The Math (and money) of Soccer Stickers - Numberphile
video description
even if they allow you to buy your last 50 the number of low probability stickers would be more than 50.
Paninni does not disclose the odds of finding a certain player. but anyone who collected them knows that there are stickers very hard to find, to the point that even if you trade with a hundred friend, maybe one of them will be lucky enough to get the rarest one.
Conclusion: Without Paninni disclosing the odds, no math formula could ever be made to predict the numbers in the video.
Date: 2022-04-08
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Comments and reviews: 9
Merch
The math used is correct, but it is missing a very important variable that makes the whole result wrong. The cost will be actually mutch much higher than projected, because of this: paninni does not make the availability of the stickers the same, meaning that the chance of finding the player Benatia are 1 in 80 and the chances of finding the player Pepe are 1 in 450 the time that the chances of finding Ronaldo or Messi or some other rare sticker are 1 in 9000. this happens every world cup, I remember in 1986 everyone was missing the Maradona sticker.
even if they allow you to buy your last 50 the number of low probability stickers would be more than 50.
Paninni does not disclose the odds of finding a certain player. but anyone who collected them knows that there are stickers very hard to find, to the point that even if you trade with a hundred friend, maybe one of them will be lucky enough to get the rarest one.
Conclusion: Without Paninni disclosing the odds, no math formula could ever be made to predict the numbers in the video.
reply
The math used is correct, but it is missing a very important variable that makes the whole result wrong. The cost will be actually mutch much higher than projected, because of this: paninni does not make the availability of the stickers the same, meaning that the chance of finding the player Benatia are 1 in 80 and the chances of finding the player Pepe are 1 in 450 the time that the chances of finding Ronaldo or Messi or some other rare sticker are 1 in 9000. this happens every world cup, I remember in 1986 everyone was missing the Maradona sticker.
even if they allow you to buy your last 50 the number of low probability stickers would be more than 50.
Paninni does not disclose the odds of finding a certain player. but anyone who collected them knows that there are stickers very hard to find, to the point that even if you trade with a hundred friend, maybe one of them will be lucky enough to get the rarest one.
Conclusion: Without Paninni disclosing the odds, no math formula could ever be made to predict the numbers in the video.
reply
Merch
The math used is correct, but it is missing a very important variable that makes the whole result wrong. The cost will be actually mutch much higher than projected, because of this: paninni does not make the availability of the stickers the same, meaning that the chance of finding the player Benatia are 1 in 80 and the chances of finding the player Pepe are 1 in 450 the time that the chances of finding Ronaldo or Messi or some other rare sticker are 1 in 9000. this happens every world cup, I remember in 1986 everyone was missing the Maradona sticker.
even if they allow you to buy your last 50 the number of low probability stickers would be more than 50.
Paninni does not disclose the odds of finding a certain player. but anyone who collected them knows that there are stickers very hard to find, to the point that even if you trade with a hundred friend, maybe one of them will be lucky enough to get the rarest one.
Conclusion: Without Paninni disclosing the odds, no math formula could ever be made to predict the numbers in the video.
reply
The math used is correct, but it is missing a very important variable that makes the whole result wrong. The cost will be actually mutch much higher than projected, because of this: paninni does not make the availability of the stickers the same, meaning that the chance of finding the player Benatia are 1 in 80 and the chances of finding the player Pepe are 1 in 450 the time that the chances of finding Ronaldo or Messi or some other rare sticker are 1 in 9000. this happens every world cup, I remember in 1986 everyone was missing the Maradona sticker.
even if they allow you to buy your last 50 the number of low probability stickers would be more than 50.
Paninni does not disclose the odds of finding a certain player. but anyone who collected them knows that there are stickers very hard to find, to the point that even if you trade with a hundred friend, maybe one of them will be lucky enough to get the rarest one.
