The problem asks us to find the probability of a full house in a well-shuffled deck of $52$ cards.
The solution in the textbook states in the first line "All of the $52\choose 5$ possible hands are equally likely by symmetry, so the naive definition [of Probability] is applicable."
Everything after this involves counting, which I understand. However, I do not see why all outcomes are equally likely. Here is why.
Say we chose a card with rank $7$. The number of $7$'s left are now less than that of other ranks. This must mean the probability of choosing a different rank must be more than that of choosing another $7$.
This means that the probability of an outcome (a hand) with ranks $(2, 3, 4, 5, 6)$ must be more than that of a hand with ranks $(2, 2, 2, 3, 3)$.
Kindly explain why the naive definition works here and why we treat every hand (all $52\choose 5$ hands) to be an equally likely outcome.
Best Answer
Your examples of (2,3,4,5,6) and (2,2,2,3,3) are not hands for the purpose of the question.
(2♦,3♣,4♥,5♠,6♦) and (2♦,2♣,2♠,3♥,3♣) are hands, and are equally likely because the chance of pulling each of the cards involved is equally likely, up to the symmetry of rearranging the order of cards in the hand.
The idea that it's more likely a card will be junk than help to form a useful hand is a good intuition for why some hands are more or less likely, and can form the basis of a different method of calculating the probability of a given hand or class of hand, but is irrelevant in this case.