Problem Solving

Question - 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

My approach :

We have four cards that are drawn. Let's call them A B C D

Now we can only have two pairs at max because four cards are drawn.

Option (i) Two cards = 1 pair : => A and B are same. C and D are different
The first card can be drawn in 12 ways.
The second card can be drawn in only 1 way because we need a pair and I am assuming that A = B
The third and the fourth can be drawn in 10, 9 ways respectively.

Therefore, total number of combinations for 1 pair = 4C2 *12*1*10*9 {4C2 is used because A and B can be permuted}.......(U)

Option (ii) 2 pairs : A and B are same. C and D are same too.... But, A != C
Using the same logic as above A B C D = 12*1*10*1 * 4C2 .............(V)

Total number of combinations = 12*11*10*9 ................(W)

Therefore, probability = (U+V)/W = 20/33

Something is wrong here. Any help is greatly appreciated.....


Thanks
Voodoo

IMO C

Because the question has the word "at least" its always easy to find the probability of what is not required and subtract it by 1.

P(getting all diff cards) = 1*(10/11)*(8/10)*(6/9)= 16/33

[1--> probability of selecting any 1 card is always 1,For second card we cannot select the pair card of 1st selection,so we are only left with 10 cards to select out of 11 cards..similarly for 3rd and 4th selection]

P(getting at least one pair card) = 1-(16/33) = 17/33

@smackmartine : ok...but what's wrong with my method ?

voodoo_child wrote:The third and the fourth can be drawn in 10, 9 ways respectively.

Therefore, total number of combinations for 1 pair = 4C2 *12*1*10*9 {4C2 is used because A and B can be permuted}.......(U)

I think you did a mistake here :
Option (i) Two cards = 1 pair :
4C2 *12*1*10*9 ,because you should not select 4th card that is same as third card (So you have one fewer option).
It should be 4C2 *12*1*10*8.

I did not try to solve through your method, but this is what I saw as an error

Can somebody please explain this question in a more elaborate way.....

Can any of the experts please help in this question??

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

Hi,
Smackmartine has already explained this. I will just elaborate this.
First we will the number of ways of picking 4 cards with distinct values.
Number of ways of picking 1st card is 12
2nd card with value distinct from 1st can be picked from 10(12 minus 1st card and card with same value as 1st) in 10 ways
Similarly, 3rd card in 8 ways
and 4th card in 6 ways.
So, number of ways is 12.10.8.6
Total number of ways of picking 4 cards one after the other i 12.11.10.9
So, probability of picking 4 cards with distinct values is 12.10.8.6/12.11.10.9 = 16/33
So, probability of getting at least 1 pair = 1-(probability of getting 4 distinct values) = 1-16/33 = 17/33

voodoo_child wrote:@smackmartine : ok...but what's wrong with my method ?

Hi,
As Smackmartine has rightly mentioned,
Option (i) Two cards = 1 pair : => A and B are same. C and D are different
The first card can be drawn in 12 ways.
The second card can be drawn in only 1 way because we need a pair and I am assuming that A = B
The third and the fourth can be drawn in 10, 8 ways respectively.
After picking your first card 'A', the remaining 3 cards A,C,D can be arranged in any order in 3! ways.
So, Number of ways is 12*10*8*3! ---(U)
I don't understand why you have used 4C2.

Option (ii) 2 pairs : A and B are same. C and D are same too....
After picking A, the remaining 3 cards(A,C,C) can be arranged in 3!/2! ways
Using the same logic as above A B C D = 12*1*10*1 * (3!/2!) .............(V)

Total number of combinations = 12*11*10*9 ................(W)

Therefore, probability = (U+V)/W = 12.10.(48+3)/12.11.10.9 = 51/99 = 17/33

I am not an expert, but I will try to explain how to solve this problem.

The basic probability formula for probability is Total favorable events/Total number of events.

Also P(event A occurred)+ P(event A did not occur) = 1 , where 1-P(event A did not occur) is composite probability of P(event A occurred). example chances that it will rain is 3/4, so probability that it will not rain is (1 - 3/4) =1/4

In the problem above, Total number of cards is 12 and only 2 suits of 6 cards each is present. The question asks to find the chances(probability) of getting at least one pair when 4 cards are drawn at random without replacement(you cannot put the card back to deck)

At least one pair means either only one pair occurs :
11XY,22XY,33XY,44XY,55XY,66XY (one pair out of 4 selection) or

OR Two pairs can occur :
1122,3344,5566,3355 ....(two pairs out of 4 selection).

Whenever you see words "at least" try to think in terms of the composite probability.

P(At least 1 Pair is selected) + P(No Pair is selected) = 1 (see the formula above)

Now calculating P(No Pair is selected) is tricky, because when you select one card you want all the other cards of different number. ex - you choose 3 first,then in next 3 selections you cannot choose 3 because you want no pair. the same rule applies to all the other selections too. Now

P(No Pair is selected) = (Probability of 1st selection)*(Probability of 2nd selection,not same as 1st selection)*(Probability of 3rd selection,not same as 1st and 2nd selection)*(Probability of 4th selection,not same as 1st, 2nd ,and 3rd selection)

(Probability of 1st selection)= 12/12 =1 (you can select any one out of 12)

(Probability of 2nd selection,not same as 1st selection) = (10/11) [out of 11 options you have only 10 options because 1 option will be same as 1st selection,and you don't want that]

(Probability of 3rd selection,not same as 1st and 2nd selection) = (8/10) [out of 10 options you have only 8 options because 2 option will be same as 1st and 2nd selection]

(Probability of 4th selection,not same as 1st, 2nd ,and 3rd selection) = (6/9) [out of 9 options you have only 6 options because 3 option will be same as 1st ,2nd and 3rd selection]

Finally, P(No Pair is selected) = 1*(10/11)*(8/10)*(6/9)= 16/33

Also, P(At least 1 Pair is selected) + P(No Pair is selected) = 1 (described above)

P(At least 1 Pair is selected)= 1- P(No Pair is selected)
= 1- (16/33)
= 17/33

Hope its clear now.

