Comments

Isn't the answer supposed to be 1/25.

The third selection is an independent event.

After the first two cards have been drawn from the deck there are only 50 cards left, two are aces, so the probablity of drawing one of them is 2/50 = 1/25.

The probablity of drawing three aces in a row is:

(4/52)(3/51)(2/50) = 0.00018 or 9/50,000

-- KDeRosa - 08 Apr 2006

wait - I think that's a typo - hang on (obviously this stuff isn't exactly becoming intuitive for me at this point)

-- CatherineJohnson - 08 Apr 2006

right - I made the change.

what about the language tells me that this is an independent event?

would it have to say something more explicit like '3 aces in a row'?

-- CatherineJohnson - 08 Apr 2006

Hey Ken - you've gotta take a look at this: What Is An Expert Student? by Robert Sternberg (pdf file)

This is the SOLE article given to our curriculum committee by our Assistant Superintendent for Curriculum.

-- CatherineJohnson - 08 Apr 2006

hmmm...

This page might be interesting: Cognition, Instruction, and the Design of Learning Environments (don't know)

-- CatherineJohnson - 08 Apr 2006

Sternberg advocates teaching wisdom in schools.

-- CatherineJohnson - 08 Apr 2006

It's not as bad as it sounds; Sternberg's work is interesting.

But for an article on 'practical intelligence' to be the SOLE reading material distributed to the curriculum committee is almost the definition of 'unwise.'

-- CatherineJohnson - 08 Apr 2006

(I should add that I haven't sat down and read this article closely, which I think I'll do before I slam it any further...)

Nevertheless, I sent all of Daniel Willingham's stuff to Ralph, and I don't see it being distributed....

-- CatherineJohnson - 08 Apr 2006

The difference is that we know what the first two cards drawn were. And those cards affect the probability of the third draw.

Let's change the problem to first we draw a card then we draw another card. What is the probability that the sedond card is an ace. The answer is: it depends because we don't know what the first card drawn is and the first card drawn affects the probability of the second draw. If we knew what the first cad drawn was, then the second draw is independent since the probablity is resolvable.

-- KDeRosa - 08 Apr 2006

One of the problems with Ed research is that it stinks. You can prove almost anything with stinky research. This article matches your curriculum director's ideology, so he likes it.

You need a Willingham type to properly dissect an article like this and expose all the shortcomings.

But I like the Engelmann response better: This research is all fine and good, but you still have to design an instructional program around it. The true test of whether your theory is any good depends on how the atudents perform after being instructed in the rpogram.

-- KDeRosa - 08 Apr 2006

We can answer Ken's revised question more precisely than "it depends." If we draw two cards, the probability that the second card is an ace is 204/2652.

-- RudbeckiaHirta - 08 Apr 2006

"First of all, choosing 3 aces in a row seems like a series of dependent events, but the answer given — 4% — is the answer you would get by assuming that the 3rd ace is an independent even."

What you are missing is that you are only really being asked about a single event. That event is the choice of the third card. The fact that two aces were chosen before is already determined just like the fact that you are using a standard 52 card deck. You can change the question to make this more evident:

A 50 card deck is made by removing two aces from a standard 52 card deck. Robert is then asked to select a card from this deck. What is the chance that the card he selects will be an ace?

In this case it's a bit more clear that we are only interested in a single event, but the question is essentially the same. The answer is number of aces in the deck/total number of cards = 2/50 = 1/25. You can change the question so that you have two events like so:

Robert was asked to select a card from a normal deck of cards. Robert is then asked to select a second card from the deck. What is the chance that both cards he selects are aces?

There you have two events, the second dependant on the first. The answer is (4/52)*(3/51) = 1/221. You can ask about independant events too:

Robert was asked to select a card from a normal deck of cards. After selecting and looking at his card, Robert replaces the card in the deck and the deck is shuffled. Robert is then asked to select a card from the deck and look at it. What is the chance that both cards he selects are aces?

Now the answer is (4/52)*(4/52) = 1/169.

Here's KDeRosa^?'s question:

Let's change the problem to first we draw a card then we draw another card. What is the probability that the second card is an ace?

Once again we have one event dependant on another. RudbeckiaHirta^? posted the solution(nicer than mine would have been) while I was writing this, so I won't repeat it. The answer is 17/221.

-- KtmGuest - 08 Apr 2006

There's a second way of doing Ken's problem, which is purely combinatorial.

There are 52*51 ways of picking two cards from a standard deck (no replacement): 52 for the first card, 51 for the second card. So that's the size of our sample space.

In order to find the probability of an event in this sample space, we need to count how many ways the event can happen. In Ken's problem, this means that we need to count how many ways there are to draw two cards such that the second card is an Ace.

There are 51 ways to have the second card be the Ace of Hearts (any of the 51 other cards can be the first card). Similarly, 51 ways to have the second card be Ace of Spades, 51 ways to have the second card be Ace of Diamonds, 51 ways to have the second card be Ace of Clubs.

