This page is outdated as I received additional information by Oliver Tacke explaining the repetition system of Dialog Cards. Read instead my new article how to calculate not repetitions but proficiency levels.
The following demonstration uses the same content as the original example at h5p.org, but with a slightly different setting:
- Repetition mode
- Maximum proficiency level1
- Allow quick progression (to facilitate the following tests.)
- For more accessible references under test conditions, I have the card names changed to numbers.
Perhaps I should add that the German translation of "Repetition" and "Maximum proficiency level" is more detailed and gives a better (hopefully correct) understanding:
- Nicht gewusste Karten wiederholen [Repeat unknown cards]
- Anzahl der Wiederholungen, nach denen eine Karte als beherrscht gilt [Number of repetitions after which a card is considered mastered]
As I understand, the idea in my example is: A card has to be known 3 times in a row.
I tried the exercise with the following data:
First example:
Every card correct in every round. Result: Programm aborts with two repetitions and says: Well done! You got all 5 cards correct 2 times in a row!
Second example:
- Round 1: Card 1 + 2 wrong, all other (3,4,5) correct.
- Round 2: Now, the program hides one or several cards from the pool 3,4,5. Card 1 wrong again, Card 2 correct.
- Round 3: One or more of the hidden cards appear. Card 1 correct. The program stops.
This is erratic behavior. Instead, it should go as follows:
Corrected first example:
Every card correct in every round. Result: Programm aborts with three repetitions and says: Well done! You got all 5 cards correct 3 times in a row!
Corrected second example:
- Round 1: Card 1 + 2 wrong, all other (3,4,5) correct.
- Round 2: Presenting all five cards! Card 1 wrong again, Card 2 correct, Card 3-5 again correct.
- Round 3: Presenting all five cards! All cards correct.
- Round 4: Presenting only Card 1 und Card 2, because Cards 3,4 and 5 was correct 3 times in a row. The user chooses the remaining two cards correctly.
- Round 5: Presenting only Card 1: Now, Card 2 was also known 3 times in a row. If now Card 1 is again correct, the program stops.
And even more complicated: If Card 2 is wrong in Round 3, then the counter for this card should be set to 0 (zero), and the learner has to restart to learn Card 2 3 times correctly in a row!
Another suggestion: I believe that the button "Next" is redundant. After all cards in a round are judged by the learner, the program could immediately present the results.