dhonu121 wrote:Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, 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

OA:17/33

The current solution of this problem uses the 1-x rule, where x is the probability that no pair is revealed.
My confusion is: While calculating x,we apply the following expression:
(12*10*8*6)/(12*11*10*9)..i.e favourable cases/total cases.
In calculating total cases why didnt we use 12C4 rather than using 12P4.
I am not able to logically get this thing.

Can anybody help ?
Thanks.

If you want to solve the question using counting techniques, then (as you mentioned), we should first recognize that P(at least one pair) = 1-P(no pairs)

P(no pairs)
First, the number of possible outcomes:
We have 12 cards and we select 4.
This is accomplished in 12C4 ways = 495 (this is our denominator)

Now we count the number of ways to select 4 different values with no pairs. In other words, we want 4 different card values. In how many ways can we do this?
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways (of course, we haven't considered different colors yet)
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2-16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

So, the probability that there are no pairs = 240/(11x5x9) = 48/99 = 16/33

So, P(at least one pair) = 1- 16/33 =17/33

Cheers,
Brent

dhonu121 wrote:Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, 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

We can also solve this question using probability rules.

First, recognize that P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33
= B

Cheers,
Brent

dhonu121 wrote:

Now we count the number of ways to select 4 different values with no pairs. In other words, we want 4 different card values. In how many ways can we do this?
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways (of course, we haven't considered different colors yet)
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2-16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

Hi Brent,
I couldn't follow through the above explanation. To calculate P,
Why didn't we simply select 1st card by 12C1 ways and then second card by 10C1 ways and 3rd card by 8C1 ways and 4th card by 6C1 ways.

Thus P=12C1*10C1*8C1*6C1/12C4
I know the above method is yielding wrong answer, but I am not able to find what am I doing wrong.

The green portion of your solution is treating the order as mattering. For example, it suggests that the selection 1,2,3,4 (same suit) is different from the selection 4,3,2,1 (same suit).

However, the red portion of your solution is treating the order as not mattering.

This mismatch is causing an incorrect answer.

If you change the red portion so that order also matters, we get 12*11*10*9
So, we get P=12C1*10C1*8C1*6C1/12*11*10*9 = 16/33

From here, P(at least one pair) = 1 - 16/33 = 17/33 = B

Cheers,
Brent

coolhabhi wrote:

dhonu121 wrote:

Now we count the number of ways to select 4 different values with no pairs. In other words, we want 4 different card values. In how many ways can we do this?
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways (of course, we haven't considered different colors yet)
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2-16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

Hi Brent,
I couldn't follow through the above explanation. To calculate P,
Why didn't we simply select 1st card by 12C1 ways and then second card by 10C1 ways and 3rd card by 8C1 ways and 4th card by 6C1 ways.

Thus P=12C1*10C1*8C1*6C1/12C4
I know the above method is yielding wrong answer, but I am not able to find what am I doing wrong.

When you are doing P=12C1*10C1*8C1*6C1/12C4, you should also consider the fact that you can choose any 1 card from the pair of 2 same numbered cards.

so it should be P=12C1*2*10C1*2*8C1*2*6C1*2/12C4
Now this will give you your answer which is 16/33

so P(at least one pair) = 1 - 16/33
= 17/33
[spoiler]Answer : B[/spoiler]

Unfortunately, the above solution (in blue) doesn't work. In fact, it yields a probability greater than 1.

Cheers,
Brent

dhonu121 wrote: Hi Brent,
Mathematically the green portion of my solution above is counting order and the red portion is not.
Agreed. Also, this is yielding probability greater than 1. So, its defintely wrong.

Your solution which says that if I want to do this without counting, I would need to select 4 cards from any pair = 6C4 and then each selected card could be selected in 2 ways. Hence 6C4*2^4 also makes a lot of sense.
However, I am logically not able to differentiate 6C4*2^4 from 12C1*10C1*8C1*6C1.

What is that I am doing wrong.
How is that in one case(yours) order is not getting counted and in my case, order is getting counted ?
Can you help me out here ?

