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September 29th, 2014, 08:52 AM  #1 
Newbie Joined: Sep 2014 From: Colorado Posts: 1 Thanks: 0  Probability of drawing a set of cards given several different drawing parameters
Hello  thank you in advance for any advice you can give me about this problem. First, let me tell you what I have figured out. My first problem was to figure out the probability of drawing at least one Ace from a deck of cards, given 7 draws from the deck (without returning the cards to the deck). 1. I believe the equation is: P(at least one Ace)=1C(48,7)/C(52,7) (is this correct?) 2. I also think that if you were returning the drawn cards to the deck each time, the equation would be: P(at least one Ace)=1(48/52)^7 (again, is this correct?) Now it starts getting more difficult. How do I figure out the probability of drawing at least 2 Aces from the deck (or at least 3 Aces), in the same 7 draws? 3. I think the equation would be: P(at least two Aces)=1C(48,7)/C(52,7)*C(48,6)/C(51,6) (is that correct? the second one has fewer possible draws so I'm using 6 instead of 7, and there are fewer possible cards in the deck, so I'm using 51 instead of 52, but there are the same number of nonAces if I successfully drew one Ace, so I'm still using 48 ) Now for the final question. I'm still drawing 7 times, but instead of drawing a single card each time, I'm drawing 2 cards each time, and each time I get to keep one card and must discard 1 card. Now I assume calculating the probability of getting at least one Ace is still the same, using equation 1 above (but with 14 cards rather than 7). But how does it change the probability of getting at least 2 Aces, or at least 3 Aces? 4. This would be equation 4 if I had any idea how to go about constructing it. If equation 3 above is correct, then here's my first stab: P(at least two Aces)=1C(48,14)/C(52/14)*C(47,12)/C(51,12) (I just know this is wrong. I think the 14 is right, since we're drawing 14 total cards. But I'm not sure where to include the probability that the second card in my first draw is an Ace I had to discard, or when I calculate at least 3 Aces where I go from there) Any help would be greatly appreciated! Last edited by Kinead; September 29th, 2014 at 08:54 AM. 
October 16th, 2014, 02:26 PM  #2 
Newbie Joined: Apr 2014 From: MÃ©xico Posts: 10 Thanks: 0 Math Focus: Probability 
Ho Kinead, at least for the first question I think you are right. When I have doubts about the solution of a particular problem I use to simulate the problem. As you will see, the aproximate solution from the simulation is very close to your answer. I think you can do something similar for the other questions. Good look. The following code is written in R (open source), I hope it is useful for you. Code: > library(prob) > deck < cards() > > #simulate the experiment of draw (random) 7 cards from deck without replacenment > draw < function(ncards = 7, from = deck){ + index < sample(1:nrow(from), size = ncards,replace = FALSE) + deck[index,] + } > > > #replicate experimento draw(7,deck) > experiment < replicate(n = 1000000, expr = draw()) > S < data.frame(matrix(unlist(experiment),nrow=1000000,byrow=TRUE)) > S < probspace(S) #add approximate probs for each result > #some results are > head(S) X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 probs 1 9 8 3 8 K K J Club Diamond Diamond Heart Diamond Heart Diamond 1e06 2 2 Q 4 9 9 7 A Heart Heart Diamond Diamond Club Diamond Club 1e06 3 8 3 K 3 7 J 9 Diamond Club Heart Spade Diamond Diamond Diamond 1e06 4 4 K 2 9 3 Q Q Diamond Diamond Spade Heart Club Diamond Spade 1e06 5 7 6 3 6 7 3 6 Club Diamond Diamond Club Heart Spade Heart 1e06 6 Q 8 7 8 K 10 9 Spade Heart Diamond Spade Club Club Spade 1e06 > #approximate probability for al least 1 ace since we took a 7 cards without replacenment > A < S[which(apply(S,1,function(row) sum(row[1:7] == "A")) >= 1),] > Prob(A) [1] 0.450524 > #Your answer > 1  choose(48,7)/choose(52,7) [1] 0.4496445 
October 16th, 2014, 10:07 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Kinead! Quote:
Quote:
$P(\text{at least 2 Aces}) \:=\:1  P(\text{no Aces})  P(\text{one Ace})$  

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