Conclusion: Without Paninni disclosing the odds, no math formula could ever be made to predict the numbers in the video.
reply
mwb911
That's still not a guarantee! If you're missing one sticker, you're not guaranteed to get it even if you buy 682/5 packs. The unluckiest person in the world could buy packs forever and still not find it the missing sticker. I think probability needs to come in to this. Many years ago I was collecting items that came in chip bags (1 per bag. Assuming a uniform distribution (an equal chance of pulling any item from any bag, I wanted to know the probability P of completing the set of N items after buying B bags. The formula I found is: P = (SUM( from: i=1, to: N-1, COMB(N, i) - -1-i - (N - i)-B ) / (N-B) where COMB(x, y) is combinations x CHOOSE y and where x-y is x to the power of y. In this case, N = 682 and B should be a multiple of 5 since there are 5 cards per pack. It takes some trials to find the desired probability, like say, 99. 9%. I never did derive the formula for finding B given N and P. If someone can, please reply.
reply
That's still not a guarantee! If you're missing one sticker, you're not guaranteed to get it even if you buy 682/5 packs. The unluckiest person in the world could buy packs forever and still not find it the missing sticker. I think probability needs to come in to this. Many years ago I was collecting items that came in chip bags (1 per bag. Assuming a uniform distribution (an equal chance of pulling any item from any bag, I wanted to know the probability P of completing the set of N items after buying B bags. The formula I found is: P = (SUM( from: i=1, to: N-1, COMB(N, i) - -1-i - (N - i)-B ) / (N-B) where COMB(x, y) is combinations x CHOOSE y and where x-y is x to the power of y. In this case, N = 682 and B should be a multiple of 5 since there are 5 cards per pack. It takes some trials to find the desired probability, like say, 99. 9%. I never did derive the formula for finding B given N and P. If someone can, please reply.
reply
Leander
There is a statistical problem I came up with while listening to a playlist of songs that I haven't been able to solve to this day. It goes like this:
So let's say you play a 100 songs playlist in alphabetical order and then you turn on the shuffle feature and play it again. My question is this: What is the probability that at least one of the songs will be played at the same rank as its alphabetical one. E. g It's the 42nd song alphabetically and it's by chance the 42nd when we play the playlist shuffled. Can you help me solve this?
Clarification: The shuffle mechanic works like this: It always picks a song at random to play as long as it hasn't been played until that point.
reply
There is a statistical problem I came up with while listening to a playlist of songs that I haven't been able to solve to this day. It goes like this:
So let's say you play a 100 songs playlist in alphabetical order and then you turn on the shuffle feature and play it again. My question is this: What is the probability that at least one of the songs will be played at the same rank as its alphabetical one. E. g It's the 42nd song alphabetically and it's by chance the 42nd when we play the playlist shuffled. Can you help me solve this?
Clarification: The shuffle mechanic works like this: It always picks a song at random to play as long as it hasn't been played until that point.
reply
KuraIthys
Hmmh. This is the same problem I was trying to work out with lego's collectible minifigures a while ago. (you can cheat a little since you can sort of tell them apart by feel, but roughly the same principle applies)
Trading card games have the same issue but with the added complication that cards are unevenly distributed. (the rarity composition of what's in a pack isn't, but the set as a whole typically has 3 rarity values - I suppose you can sort of simplify it to taking what is involved in collecting the rarest type, and assume that you'll have duplicates of the more common types in proportion to the distribution)
reply
Hmmh. This is the same problem I was trying to work out with lego's collectible minifigures a while ago. (you can cheat a little since you can sort of tell them apart by feel, but roughly the same principle applies)
Trading card games have the same issue but with the added complication that cards are unevenly distributed. (the rarity composition of what's in a pack isn't, but the set as a whole typically has 3 rarity values - I suppose you can sort of simplify it to taking what is involved in collecting the rarest type, and assume that you'll have duplicates of the more common types in proportion to the distribution)
reply
Chile
I admire Zlatan Ibrahimovic
He is an Real Tae-kwon-do Soccer Genius
He is above of Lord
No one can be instead of him and Every Soccer wants to learn from him!
