Clubtwofour, I will try to answer your question
"differentiate between these two problem types (e.g. "combinatorics problems differ from non-combinatorics probability problems because ...")?"
I do not believe there is a distinction between probability problems that are combinatorics and non-combinatorics. I believe that combinatorics is simply a tool to count the number of ways in which the event can occur and in which all the outcomes can occur. It is simply a tool set which comes to our rescue for problems that cannot be counted manually...
A problem can be solved both by using combinatorics and without combinatorics.
For example, lets consider problem number 2:
(I am also getting 17/33 as the answer and I hope this is correct.)
I will solve this problem using both approaches:
First we will understand what the problem is asking:
P(Bill finds at least one pair of cards that have the same value)
=1-P(Bill does not find any pair of cards with same value)
=1-P(All 4 cards are different value)
There are 12 cards with 2 cards each of values 1 2 3 4 5 6
or we can write the cards as: (using roman numeral to represent same value but different card)
1i 2ii 3iii 4iv 5v 6vi
Now Bill selects 4 cards one by one...
So first card can be any of the 12 cards.
For example, say Bill selected 4...Now the remaining cards are 11..so for second card to be selected, Bill has 11 options, but he has to satisfy the condition that all cards must be different, so he cannot select iv as the next card, thus he can only select from 10 cards
Now say he selects 6 as the second card, for selection of 3rd card, he has remaining 10 cards in the deck (since he has physically pulled out card 4 and 6). But now he cannot select vi or iv since he has already selected 4 and 6, so he can choose from 10-2 = 8 cards...
Now say he selects i as his third card, so he has remaining 9 cards in the deck (since he has physically pulled out card 4, 6, i). But now he cannot select iv, vi, and 1, so he can choose from 9 - 3 = 6 cards only.
Now that we understand the English and logic behind this problem, we will apply combinatorics to do the counting.
Probability is given by the ratio of x to n, where x is the number of ways in which the desired event can occur and n is the number of ways in which all outcomes can occur.
In this question, all outcomes are defined as 4 cards from 12 cards, so we apply 12P4...we are using permutation here as both selection and arrangement are important. Lets see why:
Lets consider that Bill shuffles the deck and he gets top 4 cards in the deck as 1 2 3 4...He shuffles again and now he gets the cards in the order 2 3 4 1...he shuffles again and he gets the cards as 3 4 2 1...Now when he turns over the cards, all these different shuffles will result in same layout of cards on the table, but indeed these are different shuffles which need to be counted. And hence we use 12P4 as the formula and not just 12C4 (which will not have included the different orders of the cards).
So we have our denominator as 12P4.
Next we come to the numerator...He selects 4 cards 1 by 1...
So as we discussed in our logic, he can select any 1 of 12 cards and that is given by 12C1
For second card he can select any of the 10 cards, so 10C1
For third card he can select any of the 8 cards, so 8C1 and finally for 4th card he can select from any of the remaining 6 cards, so 6C1.
Thus P(no pair is selected) = 12C1*10C1*8C1*6C1/12P4
=16/33
Final answer = 1- 16/33 = 17/33
Now we solve this question using non-combinatorics.
P(no pair is selected) = P(select 1 card)xP(select 2nd card)*P(select 3rd card)*P(select 4th card).
P(select 1st card) = 12/12 (since 12 cards are available to choose from and any one of the 12 can be selected as 1st card)
P(select 2nd card) = 10/11 (since 11 cards are left in the deck, but only 10 are available for selection)
P(select 3rd card) = 8/10 (since 10 cards are left in the deck after selecting 1st and 2nd cards, but 3rd can be selected only from 8 cards)
p(select 4th card) = 6/9
And multiplying all we get = (12*10*8*6)/(12*11*10*9)= 16/33
Giving our final answer as 1-16/33=17/33
Thus to summarize you can use combinatorics or choose to use manual counting method to solve probability questions. The key is to understand what the question is asking and then apply the toolset that you are most comfortable using.
Let me know if this helps.
