $(i,j,k)$ of open spots with $1\leq i<j<k\leq52$ has equal chance to receive cards that are of the same suit if all cards are spread out over $52$ open spots. There is no reason to think that things are different. – drhab Jan 08 '23 at 15:42
  • @Vercingetorix "but I can see the flaw now.." Nice to hear. Actually you made a giant step in combinatorics. – drhab Jan 08 '23 at 15:43
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    Elegantly minimalistic ! $;;$:) – true blue anil Jan 08 '23 at 15:53
  • @trueblueanil Thank you for the compliment tba. I tried to imitate you ;-). – drhab Jan 08 '23 at 15:54
  • @drhab: You are too kind. With sparse mathematical knowledge, I just like to try something different. I've always looked up to you and tried to absorb the portions I could understand ! $;;$ :} – true blue anil Jan 08 '23 at 16:12
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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: $$\frac{49!\times 4 \times 13 \times 12 \times 11}{52!}=\frac{22}{425}$$

KReiser
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Kandinskij
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