Hello ldd6602707,

Based on the information you provided, added up with my assumption,
Here are two options for you, all with the same cost.
You may choose whichever you want.

Option 1:
2 sets of 1 cards
1 set of 2 cards

Option 2:
If you really want to make sure you get enough,
you may need the following,
1 set of 1 card
1 set of 2 card
1 set of 5 card

Cost:
8.8 X 3 = 26.40

Relevant assumption:
5-star growth/week:        1
Reward rate:          2%

+ Tear %        3%
+ Risk buffer %        15%

1_cent_reward        60%
2_cent_reward        40%

The last but not the least,
student has to exchange their credit with you whenever it reaches 30.
This will reduce the card you need.

I have attached the model below.
You can make whatever adjustment to it according to your will.
I hope it works.
Thanks.

The card turnover rate is the key for this problem.

Formerly, I assumed the cycle time could be 2 or 3 weeks. (if so, you only need 2 sets of cards)
However, I realized that even your best student probably was not going to make enough credit for any reward.
After relevant adjustment, the model will present more reliable anticipation.

What's your opinion?
Thanks.

感谢cpamodeler，，非常感谢！
我幼儿园水平的英语，连蒙带猜，外加百度，大体了解了你推荐的2种方案。倾向于买1套1分，1套2分，1套5分的，加起来面值800，应该能周转。
想问一下，386是怎样计算出来的？
另外，当20积分就兑换掉奖品时候，理论上所有学生手中的积分达到多少？（每个学生手中的不超过20分）
如果是30分才兑换奖品的话，那有是多少呢？
注：我只有3个论坛币了，

    hw question.xlsx 这个文件就不下了。

您好ldd6602707， 不客气。
请告诉我您的电邮，给您发过去。
若您选1+2+5的话， 因您的奖励是以1和2为单位的，估计兑换会需要些精力。因1有限，会高频度的兑，工作量会增加，要有准备。当然，有大票，学生们可能会觉得更有趣些, 可征求下意见再定。

若学生努力、运气均态分布的话，估计均分在88。
如果到20就必须兑，就很难到30了，对吧？
所以，把兑换值定在30。
如果您同意发布这样的奖励政策，
那学生手中的最大未兑积分就是29。

见到模型就知386的来历了。
时间紧了些，如果我们有更多假设和透彻的讨论，会更完善些。
谢谢。

已发送，请查收。
谢谢。