A Beautiful Pair(3)-CAT4

[geenaojoseph]

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33

Please share the combinatorics way of doing this. Ans.C

[tim]

The total number of ways 4 cards can be selected from 12 is

12! / (4!)(8!) = 495

which is the denominator for the probability problem. Now let's look at what we would call a "failure" - no cards of the same value. (This is FAR FAR FAR easier than calculating successes, so we'll just subtract this value from 495 to get the successes). To get 4 cards of different values, we have to pick 4 different values from among 6:

6! / (4!)(2!) = 15

But each of these 15 sets of four cards can have each card colored either red or black, and two choices applied to 4 cards gives you 2^4 = 16 colorings for those four cards. Multiplying the possibilities for the values by the possibilities for the colorings gives you 15*16 = 240 failures, so now we have 495 - 240 = 255 successes. Total probability then is

255 / 495 = 17/33

[geenaojoseph]

The total number of ways 4 cards can be selected from 12 is

12! / (4!)(8!) = 495

which is the denominator for the probability problem. Now let's look at what we would call a "failure" - no cards of the same value. (This is FAR FAR FAR easier than calculating successes, so we'll just subtract this value from 495 to get the successes). To get 4 cards of different values, we have to pick 4 different values from among 6:

6! / (4!)(2!) = 15

But each of these 15 sets of four cards can have each card colored either red or black, and two choices applied to 4 cards gives you 2^4 = 16 colorings for those four cards. Multiplying the possibilities for the values by the possibilities for the colorings gives you 15*16 = 240 failures, so now we have 495 - 240 = 255 successes. Total probability then is

255 / 495 = 17/33

Thank you Tim! Specifically liked the strategy of considering color and number of cards separately. Gives flexibility to extend it to questions with more restrictions.

[tim]

Glad to hear it! Yes, there is plenty in this approach that you can apply to other questions.

[mcmebk]

Glad to hear it! Yes, there is plenty in this approach that you can apply to other questions.

Hi Tim

It took me a while to comprehend the logic:

Pick up 4 numbers, each one can be either red or black, so the combination of 4 fixed numbers are: 2*2*2*2=16.

Is my understanding correct?

I got this question wrong in two different ways, could you help me out?

1. Negation - no cards are paired:

First step, pick 1 out of 12, 12C1;
second step, pick 1 out of the rest 10 cards (it can not be 11 because otherwise it would be paired with the previously selected card), 10C1;
similarly, the rest 4 cards are 8C1 and 6C1

The result is 1-12C1*10C1*8C1*6C1/12C4 ---- totally wrong...

2. At least one pair of cards:

A. one pair is made
first step, pick 1 pair out of 6 pairs, 6C1
second step, pick three cards that can not be paired: 10C1*8C1*6C1
B. two pairs are made:
6C2

The result is (6C1*10C1*8C1*6C1) +6C2 ---- totally wrong...

What mistake did I make?

Thank you.

[RonPurewal]

principle number 1 of gmat probability problems: if you are mixing combinatorics and probability, then your method is too complicated.

i have never seen a GMAC problem that mixed these two things -- you have combinatorics problems, and you have probability problems, but i've never seen a problem that necessitates the use of both.
this doesn't mean it's not possible to use both -- but, if you're using both, you should look for a more efficient solution when you're reviewing the problem later.

this is a probability problem, so let's not detour into factorials and such. because factorials are yelling, and yelling is rather rude here.

you can just find the probability of zero matches:
* probability of no matches after 1 card = 1 (you clearly can't have a "match" when only 1 card has been turned over)
* probability that the second card doesn't match the first card = 10/11
* probability that the third card doesn't match the first or the second card = 8/10 (it's best not to reduce this, since it will cancel the 10 that we already saw up there)
* probability that the fourth card doesn't match the first, third, or second card = 6/9 = 2/3

so, the probability of a "fail" here -- i.e., the probability that there are no matches anywhere -- is (10/11)(8/10)(2/3) = 16/33.

so, the probability that we actually want is 1 minus that, which is 17/33.

let's keep it simple, folks

[ritwik.rocks]

I used the following logic, which it appears from the answers is incorrect, but can someone point out the fallacy:

There are 6 possible pairs. Suppose Bill draws a pair. Then he can draw the other cards in 10C2 = 45 ways.
Multiplied with 6 possible pairs, this gives 45*6 = 270 successful ways.

Total ways possible = 12C4 = 495 ways.

Hence probability = 270/495 = 6/11

[tim]

You're overcounting. For each pair, you calculate 45 possibilities for the other cards. Well, what if my pair is 2/2? Then among those 45 possibilities is 1/1, 3/3, 4/4, etc. But let's say you had a hand with a pair of 2/2 and two "other" cards of 3/3; you would end up counting that hand again when your designated pair was 3/3 and one of the 45 possibilities of the other two cards ended up being 2/2. Does that make sense?

[rustom.hakimiyan]

Hi,

Following the approach outlined above of P(one) = 1 - P(none), I can follow the steps. 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 at least 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 at least 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]

You need to find the probability of getting exactly one pair. Then you add that to the probability of getting two pairs.

If it were possible to find "the probability of at least one" in a single calculation, then there would be no need to add the probability of getting two pairs, since "at least one" already includes "two".
However, there is no way to do "at least one" with a single direct calculation. (If there were, there would be little challenge in this problem!)

[rustom.hakimiyan]

You need to find the probability of getting exactly one pair. Then you add that to the probability of getting two pairs.

If it were possible to find "the probability of at least one" in a single calculation, then there would be no need to add the probability of getting two pairs, since "at least one" already includes "two".
However, there is no way to do "at least one" with a single direct calculation. (If there were, there would be little challenge in this problem!)

Hi Ron,

If you look at my post above yours, you'll see that what you're suggesting is exactly what i'm doing but i'm not coming up with the right answer?

Am I missing something from your comment? I'm adding the two instances as mentioned above.

Thanks

[qqixiaofan]

Hi,

I'm trying to figure out how to find the probability of getting exactly one pair. Here is my calculation. Please help to confirm if i miss anything:

(1) probability of chosing 1 pair out of 6: 6C1=6
(2) ways of chosing 2 number from the remaining 5 number: 5C2=10
(3) each number has 2 color possibilities: 2^2=4
(4) probability of chosing exactly 1pair: 6*10*4=240

You need to find the probability of getting exactly one pair. Then you add that to the probability of getting two pairs.

If it were possible to find "the probability of at least one" in a single calculation, then there would be no need to add the probability of getting two pairs, since "at least one" already includes "two".
However, there is no way to do "at least one" with a single direct calculation. (If there were, there would be little challenge in this problem!)

[RonPurewal]

That looks like a correct calculation for the number of ways in which exactly 1 pair can be selected"”but it's not a probability. (Probabilities are fractions between 0 and 1.)

Did you mean "number of ways"?

[qqixiaofan]

Yes number of ways. thanks

That looks like a correct calculation for the number of ways in which exactly 1 pair can be selected"”but it's not a probability. (Probabilities are fractions between 0 and 1.)

Did you mean "number of ways"?

[RonPurewal]

Ok.