Weekly Wednesday quiz: hints & multiple choice
About 3 weeks ago I organized a contest. So far no one tried to win this contest. Most likely because it is to difficult 🤔
So I decided I will help you a lot! 😁
Multiple choice
Just take a guess and see if you got it right
A. 0.5%
B. 1%
C. 5%
D. 8%
E. 14%
F. 42%
Next you can read the original question. And last but not least I got a hint about the calculation.
The question: what is the chance that the first 28 stickers are unique if you need to collect a total of 160 stickers to complete your sticker book?
Recently supermarket Albert Hein started a new promotion where one received 4 tech stickers for spending €10. You need to collect 160 unique stickers to fully complete the sticker book. In the first days we collected 28 stickers and those were all unique. That really amazed me cause I remembered from previous sticker books that we had numerous double stickers. In the end we always had a few stickers up to 8 or even 10 times. So I decided it would be nice to know what chance I had to have the first 28 stickers all unique ones?
Calculation hint
What is the chance that your first sticker is unique?
100%
That's easy isn't it?
But what is more important is how you calculated this 100%? Now try to calculate the chance that the second sticker is unique. This can be done by the possible good outcomes sharing through the total outcomes. How many do you still need? 159. Total outcome? 160.
And then you only need the final step to calculate the chance for 28 unique stickers.
This can only be calculated only if we know total supply of stickers( how many copies each of 160 stickers has)
For this calculation we assume that each sticker occurs equally. So let's say each sticker occurs 10,000 times.
I'm not yet sure if this is really relevant? I would say it's not. The chance is equal to having each sticker 100 times. Only the deviation could or will be larger when having 100 stickers compared to 10,000.
Or is this not what you meant?
Of course that number of copies matter. If you have 1 copy of each sticker, you have 100% chance that you will get unique stickers every time. If you have 100 copies of each sticker, your chances are lower to get unique stickers. If you have 10 000, chances are way lower. Do you agree with that?
About 1 copy you are right. But with more copies there Are no differences.
The second sticker with 100 copies has the following chance to be unique:
15,900/16,000=99.38%.
With 1,000 copies you get:
159,000/160,000=99.38%.
No you are wrong 😃 only one thing is same. Amount of unique stickers represents 0.625% of all stickers. Doesn't matter if there's 1 or 10000 copies of each sticker. But if number of copies grow one thing is changing. At 1 copie of each sticker, one sticker represents 0.625% of total amount of stickers but with 10 copie of each stickers, one sticker represents 0.0625% and so on. That is why chances are not the same to find unique sticker if number of copies are growing. Do you understand? 😀
I'm afraid I still don't understand.
How would you calculate the chance that the 2nd sticker is unique?
Both for a situation where each sticker has 100 and/or 1,000 copies?
Look, when you draw first sticker, if it is only 1 copy of each sticker, there is 159 stickers left and 0 chance to pick same. If we have hundred copies, I you draw one stiker, 15999 stickers left and there is 99 more copies of sticker that you draw and there is chance to draw same sticker again. So you don't divide 15900 like you did, you should divide 15999 because there is still copies of first drawn stickers inside. With more copies of stickers, if you draw some sticker there is more copies of that sticker inside amount of left stickers and higher chance to draw it again. Do you understand now?
You are right about that I should divide by 15,999 instead of 16,000. Cause the first sticker is not anymore available.
With 100 copies that would give:
15,900/15,999=99.381%.
With 1,000 copies you get:
159,000/159,999=99.376%.
Based on this calculation I can pull the conclusion that with more copies there is a fractional smaller chance that the second card is unique. Which means there is also a slightly bigger chance that it will be the same card as the first one.
I guess that is what you mean?
Please confirm?
Lets go for E! No math behind it just a guess 😂
No clue but I was going to go with E as my name starts with it but @fullcoverbetting has that one, so lets go with F as it looks a lot like E. hahaha... :D
To listen to the audio version of this article click on the play image.
Brought to you by @tts. If you find it useful please consider upvoting this reply.
I missed out on the contest in the first place.
According to my calculations it is: D (8,12%)
I guess you could already do this without any hints?
Do I already know you day to day job?
I could, but the hints did help.
I'm a traffic and transport consultant with a special interest in data analysis.
8.12% is in case that there is unlimited number of cards, and are equally divided (same number of each card)
In case where there is limited number of cards (like in this case 10000) it should be little less, but again answer D is correct :)
Maybe a stupid question, but was there actually any prize for this quiz? Or can I only win eternal fame?
There are no stupid questions!
True, only stupid answers ;)
I was still thinking about a prize. At first I thought you guys don't earn one, since no one could answer this question 🤨
But it was just too difficult. And this time there is a lot of interaction. Which is a good thing. So a prize is a good thing.
What prize would you prefer?
An SBI-share is always nice, bu I don't know whether you are still confident about the use/ROI of that. You could also award a certain percentage of the (SBD) payout of the post.
On the other hand, eternal fame would do it for me too.
Then I should create a all time ranking containing all previous winners 😀
I don't know how SBI is right now. What actually should happen is that the Steem price goes up. And then you should repeat your calculation.
Paying out part of the SBD is not really a wise choice of someone used paid bid bots.
An all time "Hall of fame" would be awesome 👌
Not sure I saw your first post on this.
D is the right answer.
I still can't wrap my head around it. I'll go for E.