Full House in Poker -- Why do we treat all possible hands to be equally likely? 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.
 A: 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.
A: I'm going to add a second answer (highly inappropriate--except I think in this case). My answer and others are making an assumption that in a sense disregards your question. You asked "why do we treat all hands as equally likely" and the answers you received were basically "well if you treat all hands equally likely, here's what happens." (at least that describes my answer, maybe not others).
In truth the problem is far more complicated. We do have to make an assumption: the deck deals out each card with equal probability and without replacement. If we have two players, in five card draw, you get what you get (easiest to analyze), the deck deals a total of $2*5 = 10$ cards. Therefore we have $\binom{52}{10} = 15820024220$ (~$15.8$ billion combinations). For each of those combinations there are $\binom{10}{5} = 252$ possible combinations of sets of "2-hands".
So now we need to find how many of these combinations contain a full house and what's the probability that the full house ends up all on the same side (reminder they could both have full houses!).
I really don't know how to do this computation. I would start with one full house and none of the two ranks. We can figure that out. We know there are ~$3$k ways to get a full house, now we choose 5 cards from the $52 - 8 = 44$ (subtracting $8$ because we are removing the two suits from our full house), to get that we can have a single full house hand with no other possibility of $\binom{44}{5}*\#(\text{full house hands}) \approx 4066013952$ (~$4.1$ billion)!
We know that there are $252$ ways to "permute" this and that only one (two actually) puts the full house into one hand. So if we assume that a single full house is dealt within $10$ cards, the probability that "you" get it is $\frac{1}{2}\frac{2}{252}$.
But remember--the probability of you even getting into this situation is ~$\frac{4}{16} \approx \frac{1}{4}$, so you still only have a ~$\frac{1}{1000}$ chance of getting a full house (in this specific scenario). Your chances increase as we add more possibilities (but obviously not by much because the chances of having a "10-hand" that has multiple full houses is small comparatively).
A: I think your error is in the following assumption...
"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."
The chance of any specific card in a shuffled deck occupying any specific place in the deck is equal across the whole deck.
Looking at cards does not change the initial distribution determined by the original deck shuffling.  The chance that the second card is a 7 is still 1 in 52.
Shuffling is the event sparking the random distribution...not observing the cards.
