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This is a follow up to

Probability of poker pair - what's wrong with this?

The aim is to compute the probability of a pair in poker.

Consider a solution that initially counts the number of ways of picking a pair followed, in that order, by three cards that still conform to the fact that the pair will exist and be unique.

As under the original problem the pair does not need to be the first two cards to be picked, we need to multiply your answer by $C(5,2)=10$.

How to formalize the above argument using conditional probabilities?

$$P(\textrm{one pair}) = \sum P(\textrm{one pair}|\textrm{first card})P(\textrm{first card}) = P(\textrm{one pair}|\textrm{first card})$$

Now, how to continue the formal derivation?

From the above post we know that the answer is

$$ P(\textrm{one pair}|\textrm{first card}) = C(5,2) \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{44}{49} \cdot \frac{40}{48} $$

Clearly, this is also

$$ P(\textrm{one pair}|\textrm{first card}) = C(5,2) \cdot $$

$$ \cdot P(\textrm{2nd card forms pair with 1st card}|\textrm{first card}) $$

$$ \cdot P(\textrm{3nd card does not pair with 1st and 2nd card}|\textrm{first 2 cards}) $$

$$ \cdot P(\textrm{4th card does not pair with 1st, 2nd, 3rd card}|\textrm{first 3 cards}) $$

$$ \cdot P(\textrm{5th card does not pair with 1st, 2nd, 3rd and 4th card}|\textrm{all other cards}) $$

However, I was wondering if there is a more elegant way to write the above expression, making more explicit the symmetry in the problem and why $C(5,2)$ appears in the solution.

Daniel S.
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    The $~\displaystyle \binom{5}{2}~$ factor is used to adjust for the fact that otherwise, you are assuming that the first two cards drawn are the ones that pair up. Actually, there are $~\displaystyle \binom{5}{2} = 10~$ ways of choosing $~2~$ of the $~5~$ positions (i.e. 1st, 2nd, ..., 5th) and letting those $~2~$ positions represent which $~2~$ cards will be paired up. – user2661923 Feb 08 '23 at 05:17
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    Your formula does not make sense since you need at least two cards to form a pair. You could find the probabilities that there is exactly one pair in the first five cards given that you need two cards to form a pair or three cards to obtain a pair or four cards to obtain a pair or all five cards to obtain a pair. – N. F. Taussig Feb 08 '23 at 11:48
  • I think the formula makes sense but maybe it is very poorly expressed. I was wondering what would be a nice way to express, through an equation, the conditional probabilities that you are referring to above – Daniel S. Feb 08 '23 at 12:17

1 Answers1

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First, let's choose which rank our pair will be. There are $13$ ranks, and we will choose $1$. Now let's choose which suits to select for our pair. Each rank has $4$ suits, of which we will choose $2$.

This gives us $13\choose1$$4\choose2$

Now, let's choose the non-matching cards. Out of the $12$ remaining ranks, we want to choose $3$ distinct ranks. From each of the $3$ ranks, we will pick $1$ card.

This gives us $12\choose3$${4\choose1}^3$

Putting everything together with the total number of ways to choose a $5$ card poker hand from a $52$ card deck we get: $\frac{{13\choose1}{4\choose2}{12\choose3}{4\choose1}^3}{52\choose5}$

shrizzy
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  • thanks a lot for your answer, but I was wondering if there is a nice way to write the solution using conditional probabilities following the idea in the original post -- any suggestions? – Daniel S. Feb 08 '23 at 05:06
  • by the way, it seems that there are two sorts of solutions: your solution, which does not account for ordering of cards, and is kind of "atemporal" in that sense, and the solution that I've posted above, which accounts for the order, and in that sense is "temporal". how to capture the symmetry in the time events, in a nice way? – Daniel S. Feb 08 '23 at 05:07