Probability MGMAT

Mike@Magoosh wrote:

gmattesttaker2 wrote:Hi Mike,

Sorry to bother you but I am still not clear with Vodoo Child's Method 2 explanation:

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
I was trying to approach this as follows:
P(Event) = number of ways the event can occur/Total number of possible outcomes
13C1 is clear i.e. we are drawing 1 card out of 13 cards. However I am lost after that.
52C2 is clear i.e. 2 cards out of 52.
Can you please help explain this?

Sri ---
I am more than happy to help with this.
I'm not sure how familiar you are with nCr notation and the idea of combinations. Here's a blogpost that covers the basic.
https://magoosh.com/gmat/2012/gmat-permu ... binations/

In Voodoo Child's Method One, we first count the number of two-of-a-kind pairs, poker pairs, in the deck. The cards of a pair must be the same number/rank --- i.e. both 3's, or both Jack's. There are 13 numbers/ranks in the deck, so first we count that. BTW, writing that as 13C1 is, I would argue, overkill of the C notation.

Next, for the four cards of any given rank, how many different two card pairs can we create? This is a valid use of the nCr notation. We have 4 choose 2, which is

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

Again, if you are unfamiliar with the nCr formula, please read the blogpost at that link --- it will explain everything clearly.

So, the total number of pairs is the number of numbers/ranks (13) times the number of pairs within each rank (6). We'll just leave this as 13*6 --- there's absolutely no reason on Earth to multiply that out, because we know we are going to cancel factors in a fraction. Also, I know Voodoo Child wrote this as (13*6)C1, but that is just an insane use of C-notation --- Voodoo Child went a little C-happy in that write-up. The numerator of our fraction is just plain old 13*6.

Now, the denominator. We have 52 cards, and we want to choose 2, so that's 52C2, "52 choose 2". Here's the long way to calculate it.

52C2 = (52!)/[(2!)(52 - 2)!] = (52!)/[(2!)(50)!] = (52*51*50!)/(2*(50!)) = 52*51/2

The shortcut is to recognize that nC2 always equals n(n-1)/2. Anyway, that is our denominator.

Probability = (number of desired events)/(total number of events)
= (13*6)/[52*51/2] ----- dividing by a fraction means multiplying by its reciprocal

= (13*6)*(2/(52*51) ----- now, cancel the factor of 13 (because 52 = 4*13)

= (6*2)/(4*51) --- now, cancel a factor of 2

= (6)/(2*51) --- now, cancel another factor of 2

= (3)/(51) --- now, cancel a factor of 3

= 1/17

gmattesttaker2 wrote:Also, for Method 1 i.e. Permutations I am trying to understand as follows:

For First card, total choices are 52. So first card could be any of the 52 i.e.

52/52

So let us say, I draw King of Hearts. Hence, the second card should be either a King of Diamonds or
a King of Spades or a King of Clubs. So it has to be these 3 out of a possible 51 i.e.

3/51

Hence, number of ways I can get a pair = 52/52 * 3/51 = 1/17

Is this understanding correct?

Yes, that is 100% correct. I hope this is helpful. Let me know if you have any more questions.

Mike