In your solution (12C1*10C1*8C1*6C1), you are taking the task of selecting 4 cards, and you're breaking it into 4 stages:
Stage 1: Select any card. This can be accomplished in 12 ways.
Stage 2: Select any card such that this card does not have the same value as the first card. Well, there are 11 cards remaining and 1 of them matches the first card. So, there are only 10 cards that meet the given criteria.
Stage 3: Similar logic gives us 8 ways to accomplish this stage.
Stage 4: Similar logic gives us 6 ways to accomplish this stage.

Aside: You may not have actually stated these stages in your approach, but your calculation imply them.

Now, for the purposes of this discussion, let's say that the 2 suits are spades and hearts.

Big point: In your approach, it's possible to get the following as one complete selection:
Stage 1: Select a 5 of spades
Stage 2: Select a 3 of spades
Stage 3: Select a 6 of spades
Stage 4: Select a 1 of spades
This would count as one selection.

In your approach, it's possible to get this as a complete selection:
Stage 1: Select a 1 of spades
Stage 2: Select a 6 of spades
Stage 3: Select a 3 of spades
Stage 4: Select a 5 of spades
This would count as one selection AND it would count as a select different from the first selection.

Remember that these calculations are helping us determine the numerator in our probability. So, if we're going to treat these outcomes as different, then then we need to treat the outcomes in our denominator as different as well.

That is, in the denominator (counting all possible outcomes) we need to treat the selection 1, 3, 5, 6 as being different from the selection 3, 5, 6, 1. If you do that, you will arrive at the correct answer.

Cheers,
Brent

coolhabhi wrote:

Brent@GMATPrepNow wrote:

coolhabhi wrote:

dhonu121 wrote:

Now we count the number of ways to select 4 different values with no pairs. In other words, we want 4 different card values. In how many ways can we do this?
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways (of course, we haven't considered different colors yet)
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2-16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

Hi Brent,
I couldn't follow through the above explanation. To calculate P,
Why didn't we simply select 1st card by 12C1 ways and then second card by 10C1 ways and 3rd card by 8C1 ways and 4th card by 6C1 ways.

Thus P=12C1*10C1*8C1*6C1/12C4
I know the above method is yielding wrong answer, but I am not able to find what am I doing wrong.

When you are doing P=12C1*10C1*8C1*6C1/12C4, you should also consider the fact that you can choose any 1 card from the pair of 2 same numbered cards.

so it should be P=12C1*2*10C1*2*8C1*2*6C1*2/12C4
Now this will give you your answer which is 16/33

so P(at least one pair) = 1 - 16/33
= 17/33
[spoiler]Answer : B[/spoiler]

Unfortunately, the above solution (in blue) doesn't work. In fact, it yields a probability greater than 1.

Cheers,
Brent

Sorry, that was a Typo error. Actually here is how I did it. Suppose we have 6 sets of numbers. For example:
(a,A)
(b,B)
(c,C)
(d,D)
(e,E)
(f,F)

Now the probability of at least one pair of cards that have the same value = 1 - P(No pair chosen)

so P(No pair chosen) = (6*2/12)*(5*2/11)*(4*2/10)*(3*2/9)

(6*2/12) means any one of the 6 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 12.

(5*2/11) means any one of the 5 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 11.

(4*2/10) means any one of the 4 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 10.

(3*2/9) means any one of the 3 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 9.

P(No pair chosen) = 16/33

So P(at least one pair of cards that have the same value) = 1 - 16/33 = 17/33.

Perfect! Notice that your solution is pretty much the same as the approached I used in my solution using probability rules:
--------------------------------------------------
P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33
= B
--------------------------------------------------

If we break the above solution down at each step, we can see that our solutions are basically the same.

P(select any 1st card):
- I noticed that we can choose any card here, so the probability = 1
- You say: (6*2/12) means any one of the 6 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 12.
- both probabilities = 1

P(select any non-matching card 2nd):
- Me: At this point, there are 11 cards remaining. 1 of them matches the first card. So, the probability is 10/11 that the 2nd card doesn't match the first card.
- You: (5*2/11) means any one of the 5 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 11.
- both of our probabilities = 10/11

P(select any non-matching card 3rd):
- Me: At this point, there are 10 cards remaining and 2 of them are matches for the first and second cards. So, the probability is 8/10 that the 3rd card doesn't match the first or second cards.
- You: (4*2/10) means any one of the 4 pairs can be chosen and 2 because any one number can be chosen and the total number of cards are 10.
- both of our probabilities = 8/10

etc.

Cheers,
Brent