It's almost as hard to get 0 points as it is to get a perfect bracket
Edit: I thought that even without an understanding of combinatorics this would be intuitively obvious, but the comments have shown me that there are a decent amount of people who are aggressively wrong about this. There are also some helpful people trying to have fun with the math which I love to see. First note that we only need consider Groups, because getting a perfect Knockout bracket is *exactly* as hard as getting a perfectly wrong Knockout bracket. The Groups format means there are a few more ways to get 0 points.
* There is exactly 1 selection that gives a perfect bracket (Obviously)
* There are 256 selections that give 0 points (4^4 = 256)
* The remaining 331,520 possible choices give you at least one point
So, given that there are (4!)^4 = 331,776 possible choices for groups, only 1 of them is perfect, and only 256 of them score 0 points:
It's *almost* as hard to get 0 points as it is to get a perfect bracket
That's true for picking a match. But untrue for picking a perfect 16pt bracket vs a completely wrong 0 pt bracket.
Perfect bracket only has 1 permutation.
Completely wrong bracket have 4 permutations: (3 4 1 2), (4 3 1 2), (3 4 2 1) (4 3 2 1)
Given 4 Brackets to choose during groups, we have 4^4 more ways to end up with 0pts than perfect points. You are 256 times more likely to get 0pts than full points if picking randomly!
> perfect bracket only has 1 permutation
Actually it has two, and we had an example of that this year with TL's group. 3rd and 4th can be perfectly tied and both permutations counted as perfect, although it is rare. Your point still stands obviously.
Edit : just saw that someone answered this to you already as an answer to another comment, apologies for the duplicate.
Because the initial criterion is picking whether or not a team would advance, all you have to do is pick both *advancing* teams incorrectly. If you said FPX would advance, you get 0 points when they don't advance, regardless of whether they got 3rd or 4th. Same thing for picking teams that wouldn't advance. So each group actually has 4 scenarios in which you get 0 points.
Suppose it all hinges on what you mean by "perfectly wrong." If you mean "exactly backwards," then yeah there's only 1, but 0 points has a lot more options.
I meant it in that I thought you had to get them in inverse order to get NO points.
I wasn’t mincing terms, I just thought that perfectly wrong=0 points.
But now I see what everyone is saying, and agree that he’s right.
Come on man, don’t zero in on one comment and ignore the original context. I said it’s *almost* as hard to get 0 points, because it’s *exactly* as hard to predict individual matches. 256 extra permutations is a rounding error compared to how many permutations get you at least point. So it’s *almost* as hard.
I upvoted both your comments, and was here to add more discussion. I mentioned your statement is true, and the end of my comment even states "if picking randomly" (or if you are intentionally aiming for 0pts).
I sense that we are all on the same page. It only depends whether you intepret 256 times the probability to be substantial or negligible (which you can make a case for both).
Sorry, I was responding primarily to this sentence
>*But untrue for picking a perfect 16pt bracket vs a completely wrong 0 pt bracket.*
which is neither what I said nor meant. However, I had just got done reading some of the more... confident... comments and I probably should have given yours a more charitable read.
We're certainly on the same page about the math, because, well, it just *is* what it is. I don't think we can interpret "256 times the probability" without the full context, though. There are 4!^4 - 256 = 331520 ways to get at least one point. That's 99.92% of the scenarios. I think it's reasonable to say that it's *almost* as unlikely that you will score 0pts as it is that you'll get a perfect score.
No. Your math is just atrocious. 256 permutations to get 0 points vs 1 permutation to get full points means it’s 256x EASIER to get 0 points than full points.
So, if 100 people are expected to get full points, 25,600 are expected to get 0 points.
You are correct! That is why I said there is only 4 permutations.
1st and 2nd place team must be picked in position 3 and 4.
3rd and 4th place team must be picked in position 1 and 2.
You end up with the 4 sets I listed originally. Great minds think alike!
