How can card matching exercises be made easier?
With n cards in the prompt set and n in the response set, there are n2 ways one prompt card can be combined with 1 response card, and 2n2-n ways any 2 cards can be paired, if the distinction between prompt and response cards is ignored.
In the case that players have only a shaky grasp of any of the prompt-response relationships involved, to dump these 2n cards in their lap is the quickest way to lose them.
Better in this case is to give out only some of the 2n cards, and then exchange cards which match for more. If you do this, you want to ensure there is a match in all the hands the players receive to avoid frustrating them.
The way which requires least thinking by you to do this is to start with n+1 cards. In any such set, there will be at least 2 cards which match. Otherwise, finding the the cards which match with these n+1 cards would require another n+1 cards, but you only have another n-1 cards. QED
Then, when players successfully match 2 of the n+1 cards, exchange them with any 2 of the remaining cards, giving them another hand of n+1 cards in which there is again at least one successful match, for the same reasons.
But we can do better than this. It is possible to feed players a starter set of 2 prompt cards and 2 response cards in which there is only 1 correct pairing, which they have to pick from 4 possible prompt-response pairings, or 6 possible 2-card combinations.
An even easier (the easiest) starter is 3 cards, in which they have to pick from 2 possible prompt-response pairings, or a total 3 possible 2-card combinations.
In exchange for the 2 cards which form a pair, they get 2 more cards which again creates a set of 4 (or 3) prompt and response cards in which there is again only 1 correct pairing.
This process continues n-2 times until the last time, when they are fed the last 2 (or 3) cards, leaving them with 4 cards with 2 correct pairings.
But doing this in actual classes is not practical unless there is an easy way to collate cards that doesn’t require a lot of time or a degree in math to do.
One way to do this with n=4, with 8 cards on a piece of paper, is as follows.
+-----+-----+-----+-----+
| | | | |
| A | C | a | c |
| | | | |
+-----+-----+-----+-----+
| | | | |
| B | D | b | d |
| | | | |
+-----+-----+-----+-----+Cut out the 4 cards, A and B and b and d
+-----+
| |
| A |
| |
+-----+ +-----+-----+
| | | | |
| B | | b | d |
| | | | |
+-----+ +-----+-----+B and b is the only pair, leaving A and d.
Cut off C and D (or alternatively) a and c.
+-----+ +-----+-----+
| | | | |
| C | | a | c |
| | | | |
+-----+ +-----+-----+
| |
| D |
| |
+-----+D and d is the only pair, leaving A and C.
Alternatively, A and a is the only pair, leaving c and D.
The final 2 cards produce 2 pairings.
For the same n=4 case, with a 3-card starter, the cuts require even less thinking with this paper layout.
+-----+-----+-----+-----+
| | | | |
| A | C | a | c |
| | | | |
+-----+-----+-----+-----+
| | | | |
| B | D | b | d |
| | | | |
+-----+-----+-----+-----+Cut 3 cards along one side of the paper for the 3-card starter, cut the 2 cards left off the end for the second feed, and give the remaining 3 cards (which could also have been the starter) for the last feed.
The only problem left is keeping the cards separate and remembering which is the starter and which are feeds.
The n=8 case could be handled as the simultaneous playing of 2 n=4 exercises with an 8-card starter and 2 correct pairings for each feed, cutting 2 sheets of paper together, at the same time.
Alternatively, with 16 cards on 1 sheet of paper,
+-----+-----+-----+-----+
| | | | |
| A | E | d | e |
| | | | |
+-----+-----+-----+-----+
| | | | |
| B | F | a | f |
| | | | |
+-----+-----+-----+-----+
| | | | |
| C | G | b | g |
| | | | |
+-----+-----+-----+-----+
| | | | |
| D | H | c | h |
| | | | |
+-----+-----+-----+-----+The first row,
+-----+-----+-----+-----+
| | | | |
| A | E | d | e |
| | | | |
+-----+-----+-----+-----+produces the 4-card starter with E-e pairing, leaving A and d.
+-----+-----+
| | |
| a | f |
| | |
+-----+-----+produces the first (second) feed with the A-a pairing, leaving d and f.
+-----+-----+
| | |
| B | F |
| | |
+-----+-----+produces the second (third) feed with the F-f pairing, leaving d and B.
+-----+-----+
| | |
| b | g |
| | |
+-----+-----+produces the third (fourth) feed with the B-b pairing, leaving d and g.
+-----+-----+
| | |
| C | G |
| | |
+-----+-----+produces the fourth (fifth) feed with the G-g pairing, leaving d and C.
+-----+-----+
| | |
| c | h |
| | |
+-----+-----+produces the fifth (sixth) feed with the C-c pairing, leaving d and h.
+-----+-----+
| | |
| D | H |
| | |
+-----+-----+produces the sixth (seventh) and final feed with the H-h and D-d pairings.
Players get a lot of time to contemplate d. You want to limit the time any one card has to remain in players’ hands. Of course, the cards paired in the starter remain in the hand for the minimum amount of time.
Trying the 16-card layout again, with a different distribution of cards to get a more even movement of cards into and out of players’ hands,
+-----+-----+-----+-----+
| | | | |
| A | E | b | e |
| | | | |
+-----+-----+-----+-----+
| | | | |
| B | g | a | F |
| | | | |
+-----+-----+-----+-----+
| | | | |
| d | G | C | f |
| | | | |
+-----+-----+-----+-----+
| | | | |
| D | H | c | h |
| | | | |
+-----+-----+-----+-----+The feeds become:
Which is better than the 7 feeds d was in the hand, and probably the best that can be achieved seeing at least one starter card will survive three feeds.
Feeding the cards in reverse order, ie 7 to 1, looks to have the same properties as the 1 to 7 feed.
But this n=8 feed is not as robust as the n=4 feed. So I folded paper boxes to keep the feeds in order.
The n=8 case with a 3-card starter is left as an exercise for the reader.
No, I’ll do it myself. It’s easy.
+-----+-----+-----+-----+
| | | | |
| A | E | a | e |
| | | | |
+-----+-----+-----+-----+
| | | | |
| B | F | b | f |
| | | | |
+-----+-----+-----+-----+
| | | | |
| C | G | c | g |
| | | | |
+-----+-----+-----+-----+
| | | | |
| D | H | d | h |
| | | | |
+-----+-----+-----+-----+No card remains in the hand for 3 feeds.
That would require the previous feed to have been 2 cards which match. 2 matching cards do appear in the last feed, feed 7.
See my perl flash card script https://github.com/drbean/ttb/blob/master/cards/flash.pl
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