Dear
voodoo_child
Yes, there's a big problem in what you did . . .
When you are applying the Fundamental Counting Principle (FCP) (i.e. "this many possibilities for the first choice, this many possibilities for the second choice, etc.), then you have to divide by 2 (or whatever higher nCk) to account for duplicates --- that system will count both AB and BA, and you have to account for that.
By contrast, when you use nCk, that
automatically accounts for duplicates already --- they are already not counted, so the number you get, nCk, is the
real number of combination of k elements chosen from a pool of n. If AB is counted, BA automatically will not be counted, just by the way nCk is set up.
When you calculate 6C2 = 15, then there actually are fifteen distinct pairs, no duplicates. As you note, 3 of them are two-of-a-kind pairs, so they are 15 - 3 = 12 distinct non-matching pairs. There are 3 matching pairs and 12 non-matching pairs.
For clarity, I'll list them. Let's say
Set 1 = a b c
Set 2 = A B C
3 matching pairs = Aa, Bb, Cc
12 non-matching pairs = Ab, AB, Ac, AC, ab, ac, Ba, Bc, BC, bc, Ca, Cb
Probability of picking a matching pair from this set
= (number of matching pairs)/(total number of pairs) = 3/15 =
1/5
===================================
Now, move to
Set 1 = A B C D E F
Set 2 = a b c d e f
Now there are 12C2 = 66 pairs. There really are 66 unique and different pairs, no need to divide anything by two. Of those pairs, 6 are matching pairs, so the other 60 are non-matching pairs.
Probability of picking a matching pair from this set on a two card draw
= (number of matching pairs)/(total number of pairs) = 6/66 =
1/11
The harder question, the original question --- what is the probability of choosing four cards and having exactly two of them match as a pair?
You
can't just multiply the total number of matching pairs and the total number of non-matching pairs --- among other things, some of those matching pairs (e.g. Bb) will
overlap with some of the possible non-matching pairs (e.g. Bf or Db). With any given matching pair, only certain non-matching pairs are possible.
There are still six pairs. To calculate the number of non-matching pairs that can go with any particular pair, you have to, as it were, set aside that pair and
calculate anew with the remaining 10 cards. There are two ways to do that.
1) using nCk
Set aside the matching pair. Of the remaining 10 cards, there are 10C2 = 45 pairs. There really are 45 pairs, no need to divide anything by 2. Of those 45 pairs, 5 form a second matching pair, which we don't want, so there are 40 non-matching pairs.
2) using FCP
Set aside the matching pair. Of the remaining 10 cards, there are 10 choices for the third card, and if we don't want a match, there are 8 choices for the fourth man. Because we are using the FCP, and not nCk, that means we
have counted pairs twice --- eF and Fe have both been counted. Therefore, the 8*10 = 80 needs to be divided by 2 --- and again, we wind up with 40 non-matching pairs.
6 matching pairs, each with 40 non-matching pairs that can go with it ==> 240 combinations meet the condition.
That's of all possible sets of 4 ===> 12C4 = 495
So, probability = 240/495 = 80/165 = 16/33
Does all this make sense? Do you see the difference between calculating numbers with the FCP vs. with nCk? Please let me know if have any further questions.
Mike
