##### Quiz 3, First B Level Question
patrickHals
So I'm trying to find the set that corresponds to the number of ways that you can get at least 2 cards of the same value when drawing 3 cards from a 52 card deck. I thought that you would have 2 disjoint partitions: 1. {value of card, suits of 2 cards the same, value of other card that is not the same as first two, suit of other card} 2. {value of card, suits of all three cards} Would the addition of these two sets result in the number of ways you can draw at least 2 cards the same when drawing 3 cards?
Alan_Mendoza: Feb. 5, 2015, 11:08 p.m.
I don't think the suit of the 3rd card in the first partition would matter.
patrickHals: Feb. 5, 2015, 11:19 p.m.
But then the choice of the third card would be 50, choose 1, but that would mean you aren't choosing the third card at the same time you choose your other cards so you would have to do some sort of deordering right?
LauraHYu: Feb. 6, 2015, 10:55 a.m.
For the first case, when two of the values are the same and the last is not, (value of card, taking two out of four of the same value, picking a card that has any value other than the first) I didn't include suit for the third and different card because I thought it was irrelevant. But with regard to the initial question about adding what you get for the two cases, that is correct.
rachelobrien: Feb. 6, 2015, 5:21 p.m.
so for the partition of getting two cards of the same value would it be 13x4x3x48?
weisbart: Feb. 7, 2015, 11:25 a.m.
In patrickHals answer, the last two generations pick first the value of the remaining card and then the suit of the remaining card. This is the same as picking the remaining card since it is specified by the value and the suit. Does this make sense?  A quick note on notation though. In patrickHals setup, we have generations of choices. So, for example, the first set he describes would roughly look like $\{(\cdot,\cdot,\cdot,\cdot)\colon {\rm conditions}\}.$ It is a set of four tuples. It's pretty clear that this is what was meant in the writeup, but I do want to make sure this is understood so you know that there is a product structure.  For rachelobrien's post: You can choose as your last generation of choice the final card of a different value. This 48 that you get is patrickHals' $12\times 4$, 12 values and four suits for the value. Your first generation is, of course, correct since there are 13 values. But, what about your $4\times 3$? Are you sure about this?
AZobi: Feb. 7, 2015, 4:23 p.m.
For two being the same is it: (13)(4 choose 2) (12)(4 choose 1)? And for all three of them being the same, is it (13)(4 choose 3)? And therefore the answer is simply the sum of those two partitions over (52 choose 3)?
PMcelhinney: Feb. 7, 2015, 6 p.m.
To AZobi: I also got that answer except you can also write (48 choose 1) instead of (12)(4 choose 1) because there are 48 other cards to choose from and you need to choose one of them. Regardless both notations equal 48.
Alan_Mendoza: Feb. 7, 2015, 8:09 p.m.
AZobi, I also got that.
dana: Feb. 7, 2015, 8:26 p.m.
I got something a little different. For exactly 2 cards of the same value I got (13 choose 1)*(3 choose 1)*(48 choose 1) For exactly 3 cards of the same value I got (13 choose 1)*(3 choose 1)*(2 choose 1) With addition between the 2 partitions and a denominator of (52 choose 3).
Dylan: Feb. 7, 2015, 9:47 p.m.
I agree with Dana. What do you guys think?
toan123: Feb. 7, 2015, 11:12 p.m.
I got something a little different as well. Goal: Find P(getting a pair in a hand 3-card hand) Method: - Equation: $\frac{pair-from-3card-hand}{Total Outcomes}$ - Step1: figure out the total outcomes #(Total Outcomes) = (52 choose 3) ; because we are choosing 3 cards from a deck of 52 - Step2: set up generations for choices Let #(A) = pair-from-3card-hand Gen01: pair of two different values from a 13 types of cards in the deck, which would be (13 choose 2). The "2" in this aspect refers to the card types (Ace to King). So let's say I choose the 10-set and 3-set, I would get 2 sets out of the 13 types of cards) Gen02: choice of the suits for the first different card so...... (4 choose 2) [this is the 10-set] Gen03: Now, I will do the same thing for the 3-set, which is (4 choose 2) Gen04: total number of cards left, which is 44 (because we already drew 8 cards out from generations #1-3) Overall: - numerator (13 choose 2)x(4 choose 2)x(4 choose 2)x44 - denominator: (52 choose 3) I hope I did this right.
toan123: Feb. 7, 2015, 11:14 p.m.
Error Detection: I misread and thought I was supposed to choose 2 different ones. My bad. Going to redo.
iraianne: Feb. 8, 2015, 12:05 a.m.
Im not sure if I did this right but I got [(4 choose 2)*(48 choose 1)] + (4 choose 3) / (53 choose 3) The way I thought of this was that the problem asked for the same value not suit so in my head Im thinking, "well a number only repeats 4 times in a deck of cards so thats the only it would have the same value." Next the problem said at least 2 cards will be the same, hence, (4 choose 2) then you need to pick your last card and all you have remaining is 48 cards (52-4) so (48 choose 1). The next part of the numerator I said was (4 choose 3) which would satisfy having all three cards being the same value. And finally the denominator was just the total outcomes aka 52 cards you only need 3 so (52 choose 3) Did I do this right?
Alan_Mendoza: Feb. 8, 2015, 12:12 a.m.
dana, why are you only choosing from 3 suits?
Alan_Mendoza: Feb. 8, 2015, 12:15 a.m.
My logic was: (value to be doubled, suit for doubled value, other card) + (value to be tripled, suit for tripled value) all over 52 choose 3.
weisbart: Feb. 8, 2015, 10:05 a.m.
I like Azobi's answer as well. The point is that the $4\times 3$ indicates that there is a first choice of suit and a second choice of suit, where $4\choose 2$ eliminates this ordering.  iraianne, don't you also have to get the values of the repeated cards?
AZobi: Feb. 8, 2015, 3:07 p.m.
After discussing it with rrakha and marmat, another way we solved this was as follows: 1-[(13 choose 3)(4 choose 1)^3 / (52 choose 3)]
AZobi: Feb. 8, 2015, 3:07 p.m.
*marmat1