[email protected] wrote:Can any of the experts please help in this question??

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

hi.. i'm not an expert but would love to share my thought on this..!!!

out of 12 playing cards any four cards can be selected in 12C4 ways.

let suite 1 be S1: 1,2,3,4,5,6.
let suite 2 be S2: 1,2,3,4,5,6.

now we have a maximum of six possible pairs, (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),. out of these six pairs any one pair can be selected in 6C1 ways,

from remaining 10 cards, any two cards can be selected in 10C2 ways,
important thing to note here is that question asks us to find at least one pair of cards that have the same value, [i.e. 1,1 2,3 is a acceptable solution and also 1,1,2,2.]

now in 6C1*10C2 we have counted some pair twice that we need to subtract, these are,
1,1,2,2 which is same as 2,2,1,1
1,1,3,3 which is same as 3,3,1,1
1,1,4,4 which is same as 4,4,1,1
1,1,5,5 which is same as 5,5,1,1
1,1,6,6 which is same as 6,6,1,1
2,2,3,3 which is same as 2,2,3,3
2,2,4,4 which is same as 2,2,4,4
2,2,5,5 which is same as 2,2,5,5
2,3,6,6 which is same as 2,2,6,6
3,3,4,4 which is same as 4,4,3,3
3,3,5,5 which is same as 5,5,3,3
3,3,6,6 which is same as 6,6,3,3
4,4,5,5 which is same as 5,5,4,4
4,4,6,6 which is same as 6,6,4,4
5,5,6,6 which is same as 6,6,5,5

which are 15 in number
hence required probability is 6C1*10C2-15/12C4;
=17/33 hence C

Frankenstein wrote:

voodoo_child wrote:@smackmartine : ok...but what's wrong with my method ?

Hi,
As Smackmartine has rightly mentioned,
Option (i) Two cards = 1 pair : => A and B are same. C and D are different
The first card can be drawn in 12 ways.
The second card can be drawn in only 1 way because we need a pair and I am assuming that A = B
The third and the fourth can be drawn in 10, 8 ways respectively.
After picking your first card 'A', the remaining 3 cards A,C,D can be arranged in any order in 3! ways.
So, Number of ways is 12*10*8*3! ---(U)
I don't understand why you have used 4C2.

Option (ii) 2 pairs : A and B are same. C and D are same too....
After picking A, the remaining 3 cards(A,C,C) can be arranged in 3!/2! ways
Using the same logic as above A B C D = 12*1*10*1 * (3!/2!) .............(V)

Total number of combinations = 12*11*10*9 ................(W)

Therefore, probability = (U+V)/W = 12.10.(48+3)/12.11.10.9 = 51/99 = 17/33

@ Frankenstein - Thanks for responding.....

about 4C2 in case of (i) ...Let's see why...

Answer : Let's say we have A A B C cards. Then we can choose 2 out of 4 cards (order is not important) in 4C2 ways = 6 ways. Both of us concur on this.

Now my question is about A A B B. I think we can have 6 combinations for it and not 3.

Let's see : AABB ABAB ABBA BAAB BBAA BABA = 6 combinations. I am not sure why you are saying 3?

Thanks
Voodoo

voodoo_child wrote: @ Frankenstein - Thanks for responding.....

about 4C2 in case of (i) ...Let's see why...

Answer : Let's say we have A A B C cards. Then we can choose 2 out of 4 cards (order is not important) in 4C2 ways = 6 ways. Both of us concur on this.

Sorry, I still don't understand what you mean picking 2 out of 4(order not important).I will address this issue at the end.

voodoo_child wrote: Now my question is about A A B B. I think we can have 6 combinations for it and not 3.

Let's see : AABB ABAB ABBA BAAB BBAA BABA = 6 combinations. I am not sure why you are saying 3?

Voodoo

Okay..Let me explain this.
Consider your case of AABB : Let A=1,B=2
So, 1122, 1212, 1221, 2112, 2211, 2121
Let B=1,A=2
So, 2211, 2121, 2112, 1221, 1122, 1212.
So, you are counting every set twice. So, you should divide this by 2 at the end.
In my procedure, I have fixed A and arranged the other 3 numbers A,B,B in ABB,BAB,BBA
So, if A=1, B=2 the arrangements will be 1122, 1212, 1221
if A=2, B=1 the arrangements will be 2211,2121,2112.
None of the numbers will repeat in this way.

Using the same logic, I can tell you that you have intentionally/unintentionally used 4C2 in first case.
Consider AABC - Number of arrangements is 4!/2!. agreed?
Just because you count every set twice you will have to divide by 2 at the end. which is what gave you 4!/2!.2! = 4C2.

Consider my method: I have fixed A, so now I arrange the other 3(A,B,C) in 3!ways and none of the sets will be repeated.

awesome reply by smackmartine, great explanation brother!!