Clearer explanation of "middle" card, please


#1

I teach in a comprehensive, suburban, predominately white high school. My class is mostly 9th and 10th graders.

We had a difficult time defining “middle” for Challenge 4. Is it intended to be the average value of the highest and lowest cards in the set (with a bit of wiggle room for not having half-card values)? Or is it the median value of the cards in the set (with the knowledge that you would need to choose an odd number of cards?

I realize that I could choose one method or another, but given that some people have expressed that this challenge has created some frustration on the part of the students I was wondering if one is inherently easier to comprehend than the other.

I welcome any and all direction on this.


#2

Extrapolating from context in the lesson plan it means the median. The idea is that if you follow the progression of “find the smallest”, find the 2nd smallest, find the 3rd smallest and so on, and you do that for half the cards you’ll end up at the “middle” or median value.

Yes, doing this with an odd number of cards would help to start. For an odd number you can just declare whether you’ll take the higher or lower one.

Hope that helps

Baker
Curriculum Team


#3

That is a tough one. Not all of my students really get to this one in a genuine manner to really start asking those questions but for the ones that do, I say this; What do you think? Are there multiple ways to think about this problem? Are there different cases that would warrant a different solution? I try and push them back to the define step of the problem solving process to hopefully really start to see that defining a problem can be a big part of the process. And ultimately, hopefully they will come to their own understanding of what the problem is asking.

This is also a good place to start to show the students that the teacher does not always know they answer and/or is not going to just give them the solution. Students need to focus on creating their own meaning and solutions and sometimes we won’t reach that the first few times through.

Also, I would say that the lesson is not about getting to the answer, it it about processing, and coming up with that process, or trying to get to that process is more what the lesson is about which can leave some students hanging.


#4

Great, thank you for the information. It is really helpful.


#5

I don’t even know how you’d do average (arithmetic mean). I think you’d have to find the average of all of them and then it just means finding the card closest to that number. Which is really just the same as the biggest or smallest challenge.

I thought about trying to find the average of the highest and lowest, but you don’t really know what these cards are until you’ve gone through the whole deck and may have since discarded the average.

I think median is the only one that makes sense and I think it requires at least (number of cards+1)/2 safe spots to calculate. Without this many safe spots, depending on how the deck is packed, you could accidentally discard the median.

Would love to hear if there is a more elegant solution for this one.


#6

Someone at a workshop told me that you can solve this one with one safe spot. What you do is find the highest card, then discard it. Now repeat through all of the cards and find the lowest card and discard that one. Then you repeat until you only have the one card left. If there are an even number of cards, I don’t know what you do… Flip a coin as to which one is the middle?


#7

An easy test, if you are not using a large amount of cards, would be to lay them out in order. If the algorithm is correct, then the “chosen” card should simply be in the middle of the row or as close as possible in an even numbered deck. An odd amount or cards should produce the desired result.


#8

Or just count the cards out. In a stack of 15, there should be 7 of equal or lower value and 7 of equal or higher value. It would help to set rules for values such as A-K or 2-A to make sure everyone is playing by the same rules.


#9

I used the ‘median’ which the 4th grade definition is middle number in a set of ordered numbers.

Really we are looking for the median.
(though the directions do not say to place the safe spots in order of value which can be an added layer to the algorithm if one wants to be more clear for humans])
When I used median with the 7th and 8th graders with a quick reminder that it’s the middle number in a range … .and that as you lay the cards out on safe spots the range may change.

There was NO MORE confusion.

so, median. . . can only use one hand. so must create new safe spot repeatedly.

Let me know if you think that will cause a misconception in with math or computer science . . . So I can address that.


#10

the problem is, as one turns up a card . .if it is not the highest card. .then it is discarded and no longer able to be used for comparison . . .to other cards.
Therefore, one needs to use safe spots to hold that card for future reference . . .so one can see the ‘range’ of numbers’ so one can determine which one is the median.


#11

Yeah, you still have to use one safe spot, as explained to me, to find the max/min card in the process to find the middle card. Once you compare all of the cards to the one in the safe spot, then you know you have the max. Then you repeat the process but find the smallest card. Then the max, and then the min, until you have only one card left, and it should be the middle card. I haven’t actually tried it but it sounds like it would work.


#12

But then after you get the first max you can’t get to any of the other cards again because you’ve already tossed them all (they weren’t in a safe spot).


#13

Yeah. So okay, you have to play the game over and over again, which is changing the rules a bit. Rules are meant to be broken?


#14

You’re such a rebel! :wink: