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21 Sep 2016, 08:07
Kudos
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Hi guys,
I was reading some notes put together by Bunuel (in the QUANT THEORY folder of Bunuel's signature collection) on combinations / permutations and found something odd.
I assume it's a mistake, if not, I don't understand it (you can't use both permutation and combination to solve the same problems, i.e drawing cards without replacement... I assume order doesnt matter for such problems, hence the use of combinations seems justified. But why permute?) :
Inside the "Permutations to solve problems" pdf :
2. Two cards are drawn from a standard deck of playing cards without
replacement.
a) Express the number of elements in the sample space using the notation
npr. (52P2)
b) What is the probability that the two cards selected are aces?
4P2 / 52P2 = 1 / 221
c) What is the probability that the two cards are the Ace and King of
hearts in that order?
( 1P1 * 1P1 ) / 52P2 = 1 / (52 * 51 )
Three cards are drawn from a deck of 52 cards without replacement.
Ask: “ What is the probability that all are Aces?” As in the previous problem,
n(S) should be obtained. In this case, this is 52C3 as again the order of drawing
the cards is not important. The concept that there are 4C3 combinations of
ways of drawing the three Aces is not as apparent.
It should be stressed that in order to obtain n(E) we are concerned only with
the four Ace subset of the deck. The three Aces must be selected from this
group of 4. From the definition of probability, P(E) = 4C3 / 52C3 = 4/22100
I was reading some notes put together by Bunuel (in the QUANT THEORY folder of Bunuel's signature collection) on combinations / permutations and found something odd.
I assume it's a mistake, if not, I don't understand it (you can't use both permutation and combination to solve the same problems, i.e drawing cards without replacement... I assume order doesnt matter for such problems, hence the use of combinations seems justified. But why permute?) :
Inside the "Permutations to solve problems" pdf :
2. Two cards are drawn from a standard deck of playing cards without
replacement.
a) Express the number of elements in the sample space using the notation
npr. (52P2)
b) What is the probability that the two cards selected are aces?
4P2 / 52P2 = 1 / 221
c) What is the probability that the two cards are the Ace and King of
hearts in that order?
( 1P1 * 1P1 ) / 52P2 = 1 / (52 * 51 )
Three cards are drawn from a deck of 52 cards without replacement.
Ask: “ What is the probability that all are Aces?” As in the previous problem,
n(S) should be obtained. In this case, this is 52C3 as again the order of drawing
the cards is not important. The concept that there are 4C3 combinations of
ways of drawing the three Aces is not as apparent.
It should be stressed that in order to obtain n(E) we are concerned only with
the four Ace subset of the deck. The three Aces must be selected from this
group of 4. From the definition of probability, P(E) = 4C3 / 52C3 = 4/22100
21 Sep 2016, 16:25
Kudos
Bookmarks
Dear damafisch,
I'm happy to respond.
I agree that something is fishy.
First of all, I would highly recommend thinking about all counting problems primarily in terms of the Fundamental Counting Principle. All permutations and combinations are just shortcuts for the FCP, and if you understand the connection to the FCP then you really understanding the counting problems deeply.
2. Two cards are drawn from a standard deck of playing cards without replacement.
a) Express the number of elements in the sample space using the notation npr. (52P2)
What's highly anti-intuitive about this first problem is that NOTHING is clear from the statement of the scenario that order matters. In other words, common sense seems to indicate that if I pick the 5 of Heart first, then the 3 of Diamonds, this would be identical to picking them in the reverse order: I would still be holding the same pair of cards in my hands. I would say something is less than clear about the statement of the problem for them to want the permutations, because the problem does not specify that order matters.
If order doesn't matter, then we can use the FCP. On the first choice we have 52 choices. On the second choice, because there's no replacement, we get 51 choices. Since the pair would be the same if selected in either of the two possible orders, we have to divide by 2. The number would be
N = (52*51)/2
That's equal to 52C2.
If order in fact does matter, then we can use the FCP again. On the first choice we have 52 choices. On the second choice, because there's no replacement, we get 51 choices. After those two choices, we have a unique ordered pair.
N = 52*51
That's equal to 52P2.
If you confused about this, it's not your fault. The problem is unclear.
b) What is the probability that the two cards selected are aces?
4P2 / 52P2 = 1 / 221
Once again, the problem seems to be requiring order without letting the rest of us in on that particular secret. If order matters, this is 100% the correct answer, but it's not at all clear from the scenario so far or from this question that order should matter. Again, if you are confused, it is not your fault.
If order doesn't matter, the probability would be
P = (4C2)/(52C2) = 6/[(52*51)/2] = 12/(52*51) = 3/(13*51) = 1/(13*17) = 1/221
We get the same numerical answer either way, because whether order matters or not, we are looking at the same thing: getting all double-ace pairs.
c) What is the probability that the two cards are the Ace and King of hearts in that order?
( 1P1 * 1P1 ) / 52P2 = 1 / (52 * 51 )
Now, order explicitly matters, so it makes sense that we are using permutations. BTW, I think it's ridiculous to write "1P1" instead of simply "1". This the correct answer, P = 1/2652
Does all this make sense?
Mike
I'm happy to respond.
I agree that something is fishy. First of all, I would highly recommend thinking about all counting problems primarily in terms of the Fundamental Counting Principle. All permutations and combinations are just shortcuts for the FCP, and if you understand the connection to the FCP then you really understanding the counting problems deeply.
2. Two cards are drawn from a standard deck of playing cards without replacement.
a) Express the number of elements in the sample space using the notation npr. (52P2)
What's highly anti-intuitive about this first problem is that NOTHING is clear from the statement of the scenario that order matters. In other words, common sense seems to indicate that if I pick the 5 of Heart first, then the 3 of Diamonds, this would be identical to picking them in the reverse order: I would still be holding the same pair of cards in my hands. I would say something is less than clear about the statement of the problem for them to want the permutations, because the problem does not specify that order matters.
If order doesn't matter, then we can use the FCP. On the first choice we have 52 choices. On the second choice, because there's no replacement, we get 51 choices. Since the pair would be the same if selected in either of the two possible orders, we have to divide by 2. The number would be
N = (52*51)/2
That's equal to 52C2.
If order in fact does matter, then we can use the FCP again. On the first choice we have 52 choices. On the second choice, because there's no replacement, we get 51 choices. After those two choices, we have a unique ordered pair.
N = 52*51
That's equal to 52P2.
If you confused about this, it's not your fault. The problem is unclear.
b) What is the probability that the two cards selected are aces?
4P2 / 52P2 = 1 / 221
Once again, the problem seems to be requiring order without letting the rest of us in on that particular secret. If order matters, this is 100% the correct answer, but it's not at all clear from the scenario so far or from this question that order should matter. Again, if you are confused, it is not your fault.
If order doesn't matter, the probability would be
P = (4C2)/(52C2) = 6/[(52*51)/2] = 12/(52*51) = 3/(13*51) = 1/(13*17) = 1/221
We get the same numerical answer either way, because whether order matters or not, we are looking at the same thing: getting all double-ace pairs.
c) What is the probability that the two cards are the Ace and King of hearts in that order?
( 1P1 * 1P1 ) / 52P2 = 1 / (52 * 51 )
Now, order explicitly matters, so it makes sense that we are using permutations. BTW, I think it's ridiculous to write "1P1" instead of simply "1". This the correct answer, P = 1/2652
Does all this make sense?
Mike