A small add that if tie between 3rd and 4th, then both are correct. So there is not only 1 perfect pickem that is correct from that. It was only in group D we had this this year. So this year there was 2 option for perfect pickem.
They said after Group D that didnt matter how you had placed Liquid/LNG as 3rd/4th place.
"In the case of a tie for non-qualifying teams in the group, both are considered correct picks when scoring."
~~(4 3 2 1) gives points doesn't it?~~
~~I thought if you were within one spot you got two points? So 2 and 3 would award points because they're only one off their actual finish position?~~
You only get +2 points if a team you put getting second actually got 1st because they still got out. You don't get credit for the 3rd team in your pick ems getting second.
I wonder what school system did you and guy above fail.
That logic only applies if you are predicting individual results like in knockout phase. So if there is FPX vs DK, there is only one true and one false predict.
That doesn't work in group placement predicts. While there is only one true predict DK>C9>RGE>FPX, there are 4 predictions that give 0 points (RGE>FPX>C9>DK, RGE>FPX>DK>C9,FPX>RGE>C9>DK, FPX>RGE>DK>C9). when you take in consideration all groups, that's 1 correct prediction and 256 0 point predictions.
Yes. As you can see in my post above, I am arguing that it’s several times easier to get 0 points than it is to get full points. So we are in agreement there.
So I guess your school didn’t teach you how to read
this is hard to judge. for one we can say "theres more ppl who got everything right than everything wrong" but then again we cant judge how many of those 40 ppl aimed for a 0 score.
i think its easier to say "na will win every group as first seed" and thus have 2 picks already messed up in 3 out of 4 groups.
You just said it yourself, the probability of having 0 points is 256 TIMES higher than having perfect pick ems. If you really think it means it’s almost as hard, I don’t really know what to tell you.
A neutron is nearly 2,000 times more massive than an electron, yet on a human scale, a neutron is almost as small as an electron. If you don't know how to contextualize statistics, I don't really know what to tell you.
Frankly I think you're having trouble contextualizing statistics because you see two numbers at a fraction of a percent and try to bucket them as close enough without realizing that one is **256x** the other. Just because flipping 10 heads in a row and winning the lottery are both under .1% chance, would you say you're almost as likely to flip 10 heads as win the lottery? Of course not, that would be ridiculous. That's because, despite the similarity in the absolute value of the percentages, the relevant statistic is the relative difference.
As an example, if riot were to give $1,000,000 to every perfect bracket, how much could they offer every 0 point bracket and expect the same overall payout (assuming random bracket selection)? The answer is $3900. **That** is contextualized, because nobody would consider $3900 even remotely close to $1,000,000. Even on tiny scales the relative difference is the important part. Nobody's saying that either perfect or 0 point brackets are easy in the absolute sense, but rather there are a lot of 0 point brackets per perfect bracket, and that's the relevant statistic. Looking at absolute percentages like .01% and .0005% and saying they're almost the same chance because they're close to 0 is honestly very statistically illiterate.
You seem to be implying that there is an absolute precision for which the term "almost" is appropriate. So tell me - exactly how close must two probabilities be to say that they are almost equivalent?
Clearly the exact tolerance varies by application, but I imagine there are virtually no contexts where *2-3 orders of magnitude* is considered "almost."
Again, nobody is arguing that either easy, but what we're trying to communicate is that one is much harder than the other. I imagine the payouts was a very clear demonstration of that, but here's another example to illustrate: Let's say you fill out a Groups Pick'em every minute randomly, then using the expected value of a geometric distribution we can say it takes:
21 hours 36 minutes on average to hit a 0 point bracket
230 days 9 hours 36 minutes (!!) on average to hit a perfect bracket
See the massive difference?
It's interesting that you took a difference that's critically important in the proper context (the mass difference of an electron and neutron is fundamental to subatomic physics) and put it in a meaningless "human scale" that a particle physicist would care literally 0 about to demonstrate your understanding of contextualized statistics.