Add these all up, and there are 51+51+51+51 = 204 ways to have the second card be an Ace.

So again, the probability is 204/2652.

-- RudbeckiaHirta - 08 Apr 2006

This is a good example why you won't find me at KTM poker night.

Let's clarify for the non-nerds.

Rudbeckia's solution(s) gives the probability that the second card drawn is an ace before any cards are drawn. p = .077

Once the first card is drawn, however, and placed face down, now there is a new probability that the second card is an ace. And that probability depends on the whether the first card was an ace or not.

If the first card drawn was an ace, the probability of drawing a second ace is 3/51.

If the first card drawn was not an ace, the probability of drawing an ace is 4/51.

We just don't know which probability applies until we know whether the first card was an ace or not. This is why the second event (card draw) is dependent upon the first event (card draw). You can see this play out in Rudbeckia's decision tree.

It is this second probability that I was referring to in my hastily worded example. But, as Rudbeckia's analysis shows you can work out the probability of doing pretty much anything with a deck of cards in advance even when you don't know the order the cards are drawn since we know the probability space.

-- KDeRosa - 09 Apr 2006

Ken writes: Once the first card is drawn, however, and placed face down, now there is a new probability that the second card is an ace. And that probability depends on the whether the first card was an ace or not.

As long as the first card is face down, the probability is 204/2652. The probability will change if you PEEK at the upside down card, though.

This is usually easier to see in the combinatorial argument than in the tree argument. There are 2652 ways to draw two cards (without replacement). There are 204 ways for a pair of two cards to have second card as an Ace. It doesn't matter if they're right side up, upside down, drawn sequentially, drawn simultaneously, etc. So you have a 204/2652 chance that you are in one of these cases.

Probability is one of the hardest topics taught in school mathematics. To really understand all the permutations, combinations, and whatever else involved in probability, it really does require a serious remodeling of your brain.

Textbooks are written very "flat" without any cues as to what's intrinsically hard and what's easier. They'll present the probability chapter with much the same tone as everything else.

For me, my ability to do probability problems varies widely. If it's part of the classic collection of cards, dice, and coin problems, then I can do it off the top of my head (because I've done all the classic problem types a zillion times). If I'm teaching a course and it's a problem from the textbook, then I can do it off the top of my head (because I know that there are a limited set of rules that can apply). For an unfamiliar problem out of the blue, I'd have to slowly and carefully work it out on paper and double-check my answer before talking about the problem. Probability is sneaky.

-- RudbeckiaHirta - 09 Apr 2006

Okay, you math heads are scary.

I'm presently working on this book myself and nothing has been presented that complicated. (Well, complicated for us phobics). I'm not at that particular chapter yet, but I'm close. There was one chapter on compound probability where the diagrams look like what you all are doing, but most have been more simple.

The solution manual's answer for this problem is: 2/50=1/25=*4%*

-- SusanS - 09 Apr 2006

One of my avocations is game design, which tends to foster an interest in probability theory and combinatorics. It can be interesting to look at the problems from the other side too.

For instance, when designing rules for a (Hollywood-style) wild west shootout game, I wanted a skewed distribution for wound severity, with most of the wounds being "just flesh wounds", but still allowing for the occasional catastrophic damage.

In the end, I used a system where the damage was the product (rather than the sum) of two dice. The result is a plot where the mean and median can be quite far apart (depending on the type of dice used).

-- DougSundseth - 09 Apr 2006

whoa!

i have my work cut out for me!

THANK YOU!

-- CatherineJohnson - 09 Apr 2006

Carolyn if you're around - is Rudbeckia's blog not on our sidebar?

I thought it was - but I'm not seeing it (I don't think) when I'm in edit mode....

-- CatherineJohnson - 09 Apr 2006

oh that's funny

it's there now, but not there in the edit mode

!

-- CatherineJohnson - 09 Apr 2006

Learning Curves is usually on the sidebar.:)

-- SusanS - 09 Apr 2006

In the interest of clarity to those who sort of skim through the math, my posted solution was to Ken's adaptation of the problem.

The original middle school level problem in Saxon really does have the solution 2/50.

-- RudbeckiaHirta - 09 Apr 2006

oh good lord - thanks (I've just skimmed so far)

-- CatherineJohnson - 10 Apr 2006

I feel better now, too. I thought I'd be hitting some horrific 7th grade wall in a few chapters.

-- SusanS - 10 Apr 2006

There's an easier way to get to 204/2652 (that would be 1/13), of course. With a normal deck of cards (no jokers), the probability that an arbitrary card will be an ace is 4/52 or 1/13. The second card in the deck (or the 37th card in the deck) is an arbitrary card. Removing a random card (say the first card) without looking at it doesn't change that probability.

-- DougSundseth - 10 Apr 2006

Attachment	Action	Size	Date	Who	Comment
probability.jpeg	manage	45.2 K	08 Apr 2006 - 23:33	RudbeckiaHirta