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?

(A) \(\frac{8}{33}\)

(B) \(\frac{62}{65}\)

(C) \(\frac{17}{33}\)

(D) \(\frac{103}{165}\)

(E) \(\frac{25}{33}\)

I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?

First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4

Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4

It looks like I have done a permutation on the numerator. Could you help me with the error in the above?

Thanks Zarrolou

I understood the solution, but I would need a little more help on some concepts.

PERMUTATIONS OF 'n' items having some identical elements:
Number of permutations of 'n' items out of which p1 are alike, p2 are alike, p3 are alike and the rest are different is n!/p1!p2!p3!

Suppose, we have a set of 8 alphabets - {A, A, B, B, C, C, D, D}

The number of ways we can arrange the 8 alphabets is 8!/2!*2!*2!*2! (where we divide by 2! because we cannot distinguish between AA and its combination reversed, AA )

So, just the way we have the above formula, what is the formula to get all the possible combinations of 3 elements in the above case.

Will it be? - 4c3*(2)*(2)*(2)/8c4

where 4c3 is choosing combination of 3 elements among 4 unique elements.
and (2)*(2)*(2) is for each of the 3 elements in the combination there are two options (two identical elements)

Is there a general formula used to determine combinations in cases such as above?

The other doubt that I have is in the above set of alphabets {A, A, B, B, C, C, D, D}

how do I get the number of permutations of 3 elements? Is it 3!/{what?} because I don't know how many elements are repeated among those 3.

\frac{17}{33} is the answer.
You cant also do it using combinations.
select any pair(i.e. 2 cards) and the other two + select any two pairs and divide the total by 12C4.
this thought process leads to :-

(6C1*8*5 + 6C2)/12C4

for the first term in the Numerator - first select any one pair(6C1 ways), the we are left with 5 of black and 5 of red. Now select any one from one of these groups (5 ways). Lastly we are left with 9 cards, but in this group, one of them is the one with the same value in another color of the previously selected. So we are left with 8 to select one from(8 ways). Multiplying together -> 6C1*8*5.

Here is my calculation:

Answer= 1- ( number of ways to select 4 different cards/ total number of ways to select 4 cards)


( number of ways to select 4 different cards/ total number of ways to select 4 cards)=

denominator= 12C4 so I haven't got any problem with calculating denominator

numerator= 12*10*8*6 Obviously I have a problem in calculating the numerator. My logic is that for the first card we can select one of the 12 cards. Then there are 10 cards left because we have already selected one of them and have discarded the pair of what we have selected previously. So there are 10 different cards for our second selection. Regarding previous way, we have 8 cards left to select for our third selection. Finally, we have 6 cards left to select four our fourth selection. Therefore total number of ways to select 4 different cards is 12*10*8*6.

I don't know what is wrong with my calculations but numerator should not be 12*10*8*6. Can anybody help please ?

Probability Method:

Theory: Prob of at least 1 pair + Prob of 2 pairs

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{9}{9}\) * \(\frac{4!}{2!2!}\)

+

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{1}{9}\) * \(\frac{4!}{2!2!}\) = 20/33.

Can you explain why r we multiplying by 4C2 ?

I get that we have 4 outcomes to be arranged. What is it divided by in the form of 2! * 2! ?

Hi Guys:

Had a doubt with the following question:

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?

Correct Answer: 17/33

I used the following way:

No. of ways of selecting one pair (AAXY)= 6C1 * 10C2

No. of arrangements= 12 * 6C1 * 10C2


No. of ways of selecting two pairs (AABB)= 6C2

No. of arrangements= 6C2 * 6

Total number of possible favorable arrangements= (6C1 *10C2 *12) + (6C2 * 6)

Total number of arrangements= 12C4

Hence probability = { (6C1 *10C2 *12) + (6C2 * 6) } / 12C4

= 37/132

Can someone please explain me the flaw in this.

Thanks and Regards

This is not correct. 6C1 * 10C2 includes the cases in which you get both pairs. Say you select the 3 black - 3 red pair from 6C1. Then when you select 2 of the remaining 10 cards, you could get both 4s. In that case, you will have two pairs.
Also, no arrangement is necessary. You just have to select the 4 cards.

Check out the correct solution at the link given above.
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Hi there, when you multiply 10*12*8*6 you are also including the arrangements of the four cards. The number of arrangements for 'each' possible selection is 4!. Since in this question we only need to select cards, and not arrange them, we divide the total number of arrangements by 4! so that only the selections remain.

Hope that helps.

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My Solution:

We can solve it by find the opposite probability when bill finds no pair of cards that have the same value:

P = 1 -(none)

We have total 12 cards, then we have

12/12 * 10/11* 8/10* 6/9 = 16/33

Now subtract this from 1 we get

P = 1-16/33 = 17/33 Answer is Option C
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