Doubts have been expressed as to what exactly "order matters/is important" means in relation to the following question on this site here

"In a deck of 40 cards there are 4 aces. What is the probability that when drawing two cards only one ace is drawn."

There can be two cases: first card is the ace, or second card is the ace.

One view expressed is that since both cases refer to one event, order is not important.

The other view is that since we are considering two possible permutations, order is important.

Or put differently, is "order important/matters" if

(a) the order is specified, e.g. first card is an ace, or

(b) the order is unspecified, i.e. all possible orders are to be considered

I have the feeling that both the interpretations are being used on this site by different people.

It would be good if some one could give a definitive opinion on this matter.

  • $\begingroup$ @turkeyhundt: The question isn't w.r.t the particular problem. I have edited the question to make the dichotomy clearer: specified order = ?, unspecified order = ? $\endgroup$ – true blue anil Jan 2 '17 at 16:19
  • $\begingroup$ @turkeyhundt As you say, order is important(/specified) in each case, but as either case is part of the event, then order is not-important(/unspecified) in the event. $\endgroup$ – Graham Kemp Jan 2 '17 at 16:28
  • $\begingroup$ @Graham Kemp: The precise question I am asking (forget about the particular problem, and the ways available to solve such problems is) : is order important because it has been specified, therefore must be important; or is order important when left unspecified, because then all possible orders must be considered. $\endgroup$ – true blue anil Jan 2 '17 at 16:53
  • $\begingroup$ All possible orders must be considered when the order is unspecified because the order is not important and it does not matter which of the possible orders occurs for the event to happen. However, if the order were important then only the specified order would need be considered because that is the order that matters. (See the posted answer for further clarification.) $\endgroup$ – Graham Kemp Jan 2 '17 at 17:08
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    $\begingroup$ @Graham Kemp: It seems that you have taken the opposite view of the meaning of "order matters" at math.stackexchange.com/questions/1489934/… $\endgroup$ – true blue anil Jan 2 '17 at 18:58

For the experiment of drawing two cards from the specified deck, define the random variable, $A$ as the count of aces drawn, and the event $A_1, A_2$ as "an ace is drawn first" and "... second" respectively.

In the event of $A=1$, the order is not important.   It does not matter whether the single ace is the first or second card drawn.   The order is not specified by the event of "exactly one ace is drawn".

One solution is that $\mathsf P(A=1)={\tbinom 41\tbinom {36}1}\bigm/{\tbinom {40}2}$, which is the probability for selecting 1 from 4 aces and 1 from 36 non-aces given that 2 from 40 cards are selected.   This solution does not concern itself with the order the cards are selected –in either numerator nor denominator–.

Another solution is that $\mathsf P(A=1)= \mathsf P(A_1\cap A_2^\complement)+\mathsf P(A_1^\complement\cap A_2) = (\dfrac 4{40}\cdot\dfrac{36}{39})+(\dfrac {36}{40}\cdot\dfrac 4{39})$

This solution partitions the event into two subsets, each an event in which the order the cards are drawn is important.   For each part, it matters that the ace is drawn first or second.   $A_1\cap A_2^\complement$ specifies the order of the cards, as does the other part, $A_1^\complement\cap A_2$.

The solution to each part clearly concerns itself with the order of the draw.   The relevant terms are obtained by evaluating, $\mathsf P(A_1\cap A_2^\complement)=\mathsf P(A_1)\mathsf P(A_2^\complement\mid A_1)$ and $\mathsf P(A_1^\complement\cap A_2)=\mathsf P(A_1^\complement)\mathsf P(A_2\mid A_1^\complement)$

However, as the event is the union of the parts — either part satisfies the event — then, as a whole, the order is not important.   That is, because the order is not important, we may measure the probability for each ordering, then add these results to obtain the measure for the event.


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