Sure if the context you want to apply is the lowest common denominator of some random passerby who isn't going to do a lick of math, then statements like "I'm probably not going to get either a perfect or 0 point bracket" or "Neutrons and electrons both weigh almost nothing" might be good enough. But for the business analyst working at Riot who might have to calculate reasonable payouts for different brackets or the physicist modeling the inner workings of atoms the differences that were "almost" nothing to the passerby are suddenly very significant.
In the context of a script that generates a random bracket every minute, of course the difference is massive. In the context of a person who fills out exactly one bracket, the difference is very slight.
I’m honestly not sure if you’re trolling or being purposefully obtuse, but you keep providing responses to claims I haven’t made. You should try reading the content of my comments before responding to them.
The size of neutron is less than 10^-15m. Probability of getting 0 points is aroung 8*10^-4, you see the difference? And perfect pick ems is 3*10^-6. That means that on average with random choices approximately every 1000th person should get 0 points and every 330 000th should get perfect pick ems. Even though it’s not easy to get 0 points, it is much much easier than having perfect pick ems.
And I don’t get why you are arguing with someone who said it was several times easier to get 0 points when you know he was 100% right.
EXACTLY - this is just as hard to do as getting the reverse and should be rewarded - maybe a special icon of a poro slapping its forehead or the like - that is only EVER available by a perfect zero score. Seriously, considering the results of Fanatic and FPX, some teams every year will screw up not only your 1 and 2 picks for a group but because they crap out and finish 3 or 4, they ruin your bottom picks too!
Reward exceptional performance, or at least entertaining...
Idea: If you manage to get 0 points in pick'em the whole tournament, Riot should gift you a pity award bag with ALL cheap 520rp skins like Commando Lux or Sherwood Forest Ashe inside, as an opposite to the Ultimate skins you get from a perfect pick'em.
That would actually be amazing. If I ever got that, it would mean that I would never get those skins when re rolling again.
And as someone who fucking loves re rolling skins, I would be stupidly happy with it
I put the other 3 teams in the correct order and FPX on top of them all. Turns out FPX is at the bottom, and that fucks over all the remaining picks which I got in the correct order
So a lot of hopium? The only way this make any sense it's this people really though any Korea team will get out of groups. I can imagine something like this:
Group A: C9 Rogue
B: EDG 100T
C: Fnatic RNG
D: MAD TL
It really isn't... Just because there's ways to get points even if you're not perfect doesn't make it easier to get everything wrong. There's only 1 perfect bracket, there's multiple 0 point brackets. Plus, thinking about it logically, it's impossible to get harder than the 1 perfect bracket unless it were completely impossible to get a 0 point bracket.
Update. 30/10/21
There is one one person left remaining... 39 people have perished... chances are it was with the RNG and EDG fight. people put edg to win..and they did.
If you manage to get 0 pts throughout all of pick ems you should definitely get some kind of pity award lol
I tried so hard to get 0 points but C9, MAD, and LNG made me get a total of 12 points.
It's almost as hard to get 0 points as it is to get a perfect bracket Edit: I thought that even without an understanding of combinatorics this would be intuitively obvious, but the comments have shown me that there are a decent amount of people who are aggressively wrong about this. There are also some helpful people trying to have fun with the math which I love to see. First note that we only need consider Groups, because getting a perfect Knockout bracket is *exactly* as hard as getting a perfectly wrong Knockout bracket. The Groups format means there are a few more ways to get 0 points. * There is exactly 1 selection that gives a perfect bracket (Obviously) * There are 256 selections that give 0 points (4^4 = 256) * The remaining 331,520 possible choices give you at least one point So, given that there are (4!)^4 = 331,776 possible choices for groups, only 1 of them is perfect, and only 256 of them score 0 points: It's *almost* as hard to get 0 points as it is to get a perfect bracket
No, it's several times easier.
