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Kinead September 29th, 2014 08:52 AM

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)=1-C(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)=1-C(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 non-Aces 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)=1-C(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!

mandradebs October 16th, 2014 02:26 PM

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 1e-06
2  2  Q  4  9  9  7  A  Heart  Heart Diamond Diamond    Club Diamond    Club 1e-06
3  8  3  K  3  7  J  9 Diamond    Club  Heart  Spade Diamond Diamond Diamond 1e-06
4  4  K  2  9  3  Q  Q Diamond Diamond  Spade  Heart    Club Diamond  Spade 1e-06
5  7  6  3  6  7  3  6    Club Diamond Diamond    Club  Heart  Spade  Heart 1e-06
6  Q  8  7  8  K 10  9  Spade  Heart Diamond  Spade    Club    Club  Spade 1e-06
> #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


soroban October 16th, 2014 10:07 PM

Hello, Kinead!

Quote:

Find the probability drawing at least one Ace from a deck of cards,
given 7 draws from the deck (without replacement).

1. I believe the equation is: P(at least one Ace) = 1 - C(48,7)/C(52,7)
(is this correct?)$\;\; \color{blue}{\text{Right!}}$

2. I also think that if you were replacing the cards to the deck each time,
the equation would be: P(at least one Ace) = 1 - (48/52)^7
(again, is this correct?)$\;\;\color{blue}{\text{Yes!}}$

Quote:

Now it starts getting more difficult.
Find the probability of drawing at least 2 Aces in the same 7 draws?

$P(\text{no Aces}) \:=\:\dfrac{_{48}C_7}{_{52}C_7} \qquad P(\text{one Ace}) \:=\:\dfrac{4(_{48}C_6)}{_{52}C_7}$

$P(\text{at least 2 Aces}) \:=\:1 - P(\text{no Aces}) - P(\text{one Ace})$



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