[For a detailed conceptual discussion of several methods, read the excellent comments by Mike and novemberfive.]

It's 'probably' too early in the year for this, but...

This is one of those typical probability questions that all students struggle with until they have enough experience to feel comfortable. When away from these for awhile, most everyone needs to revisit this kind of question and work through the theory again. It's so elusive...

Here's a typical version of this question:

Three cards are drawn at random from an ordinary deck of 52. What is the probability that they will consist of a king, a queen and an ace?

Do you feel totally comfortable when seeing these? Imagine how most students feel. They either have a strong concept here or they get that queasy feeling and express, "I don't like these!"

So here's the issue:

Is there one method you have found for these that makes sense to most students or should multiple methods be demonstrated (and shown to be equivalent)?

To get you started:

Method One: For the denominator, select all 3 cards 'simultaneously' in C(52,3) ways. The numerator is more interesting, so I'll leave that to you.

Method Two: Analyze one card at a time without replacement. Thus, the probability that the first card will be a king is 4/52, etc. Of course, most of you know the 'catch' here, but the method can still work if...

Method Three: Yours!

## Friday, August 31, 2007

### Drawing 3 cards one at a time vs. Drawing 3 cards simultaneously - Those nasty probability questions!

Posted by Dave Marain at 6:56 AM

Labels: combinatorial math, probability

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## 7 comments:

I may well be wrong on this one, but my method is this: First card, you have 12/52 probability of picking Q,K, or A. Second card, you have 8/51 probability of picking, out of Q,K, or A, one of the two types you didn't pick on first card. Third card, you have 4/50 probability of picking the type of card you haven't yet picked. Multiplying the three probabilities, you have a probability of 16/5525, or about 0.002896 probability. Yes? No?

I like your method, mike!

For me, Dave's 2nd method is the "true" way to solve it... that is, it's the way that corresponds most closely to my intuition of the situation and seeing it solved that way gives me the most confidence that the answer is correct.

Following the 2nd method, P{A,K,Q}) = P(<A,K,Q>) + P(<A,Q,K>) + P(<K,A,Q>) + P(<K,Q,A>) + P(<Q,A,K>) + P(<Q,K,A>). (Notation: {

A,B,C} denotes drawing the cardsA,B,Cin any order, while <A,B,C> denotes drawingA,B,Cin that order, i.e. firstA, thenB, thenC.)Now P(<A,K,Q>) = (4/52) * (4/51) * (4/52). And indeed, each term in the sum is equally likely. There are 6 terms, so the answer = 6 * P(<A,K,Q>) = 16/5525.

I gather that the numerator in Dave's 1st method is 4 * 4 * 4, but my intuition as to why this is a correct way to set up the problem is not as strong as it is for the 2nd way of solving it.

Hey Dave, "are [your students] still doing [their] algebra homework the OLD way?" Welcome to Algebra Buster homework solver. "It solves your algebra homework just like your math teacher!"

The existence of programs like this is one of the reasons I like non-routine problems: it's nice to emphasize the (ever-narrowing range of) things which people can do but computers can't.

mike--

Along with novemberfive, I really like your 'Method 3'. Very instructive, although, I probably prefer novemberfive's modification of Method Two.

Showing all these methods are equivalent can easily be done numerically, but not so easily explained on a conceptual level.

Thus your (12)(8)(4) can be rewritten as (4)(4)(4)(3)(2)(1) which makes it look like Method Two! The (3)(2)(1) or 3! is the number of ways each 'hand' can be rearranged which novemberfive explicitly showed us.

Now for a discussion of Method One, which is usually more problematic for many, but one I still like and preferred by many students and texts:

If C(52,3) is used in the denominator, then we want to count only ONE possible arrangement or 'hand' in the numerator. This is why we use (4)(4)(4) as novememberfive indicated. Let me clarify:

(4)(4)(4) counts, for example, the 'hand': King(Hearts),Queen(Clubs) and Ace(Hearts) only once, not SIX times. That's because the denominator is counting 'hands' only once as well (any rearrangment of a 'hand' is the SAME 'hand' when playing cards!).

Numerically, this works since

(4)(4)(4) divided by

(52)(51)(50)/((3)(2)(1)) is equivalent to the Method Three result!

By moving the 3! up to the numerator, we see there is an alternate form for Method I:

P(3,3)(4^3) divided by P(52,3), since P(3,3) equals 3!. This is an important but subtle point for probability questions regarding selection of more than one object. I usually tell my students that if you're using permutations in the denominator, you must also use permutations in the numerator (same rule for combinations). The extra factors will cancel anyway but it shows the student how many options there are. Seems confusing? Absolutely! That''s why I decided to pose this question. Now, here's the rub:

If a student catches on to this with a look of understanding, just give her/him another one for homework or on an assessment which is slightly different and see what happens. One cannot memorize solutions to these. One needs to practice many variations on a regular basis. If I don't solve a few of these for several months, I feel like I'm starting over again! Fortunately, the methods do come back. I find, I usually need to solve it by more than one method to validate myself!

By the way, you want students to feel they really understand these kinds of combinatorial problems. Forget the probability piece and just have them solve the poker hands combinations, you know, how many 'hands' have one pair, two pair, three of a kind, straight, flush, etc.

From my experience with students, some do prefer Method I, which is the preferred method in many texts. Others like Method Two, because it's easier for them to handle one card at a time rather than three. They often mess up the issue of 'order' however. In the end, perhaps it all comes down to what should have been the title of this post:

DOES ORDER COUNT?Perhaps this lengthy reply should appear in the post itself, but, hopefully readers get to see all of these. Thanks, Mike and novemberfive for motivating me to type this much...

P(AKQ, any order)

= # of rearrangements * P(A|K|Q)

= 3!/(1!1!1!) * 4/52 * 4/51 * 4/50

I like this best, because each number above has some direct physical meaning.

I like Mike's, but I am not certain we can reapply it to P(AAK, any order), for me the lack of generality is a drawback. Still very cute.

In NYS many teachers were introduced to probability through the introduction of Sequential Math 25 years or so ago, and they all picked up

[C(4,1)*C(4,1)*C(4,1)]/C(52,3)

I understand the value of this approach, but its inflexible application is unfortunate.

Late this fall in combinatorics I will do some poker counting and probabilities, and we will contrast the methods (and hybrids) and let the kids select for each problem. Last year I spent I think 2 days throwing out problems and then getting kids to tell the class which way they did it, and why. They really listened to each other.

Dave, your work with moving the 3! 'up' is what I would use to show the equivalence between methods.

I do maintain the distinction:

Combinatorial approach (all desired outcomes over all possible outcomes)

Synthetic or conditional approach for one ordering (conditional probabilities with suitable multiplication of fractions, times the number of orderings

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