Reason why these two probabilities are equal (Picking Balls in exact order without Replacement) I realize the probability of the following two events are equal. I am curious: is there a reason, besides coincidence, that the probabilities are equal?
Suppose there are five balls in a bucket. 3 of the balls are labelled A, and 2 of the balls are labelled B. There is no way to distinguish between balls labelled A. There is no way to distinguish between balls labelled B.
Suppose I draw balls at random, without replacement. The event $\{AAABB\}$ means I pick 3 balls labelled A, then 2 balls labelled B (in that exact order). Then,
$$ P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1 $$
Also,
$$ P(\{AABAB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \times 1 = 0.1$$
As you can see, the probability of the events $\{AAABB\}$ and $\{AABAB\}$ are exactly the same. I have seen the claim that any possible order of 3 A's and 2 B's is the same. Why is this true (if it indeed true)? If the claim is true, then I don't have to multiply out individually for every conceivable event.
Thanks.
 A: The claim indeed seems true.
For a way to see this.
Consider the first probability
$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1$
This can actually be written as
$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = 0.1$
Now notice the denominators: This will always be $5 \times 4 \times 3 \times 2 \times 1$.
Now the numerator will just be $3 \times 2 \times 1 \times 2 \times 1$, in some order.
The $3$ will appear when you pick the first A, one of the $2$ will appear when you pick the second A or first B etc.
Thus the probability is $\frac{12}{120} = 0.1$.
A: I will describe a problem that has the same flavour as yours, since it may throw some additional light on your observation.
A standard deck of cards is thoroughly shuffled.  Someone lifts up the top $5$ cards, without looking at them, and looks at the sixth card.  What is the probability that this sixth card is a Queen?  It is probably intuitively obvious that this probability is $4/52$, since all orderings are equally likely.
Now change the problem a tiny bit, we deal out the cards one by one, looking at each one.  What is the probability that the sixth card dealt is a Queen? (We are allowing one or more of the first five cards to be a Queen.)
Whether we looked or not obviously does not change the probability, so the probability is $4/52$.
But we are accustomed to solving such problems by a "tree diagram" procedure, so we might want to trace out all ways in which we can get a Queen on the sixth draw. One of them, for example, could be QQNNNQ, another could be NNNNNQ. Calculate the probabilities for each path (they are not all the same), and add up.  Not too hard, but it involves some work.  After the smoke clears, the mess will simplify to $1/13$.  
After performing the computation and noting that the answer is (equivalent to) $4/52$, which is the same as the probability that the first card drawn is a Queen, one might ask a question very similar to the one you asked.  
The answer to the cards question has already been given in the first few paragraphs of this answer.
Now to your problem.  Let's change your description a little.  We have $5$ cards, the $10$, Jack, Queen, King and Ace of spades.  Let's label the first three A, with invisible ink.  Let's label the last two B, again with invisible ink.  Shuffle the $5$ cards.  Note that all orders of the cards are equally likely. 
Now reveal the labels, using a magic light. It is I hope obvious that the order AAABB has the same probability as (say) the order ABABA.  
Sometimes, as in this case, even when we are told that certain objects are identical, one gets a clearer analysis by imagining them distinct, with any identicalness temporarily "secret."  
