# How are all possible poker hands of equal probability?

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.

• 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. Mar 12, 2021 at 3:10
• @saulspatz Would be a good answer, imo.
– jMdA
Mar 12, 2021 at 3:11
• @jMdA Okay, I posted it. Mar 12, 2021 at 3:12
• Is there an intuitive way to understand why these possibilities are equal? I can calculate each of these combinations to be equal in probability, but intuitively I don't see why they have to be. Mar 12, 2021 at 3:36
• @VictorM Does the example in my answer make sense to you? The poker hands have equal probability because the card shuffle is assumed to be fair. Of course, someone could always cheat and manipulate the shuffle to increase their chance of (say) a pair of aces. Mar 12, 2021 at 12:25

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".
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 $$\frac{1}{\binom{52}{5}}$$.