mscroggs.co.uk
mscroggs.co.uk

subscribe

Comment

Comments

Comments in green were written by me. Comments in blue were not written by me.
@Matthew: Here is how I calculated it:

You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
on /blog/56
               
@Matthew: Thank you for the calculations. Good job I ordered the stickers I wanted #IRN. 2453 stickers - that's more than the number you bought (1781) to collect all stickers!
Milad
on /blog/56
               
@Matthew: Here is how I calculated it:

You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
on /blog/56
               
@Matthew: Thank you for the calculations. Good job I ordered the stickers I wanted #IRN. 2453 stickers - that's more than the number you bought (1781) to collect all stickers!
Milad
on /blog/56
               
@Matthew: Here is how I calculated it:

You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
on /blog/56
               

Archive

Show me a random blog post
 2024 

Feb 2024

Zines, pt. 2

Jan 2024

Christmas (2023) is over
 2023 
▼ show ▼
 2022 
▼ show ▼
 2021 
▼ show ▼
 2020 
▼ show ▼
 2019 
▼ show ▼
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

data visualisation christmas card logs sound final fantasy estimation newcastle boundary element methods christmas approximation light realhats bodmas matrices asteroids go anscombe's quartet arithmetic tmip sorting php fractals dinosaurs hyperbolic surfaces fonts javascript martin gardner python manchester science festival mathsjam cross stitch flexagons gaussian elimination weather station gather town interpolation tennis edinburgh reuleaux polygons ucl fence posts gerry anderson royal institution football propositional calculus squares world cup numerical analysis determinants menace radio 4 news phd noughts and crosses mean game show probability pythagoras simultaneous equations sobolev spaces standard deviation geogebra geometry stickers advent calendar chalkdust magazine pizza cutting london underground oeis london errors crossnumber draughts runge's phenomenon sport talking maths in public nine men's morris manchester computational complexity correlation hannah fry plastic ratio signorini conditions puzzles misleading statistics probability chebyshev games convergence logic coins graph theory books rhombicuboctahedron inline code matt parker golden spiral countdown rugby bempp folding paper royal baby captain scarlet map projections mathsteroids databet national lottery pi approximation day data matrix of minors cambridge chess statistics numbers folding tube maps bubble bobble ternary wool guest posts graphs recursion frobel electromagnetic field triangles trigonometry turtles hats palindromes people maths zines 24 hour maths weak imposition craft polynomials youtube machine learning exponential growth inverse matrices dragon curves programming matrix multiplication dates logo accuracy pascal's triangle curvature raspberry pi binary quadrilaterals platonic solids finite group latex game of life european cup hexapawn live stream harriss spiral the aperiodical matrix of cofactors crochet preconditioning error bars datasaurus dozen finite element method stirling numbers pac-man wave scattering a gamut of games big internet math-off speed mathslogicbot braiding video games pi reddit dataset golden ratio

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2024