Problem Solving

Jim@StratusPrep wrote:your logic is wrong on the last two. Even though only 3 cards satisfy the pair you are not choosing from 3 cards you are choosing from 51. Even your logic in number 2 is wrong (the denominator should be 52*51/2) - the math doesn't work out to 1/17.

Stick with the first way!

voodoo_child wrote:You have to choose 2 cards from a standard 52 deck card. What is the probability that you find a pair? (pair = same number or rank)

a) 1/26
b) 1/17
c) 1/13
d) 4/51
e) 1/4

OA -B

I was able to arrive at the correct answer using two methods. However, initially I got stuck with Method3 and Method4. Can someone please guide me esp. Method3 and Method4?



Method 1 : Using permutations:

First card => 52/52
Second card => 3/51

Probability = 3/51= 1/17

Method 2 : Combinations:

There are 13C1 * 4C2 pairs = 13*6 pairs.
Total number of combinations to choose two cards = 52C2

PRobability = (13*6)C1/52C2 = 13*6*2/52*51 = 1/17

Method 3: (This is where I need help)

Let's think of 13X4 cards as matrix.

For instance,

[HEART SPADES DIAMOND CLUBS]
[ 1 1 1 1 ]
[ 2 2 2 2 ]
[ 3 3 3 3 ]
AND SO ON........


One can choose any one card out of 13 possible unique numbers/ranks (i.e. rows in the matrix). (I am ignoring difference in face card for now) => 13C1
Now, corresponding to each of these numbers, there are three other same number/rank card but with a different face. (i.e. column in the matrix) => second card can be chosen in 3C1 ways

Therefore, total number of combinations = 13*3/52C2 = 13*6/52*51 = 1/34. Incorrect.

Method 4:
I thought to myself : May be there is an issue with choosing *the first* card (i.e. considering all elements in the matrix). What if I choose the first card in : 52C1 number of ways
The second card can be chosen in only : 3C1 ways because there are only three other cards left with the same rank/number but different face.

Probability = 52 * 3 / 52C2 = 3*6/51 = 6/17...What????

I am stuck in black hole esp. with Method 3 and Method 4.

Can anyone please help me?

Thanks
Voodoo

gmat_and_me wrote:Your first solution is very nice. I can not make sense of
2 and 3.

Mike@Magoosh wrote:Hello voodoo_child,



On Method Three ---
What's bizarre about this --- you start with, "One can choose any one card out of 13 possible unique numbers/ranks" --- but picking just the row (i.e. the number/rank) is not the same as picking an individual card. 13*3 = 39 is not particularly the number of anything. ----- This would be the correct approach if the question were something like "What's the probability that, on a two-card draw, you pick a two-of-a-kind pair in which one of the two cards is a heart?" That way, just the choice of the suit would be enough to determine a unique card, the heart.
If for each of the 13 rows, you pick the numbers of two-of-a-kind pairs you can find on that row, 4C2 = 6, then we are back to something that looks very much like Method Two.

Mike@Magoosh wrote:

dhonu121 wrote:Hi Mike,
If that is the definition of poker pair,
the what is wrong with the following approach to calculate the desired probability
Number of ways to draw two cards of same pair wthout replcement = 13C1*4C2.
Total number of ways = 52C2.
Prob = (13C1*4C2)/52C2

Dear dhonu121:
Nothing is wrong with that. That gives the correct answer of 1/17, and in fact, it's identical to voodoo child's "Method One" in the top post in this thread. Just so you see how the calculation simplifies ---

13C1 = 13

4C2 = (4*3*2*1)/[(2*1)(2*1)] = (4*3)/(2*1) = 6

52C2 = (52*51)/2

(13C1*4C2)/52C2 = (13)(6)/[(52*51)/2] = (2)(13)(6)/(52*51) = (2)(6)/(4*51)

= (6)/(2*51) = 3/51 = 1/17

Does all that make sense? Let me know if you have any further questions.

Mike