Uruguay won 2018 Free China Cup for Luis Suarez, Edinson Cavani (This kick look like Zlatan Ibrahimovic) and Goalkeeper Fernando Muslera)
Last: Cristiano Ronaldo is So rock this is because he paints his toenails
Actually, Cristiano Ronaldo is friendly with Messi this is because he and his son are older than Messi 869 days
This is because C. Ronaldo take care of Messi's son and Messi take care of C Ronaldo Sometimes
reply
I admire Zlatan Ibrahimovic
He is an Real Tae-kwon-do Soccer Genius
He is above of Lord
No one can be instead of him and Every Soccer wants to learn from him!
Uruguay won 2018 Free China Cup for Luis Suarez, Edinson Cavani (This kick look like Zlatan Ibrahimovic) and Goalkeeper Fernando Muslera)
Last: Cristiano Ronaldo is So rock this is because he paints his toenails
Actually, Cristiano Ronaldo is friendly with Messi this is because he and his son are older than Messi 869 days
This is because C. Ronaldo take care of Messi's son and Messi take care of C Ronaldo Sometimes
reply
James
I remember working something like this out a long time ago and my formula had a value for 'odds of getting a full set'. If you just wanted a 50% chance then I seemed to recall you needed to get about 5x the number of stickers. If you wanted a 99% chance you needed quite a bit more.
The odds of getting a full set are not in this formula. I have to wonder what they are here -- (I'm guessing it's about 1 standard deviation. You can theoretically buy 10000 stickers and STILL not complete a set if you're unlucky enough.
I never did work out a 'trading with friends' version however.
reply
I remember working something like this out a long time ago and my formula had a value for 'odds of getting a full set'. If you just wanted a 50% chance then I seemed to recall you needed to get about 5x the number of stickers. If you wanted a 99% chance you needed quite a bit more.
The odds of getting a full set are not in this formula. I have to wonder what they are here -- (I'm guessing it's about 1 standard deviation. You can theoretically buy 10000 stickers and STILL not complete a set if you're unlucky enough.
I never did work out a 'trading with friends' version however.
reply
Xavier
I just realized in the formula, the numerator of the quotient is the (f-1)th Maclaurin polynomial for e-x. This means as your number of friends gets bigger, the quotient approaches 1 and the integrand itself approaches 1.
This raises a little problem though, at least in terms of pure math. If you have an infinite number of friends, then, as mentioned, the integrand is 1. But then the improper integral diverges. On top of that, the fraction outside the integral approaches zero.
Can anyone say what this would translate to in real life?
reply
I just realized in the formula, the numerator of the quotient is the (f-1)th Maclaurin polynomial for e-x. This means as your number of friends gets bigger, the quotient approaches 1 and the integrand itself approaches 1.
This raises a little problem though, at least in terms of pure math. If you have an infinite number of friends, then, as mentioned, the integrand is 1. But then the improper integral diverges. On top of that, the fraction outside the integral approaches zero.
Can anyone say what this would translate to in real life?
reply
Eldmyra11
You didn't account for the fact that there are no duplicates in one pack. So for your first five unique cards you would always need exactly one pack. And if you open a pack with the first four cards all already in your collection, the fifth card has a greater chance than the first of being a new one, as it now can't be four of those you already have. Although the packs are independent, the cards within a pack are not. Due to this effect the pack size should affect the expected number of packs and therefore be visible in the formula.
reply
You didn't account for the fact that there are no duplicates in one pack. So for your first five unique cards you would always need exactly one pack. And if you open a pack with the first four cards all already in your collection, the fifth card has a greater chance than the first of being a new one, as it now can't be four of those you already have. Although the packs are independent, the cards within a pack are not. Due to this effect the pack size should affect the expected number of packs and therefore be visible in the formula.
reply
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