mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

Big Internet Math-Off stickers 2019

 2019-07-03 
This year's Big Internet Math-Off is now underway with 15 completely new contestants (plus one returning contender). As I'm not the returning contestant, I haven't been spending my time preparing my pitches. Instead, I've spent my time making an unofficial Big Internet Math-Off sticker book.
To complete the sticker book, you will need to collect 162 different stickers. Every day, you will be given a pack of 5 stickers; there are also some bonus packs available if you can find them (Hint: keep reading).

How many stickers will I need?

Using the same method as I did for last year's World Cup sticker book, you can work out that the expected number of stickers needed to finish the sticker book:
If you have already stuck \(n\) stickers into your album, then the probability that the next sticker you get is new is
$$\frac{162-n}{162}.$$
The probability that the second sticker you get is the next new sticker is
$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$ $$=\frac{n}{162}\times\frac{162-n}{162}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next new sticker is
$$\left(\frac{n}{162}\right)^{i-1}\times\frac{162-n}{162}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:
$$\sum_{i=1}^{\infty}i \left(\frac{162-n}{162}\right) \left(\frac{n}{162}\right)^{i-1} = \frac{162}{162-n}$$
Therefore, to get all 162 stickers, you should expect to buy
$$\sum_{n=0}^{161}\frac{162}{162-n} = 918 \text{ stickers}.$$
Using just your daily packs, it will take you until the end of the year to collect this many stickers. Of course, you'll only need to collect this many if you don't swap your duplicate stickers.

How many stickers will I need if I swap?

To work out the expected number of stickers stickers you'd need if you swap, let's first think about two people who want to complete their stickerbooks together. If there are \(a\) stickers that both collectors need and \(b\) stickers that one collector has and the other one needs, then let \(E_{a,b}\) be the expected number of stickers they need to finish their sticker books. The next sticker they get could be one of three things:
Therefore, the expected number of stickers they need to complete their sticker books is
$$E_{a,b}=1+\frac{a}{162}E_{a-1,b+1}+\frac{b}{162}E_{a,b-1}+\frac{162-a-b}{162}E_{a,b}.$$
This can be rearranged to give
$$E_{a,b}= \frac{162}{a+b}+ \frac{a}{a+b}E_{a-1,b+1} +\frac{b}{a+b}E_{a,b-1} $$
We know that $E_{0,0}=0$ (as if \(a=0\) and \(b=0\), both collectors have already finished their sticker books). Using this and the formula above, we can work out that
$$E_{0,1}=162+E_{0,0}=162$$ $$E_{1,0}=162+E_{0,1}=324$$ $$E_{0,2}=\frac{162}2+E_{0,1}=243$$ $$E_{1,1}=\frac{162}2+\frac12E_{0,2}+\frac12E_{1,0}=364.5$$
... and so on until we find that \(E_{162,0}=1269\), and so our collectors should expect to collect 634 stickers each to complete their sticker books.
For three people, we can work out that if there are \(a\) stickers that all three need, \(b\) stickers that two need, and \(c\) stickers that one needs, then
$$ E_{a,b,c} = \frac{162}{a+b+c}+ \frac{a}{a+b+c}E_{a-1,b+1,c} +\frac{b}{a+b+c}E_{a,b-1,c+1} +\frac{c}{a+b+c}E_{a,b,c-1}. $$
In the same way as for two people, we find that \(E_{162,0,0}=1572\), and so our collectors should expect to collect 524 stickers each to complete their sticker books.
Doing the same thing for four people gives an expected 463 stickers required each.
After four people, however, the Python code I wrote to do these calculations takes too long to run, so instead I approximated the numbers by simulating 500 groups of \(n\) people collecting stickers, and taking the average number of stickers they needed. The results are shown in the graph below.
The red dots are the expected values we calculated exactly, and the blue crosses are the simulated values. It looks like you'll need to collect at least 250 stickers to finish the album: in order to get this many before the end of the Math-Off, you'll need to find 20 bonus packs...
Of course, these are just the mean values and you could get lucky and need fewer stickers. The next graph shows box plots with the quartiles of the data from the simulations.
So if you're lucky, you could complete the album with fewer stickers or fewer friends.
As a thank you for reading to the end of this blog post, here's a link that will give you two bonus packs and help you on your way to the 250 expected stickers...

Similar posts

Runge's Phenomenon
World Cup stickers 2018, pt. 3
Mathsteroids
World Cup stickers 2018, pt. 2

Comments

Comments in green were written by me. Comments in blue were not written by me.
 2019-07-06 
@Pat Ashforth: Thanks, fixed
Reply
Matthew
 2019-07-04 
Link to sticker book, in the first paragraph, does not work. It points to mathoffstickbook.com
Reply
Pat Ashforth
 2019-07-04 
@Road: Thanks, fixed
Reply
Matthew
 2019-07-03 
minor typo for the 2 collector case


> and so our collectors should expect to collect 364 stickers

should be 634.
Reply
Road
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "tnemges" backwards in the box below (case sensitive):

Archive

Show me a random blog post
 2019 

Sep 2019

A non-converging LaTeX document
TMiP 2019 treasure punt

Jul 2019

Big Internet Math-Off stickers 2019

Jun 2019

Proving a conjecture

Apr 2019

Harriss and other spirals

Mar 2019

realhats

Jan 2019

Christmas (2018) is over
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

hexapawn realhats dragon curves martin gardner golden ratio golden spiral reddit estimation statistics asteroids cross stitch graph theory royal baby reuleaux polygons gerry anderson sport mathsteroids pythagoras pizza cutting python arithmetic chess nine men's morris light radio 4 game of life world cup cambridge map projections palindromes logic dates php propositional calculus bodmas misleading statistics electromagnetic field polynomials error bars books menace hats approximation plastic ratio craft football a gamut of games stickers dataset latex fractals flexagons harriss spiral oeis geometry matt parker draughts probability speed weather station ternary final fantasy european cup accuracy captain scarlet braiding machine learning folding tube maps interpolation trigonometry javascript platonic solids wool noughts and crosses puzzles programming raspberry pi video games frobel manchester science festival rugby mathsjam inline code london sorting sound chalkdust magazine news manchester game show probability go christmas curvature national lottery games tennis pac-man tmip mathslogicbot the aperiodical rhombicuboctahedron folding paper twitter people maths triangles talking maths in public data london underground big internet math-off binary chebyshev bubble bobble coins countdown christmas card

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2019