A Beautiful Pair(3)-CAT4

[rustom.hakimiyan]

Hi Ron,

Reposting as it was probably lost in the abyss. I've correct this based on your comments above:

For the sake of learning, I tried doing this a different way (where I do combine prob and combinatorics in one method, again, just trying to learn) but I can't seem to get the answer.

Probability Method:
Theory: Prob of 1 pair + Prob of 2 pairs

(12/12)(1/11)(10/10)(9/9) * (4!/2!2!) + (12/12)(1/11)(10/10)(1/9) * (4!/2!2!) = 20/33

I sense that my permutation is a little off here?

Combinatorics Method
Theory: Prob of 1 pair + Prob of 2 pairs

(6c1)(6c1)/(12c4) + (6c2)(6c2)/(12c4) = 261/495 - I get a completely wrong answer.

What am I doing wrong here?

Thanks in advance.

[RonPurewal]

Hi Ron,

Reposting as it was probably lost in the abyss. I've correct this based on your comments above:

For the sake of learning, I tried doing this a different way (where I do combine prob and combinatorics in one method, again, just trying to learn) but I can't seem to get the answer.

Probability Method:
Theory: Prob of 1 pair + Prob of 2 pairs

(12/12)(1/11)(10/10)(9/9) * (4!/2!2!) + (12/12)(1/11)(10/10)(1/9) * (4!/2!2!) = 20/33

I sense that my permutation is a little off here?

You need 2 fixes here.

"- In the left-hand part, you need to change the 9/9 to 8/9, since one of those 9 cards would make a second pair. (Remember, this is a calculation for exactly one pair.)

"- In your calculation for 2 pairs, each set of pairs (= each pair of pairs!) is actually going to show up twice"”once in each possible order. So, you need to multiply that whole expression by 1/2.
(This is not an issue in the left-hand expression, which only accounts for 1 pair.)

If you make both of these corrections, you’ll arrive at the correct value.

Although this solution involves a completely unnecessary amount of blood, sweat, and tears, at least you’ve discovered something you don’t get from the simpler method: namely, the specific probabilities of getting exactly 1 pair and of getting 2 pairs. Those are, respectively, 16/33 and 1/33 (the left- and right-hand fractions).

[RonPurewal]

Combinatorics Method
Theory: Prob of 1 pair + Prob of 2 pairs

(6c1)(6c1)/(12c4) + (6c2)(6c2)/(12c4) = 261/495 - I get a completely wrong answer.

What am I doing wrong here?

Thanks in advance.

OK, now this one needs a more complete re-work.

- The denominators (12c4) are legitimate.

- Two pairs is less complicated, so let’s get that out of the way first.
You’re approaching these as combinations in which order doesn’t matter. So, the number of ways to get two pairs is exactly that"”the number of different ways to select 2 pairs.
There are 6 pairs.
You want 2 of them.
So, the top of the right-hand fraction is just 6c2. The fraction is 6c2/12c4, which should reduce to 1/33.

- One pair is trickier. Here’s the first way that comes to my mind. (It’s certainly possible that something simpler exists.)
"”The number of ways of picking 1 pair is 6c1 = 6.
"”There are 10 cards remaining. You want 2 of them. That can be done in 10c2 = 45 ways.
"”Of those 45 ways, though, there are 5 you don’t want: namely, the 5 remaining pairs. So, there are actually 45 - 5 = 40 ways to select the 2 remaining cards.
So, the left-hand fraction should be 6*40/(12c4), which is 16/33 as required.

[TooLong150]

Can anyone explain where I am going wrong in my approach?

P(at least one pair) = 1 - P(no pairs)

P(no pairs)=
(12 possible cards for first card * 10 possible cards for second card * 8 * 6)/( [12 C 4] = number of ways to choose 4 cards from 12 cards) =
(12*10*8*6)/(12 C 4)
However, this value is greater than one, so I overcounted, but I don't know where.

Thanks.

[RonPurewal]

If there's a "first" card, a "second" card, etc., then the "C" formulas don't work. Those formulas are only for situations in which there is no order to the items being selected.

You'd have to use a "P" formula instead.

Personally, I try to avoid the use of either; I'd rather write explicit products and/or fractions that actually mean something to my intuition.

[danr969]

Hi Ron

I wanted to try this problem using a combinatoric approach just to see how it works out. I think the method of counting the number of hands with no match and subtracting from the total number of hands is the easiest. But, I wanted to try another approach.

The number of hands with at least one pair is:

P(1) + P(2) + .... P(6)
- (P(1 and 2) + P(1 and 3) + ...)

the first line accounts for all hands that contain a pair of 1's, a pair of 2's, etc. This overcounts the answer because it doubly counts the hands that contain 2 pairs. Thus, the second line.

P(1) means the number of hands that contain a pair of 1's. This means the hand contains two 1's, plus 2 more cards from the 10 remaining. This is 10C2 = 45. Same for P(2), P(3),etc.

P(1and2) means the number of hands that contain a pair of 1's and a pair of 2's, which is just 1 way. So, there are 6C2 ways to have 2 pairs, or 15 hands with two pairs.

Putting these ideas together gives 6 * 10C2 - 15 = 255.

So, there are 255/495 ways, or 17/33 ways to have at least one pair.

The actual work is (10C2 * 6 - 6C2)/495.

what I like about this method is that the the cases you have to consider are relatively plain: how many hands have a pair of 1's? no need to worry about overcounting at this level, the method directly takes care of the overcounting in the next step.

The method would probably not be useful unless it was well-practiced, but I wanted to get your feedback. Thanks in advance!

[tim]

This method works, but as you can see it's pretty complicated. It's always nice to have a backup plan, but if you can solve the problem an easier way on the test you should always do that. :)

[RonPurewal]

The method would probably not be useful unless it was well-practiced, but I wanted to get your feedback. Thanks in advance!

More methods = better. Be greedy and hoard as many methods as possible.

Don't spend too much brainpower trying to "rank" methods. If you do, the most likely consequence is that you'll develop a prejudice against your "backup" methods--i.e., you'll become disinclined to use anything except for your "#1" method.

[guilopes]

Hello

How about the following approach, is that correct?

Suppose we do not want one pair of cards numbered 2:

2 / Not 2 / Not 2 / Not 2

(1/12) . (10/11) . (9/10) . (8/9) = 2/33

Multiply by 4 (ordering): 8/33

Multiply by 2 (black or white): 16/33

1 - (16/33) = 17/33

[Sage Pearce-Higgins]

I think that your approach leaves out a few issues. First of all, why are you focusing only on "2"? Presumably any 2 cards with the same value would work. So we'd have to repeat the process for the other numbers. Second, why multiply by 4 for the ordering? If we're putting 4 cards in order, then there are 4x3x2x1 different orders (if all the cards are different, as they are here). If we try to take these things into account, then things get really complicated. We can't just add the possibility of not getting a pair of 2s to the possibility of not getting a pair of 3s, as there is a (significant) overlap in these possibilities. So I think that you might have stumbled across the right answer by chance.

Reading through this thread, I find Tim's explanation rather complicated (I'm not sure how he derives his "four different values from six" formula). Ron's approach seems much simpler.

As a challenge, you might consider the total number of outcomes (495: see TIm's post above) and try to count up the number of desired outcomes. This is trickier than Ron's approach, but may provide some insight into combinatorics.