How are all possible poker hands of equal probability?

Question

Please pardon the elementary question, for some reason I'm not grocking why all possible poker hand combinations are equally probable, as all textbooks and websites say. Just intuitively I would think getting 4 of a number is much more improbably than getting 1 of each number, if I were to draw 4 cards. For example, ignoring order, to get 4 of a single number there are only $4 \choose 4$ distinct possibilities, whereas for 1 of each number I would have ${4 \choose 1}^4$ distinct possibilities.

Answer 1

Yes, that's true, but they mean that any particular hand of 5 cards has the same probability as any other hand of 5 cards. Once you start talking about the probability of a pair or four of a kind, you're talking about the probability of getting one of a number of hands. To put it another way, the probability of drawing a royal flush in spades is exactly the same as the probability of drawing the 2,3 of diamonds, the 6,8 of clubs, and the Jack of hearts.

Answer 2

Often the books and websites speak a little too loosely. I would say that "getting 4 of a kind" is not a poker hand. It is a set of many different poker hands. A "possible poker hand" is completely specific, e.g. the hand 4 of spades, 4 of hearts, 4 of clubs, 7 of clubs, 8 of diamonds.

Answer 3

This is just like flipping a coin. You are just as likely to get exactly $$HHTHTTH$$ as you are to get $$HHHHHHH$$ You are intuitively grouping poker hands into their categories, and you are right that for example four of a kind is less likely than high card. Notice for example, that the wikipedia page is careful to distinguish poker hands from "hand-ranking categories".

Answer 4

I can try deriving it. Imagine a classic case of a dealer drawing 5 cards from the top of a shuffled deck one at a time.

Assume that there exists an ordering in a hand, so JQK12 is different from 1JQK2.

This means that the probability of choosing a hand is $\frac{1}{52} \times \frac{1}{51} \times \frac{1}{50} \times \frac{1}{49} \times \frac{1}{48}$. However, the multiplication operation assumes an ordering. There are $5!$ ways of drawing the "same" hand. This means that we can multiply our previous product by $5!$. After some simplification, you will notice that it is equivalent to $\frac{1}{\binom{52}{5}}$.