MGMAT Combinatorics and Probability Drill Q#27

[isaMD]

Hello, I am having trouble withe the explanation of question #27 in the Combinatorics and Probability Drill.

A standard 52-card deck of cards has 4 suits (spades, hearts, clubs, and
diamonds). Each suit has 13 cards (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King).
What is the probability that the two cards you are dealt are a pair (the same number or the
same rank of face card)?
A) 1/26
B) 1/17
C) 1/13
D) 4/51
E) 1/4

I get the part of how many ways you can choose 2 cards (denominator), but I don't get the part where you have to divide 52x3 into 2 (when choosing a matching card).

I will really appreciate if you could explain to me this part.

Thanks!

[RonPurewal]

hi,

I get the part of how many ways you can choose 2 cards (denominator), but I don't get the part where you have to divide 52x3 into 2 (when choosing a matching card).

the red thing looks like a reference to an answer key.

if that's what the red thing is, then please post the contents of that answer key. thanks. (otherwise there's no way to tell what you are asking about!)

[RonPurewal]

in any case, this problem is not a great candidate for using combinations. it's much easier (and much more intuitive) to think about probability directly.

* you get your first card... okay, you have a card.
* now you get your second card. at this point, the only thing under consideration is whether your second card is equal in value to your first card.
no matter what your first card is, there are exactly 3 cards left in the deck with the same value, out of 51 remaining cards.
since this probability is always the same, regardless of your first card, the probability is 3/51 = 1/17.

[isaMD]

Thanks a lot Ron! This probability approach is a lot easier to understand than the combinatorics one and, as you said, much more intuitive. Now I get how to answer this question!

Muchas gracias!!

[RonPurewal]

you're welcome.

[RonPurewal]

MORE IMPORTANTLY—of note here:

the official problems will NEVER require the use of both probability concepts and combinatorial concepts.
…never ever ever.

if you look through the entire canon of GMAC’s combinatorics and probability questions, you’ll find that NONE of them— not a single one— uses both.

• if GMAC gives you a probability question, you may need to manipulate probabilities (e.g., multiplying consecutive probabilities). alternatively, you may just be able to count things. it will NEVER be necessary, though, to use combinatorics to count things.

• if GMAC gives you a combinatorics question, you will NEVER have to use probability concepts on top of it.
(also, as i’ve remarked quite often on this forum, you’ll seldom have to use fancy combinatorial things. as long as you just “pick up the shovel and dig”— i.e., get to work quickly, without self-defeating thoughts— you can actually solve MOST of GMAC’s combinatorics problems by merely making lists and counting things.)

[RonPurewal]

^^ in case you're wondering, GMAC has not actually declared anything to this effect; our evidence is all empirical.

however, for this particular claim ("you'll never have to use probability concepts and combinatorial concepts together"), the empirical evidence is VERY strong.
specifically, since these two concepts are very closely related, it would be very, very easy to write a problem that combines them. yet, of the many probability/combinatorics problems in the GMAC canon, not one requires the use of both sets of concepts.
this is a very, very, very conspicuous omission.
(imagine if there were 20 problems with roots and another 20 problems with exponents, but no problems with BOTH roots AND exponents... you see where i'm going with this. very, very noticeable.)

so, i can feel secure coming on here and saying "you'll never ever ever have to do this."

(this doesn't mean that you CAN'T find solutions that use both sets of concepts; i mean only that the problems will not REQUIRE such a thing.)