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Big Problem...No Comibatorics Please

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by knight247 » Wed Oct 05, 2011 4:28 am
In the poker game Five-Card Draw, each player is dealt a hand consisting of 5 cards from a deck of 52 cards. Each card in the deck has a suit (clubs, hearts, diamonds, or spades) and a value (A, 2,..., 10,J,Q,K).

A. What is the probability that we are dealt a four-of-a-kind?

B. What is the probability that we are dealt full house? (A full house is a hand with both a three-of-a-kind and a two-of-a-kind.)

C. What is the probability that we are dealt two pairs? (Fifth is different)

D. What is the probability that we are dealt three of a kind? (Others are different)

E. What is the probability that we are dealt one pair? (Others are different)

F. What is the probability that we are dealt all five of different ranks?

G. What is the probability that we are dealt hands with every suit?


Don't have the OAs for this. Its pretty easy to solve using combinatorics but am looking to solve using only regular probability fundamentals. NO COMBINATORICS PLEASE!!!
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by Brent@GMATPrepNow » Wed Oct 05, 2011 5:30 am
knight247 wrote:It's pretty easy to solve using combinatorics but am looking to solve using only regular probability fundamentals. NO COMBINATORICS PLEASE!!!
The great thing about probability question on the GMAT is that you can often solve them by using two different approaches:
1) probability rules
2) counting techniques
Of course the big proviso here is "questions on the GMAT." Solving the above question using probability rules alone is very cumbersome (and not to be attempted if you want to retain your sanity :-))
The problem is that, if we don't use counting techniques, we need to account for all of the combinations when using our probability rules.

The easiest question here is A, but it's still a pain to solve without counting techniques.

A. What is the probability that we are dealt a four-of-a-kind?

P(4 of a kind)= P(first 4 cards are same and 5th card is different) + P(1st, 2nd, 3rd, 5th cards are same and 4th card is different) + P(1st, 2nd, 4th, 5th cards are same and 3rd card is different) + P(1st, 3rd, 4th, 5th cards are same and 2nd card is different) + P(2nd, 3rd, 4th, 5th cards are same and 1st card is different)

I'll just tackle the first part: P(first 4 cards are same and 5th card is different)
= (1)(3/51)(2/50)(1/49)(1)
= 6/[51x50x49]
= 1/[17x25x49]

The other four probabilities are the same as the above.
So, P(4 of a kind) = 5/[17x25x49] = 1/[17x5x49] = 1/4165

The other questions are a much bigger pain than that, so they're definitely not GMAT questions.

Aside: Card questions, in general, are not on the GMAT, since those kinds of questions would be culturally biased

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Brent
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by knight247 » Wed Oct 05, 2011 6:03 am
Hey Brent,
Thanks for ur response Brother. I managed to find an OA. Following is the link. Look at post#2 which has the answers solved using combinatorics. The answer to (A) which u solved is apparently 5 times more than the actual answer. Hoping u could look at it. Cheers

https://gmatclub.com/forum/let-s-play-poker-84956.html
Last edited by knight247 on Wed Oct 05, 2011 6:31 am, edited 1 time in total.
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by Brent@GMATPrepNow » Wed Oct 05, 2011 6:06 am
knight247 wrote:Hey Brent,
Thanks for ur response Brother. I managed to find an OA. Following is the link. Look at post#2 which has the answers solved using combinatorics. The answer to (A) which u solved is apparently 5 times more than the actual answer. Hoping u could look at it. Cheers
Where's the link?

Cheers,
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by GmatMathPro » Wed Oct 05, 2011 6:17 am
I'll take a shot at the full house one:

The hand should be of the form AABBB or some rearrangement of that.

Looking at this specific sequence:

1st card can be anything: P=1
2nd card matches first: P=3/51
3rd card is something not matching A: 48/50
4th card matches 3rd: 3/49
5th card matches 3rd: 2/48

Multiplying them all together= 3/20825

Now, how many ways can we rearrange AABBB?
Use the anagram method: 5!/2!3!

or, reason that once you decide where the A's go, the sequence is defined because the B's have to fill the remaining slots. 5 choices for where to put the first A, 4 choices for the second. Divide by two to eliminate repeats: 5*4/2=10

(3/20825)*10=6/4165
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by Brent@GMATPrepNow » Wed Oct 05, 2011 6:33 am
F. What is the probability that we are dealt all five of different ranks?

= (1)(48/51)(44/50)(40/49)(36/48)

Cheers,
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by Brent@GMATPrepNow » Wed Oct 05, 2011 6:37 am
knight247 wrote:Hey Brent,
Thanks for ur response Brother. I managed to find an OA. Following is the link. Look at post#2 which has the answers solved using combinatorics. The answer to (A) which u solved is apparently 5 times more than the actual answer. Hoping u could look at it. Cheers

https://gmatclub.com/forum/let-s-play-poker-84956.html
My solution of 1/4165 matches the solution on that link.

Cheers,
Brent
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by knight247 » Wed Oct 05, 2011 7:12 am
Yes ur right Brent. I was looking at the wrong thing. Sorry about that.
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