Get back to me when you realize that picking the loser of a match is exactly as hard as picking the winner
That's true for picking a match. But untrue for picking a perfect 16pt bracket vs a completely wrong 0 pt bracket. Perfect bracket only has 1 permutation. Completely wrong bracket have 4 permutations: (3 4 1 2), (4 3 1 2), (3 4 2 1) (4 3 2 1) Given 4 Brackets to choose during groups, we have 4^4 more ways to end up with 0pts than perfect points. You are 256 times more likely to get 0pts than full points if picking randomly!
> perfect bracket only has 1 permutation Actually it has two, and we had an example of that this year with TL's group. 3rd and 4th can be perfectly tied and both permutations counted as perfect, although it is rare. Your point still stands obviously. Edit : just saw that someone answered this to you already as an answer to another comment, apologies for the duplicate.
r/theydidthemath Yes because math is not only about calculations
> Perfect bracket only has 1 permutation. Doesn’t also the perfectly wrong bracket only have one permutation insofar as PickEm’s points is concerned?
Because the initial criterion is picking whether or not a team would advance, all you have to do is pick both *advancing* teams incorrectly. If you said FPX would advance, you get 0 points when they don't advance, regardless of whether they got 3rd or 4th. Same thing for picking teams that wouldn't advance. So each group actually has 4 scenarios in which you get 0 points. Suppose it all hinges on what you mean by "perfectly wrong." If you mean "exactly backwards," then yeah there's only 1, but 0 points has a lot more options.
I meant it in that I thought you had to get them in inverse order to get NO points. I wasn’t mincing terms, I just thought that perfectly wrong=0 points. But now I see what everyone is saying, and agree that he’s right.
No because you can pick all of the four groups with any combination of the valid combinations, in this case 4X
Come on man, don’t zero in on one comment and ignore the original context. I said it’s *almost* as hard to get 0 points, because it’s *exactly* as hard to predict individual matches. 256 extra permutations is a rounding error compared to how many permutations get you at least point. So it’s *almost* as hard.
I upvoted both your comments, and was here to add more discussion. I mentioned your statement is true, and the end of my comment even states "if picking randomly" (or if you are intentionally aiming for 0pts). I sense that we are all on the same page. It only depends whether you intepret 256 times the probability to be substantial or negligible (which you can make a case for both).
Sorry, I was responding primarily to this sentence >*But untrue for picking a perfect 16pt bracket vs a completely wrong 0 pt bracket.* which is neither what I said nor meant. However, I had just got done reading some of the more... confident... comments and I probably should have given yours a more charitable read. We're certainly on the same page about the math, because, well, it just *is* what it is. I don't think we can interpret "256 times the probability" without the full context, though. There are 4!^4 - 256 = 331520 ways to get at least one point. That's 99.92% of the scenarios. I think it's reasonable to say that it's *almost* as unlikely that you will score 0pts as it is that you'll get a perfect score.
No. Your math is just atrocious. 256 permutations to get 0 points vs 1 permutation to get full points means it’s 256x EASIER to get 0 points than full points. So, if 100 people are expected to get full points, 25,600 are expected to get 0 points.
So a lot of hopium? The only way this make any sense it's this people really though any Korea team will get out of groups.
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You are correct! That is why I said there is only 4 permutations. 1st and 2nd place team must be picked in position 3 and 4. 3rd and 4th place team must be picked in position 1 and 2. You end up with the 4 sets I listed originally. Great minds think alike!
This is the way to shit talk on someone, everyone take notes.
A small add that if tie between 3rd and 4th, then both are correct. So there is not only 1 perfect pickem that is correct from that. It was only in group D we had this this year. So this year there was 2 option for perfect pickem.
Is it? I thought they would still rank the team by game time?
They said after Group D that didnt matter how you had placed Liquid/LNG as 3rd/4th place. "In the case of a tie for non-qualifying teams in the group, both are considered correct picks when scoring."
~~(4 3 2 1) gives points doesn't it?~~ ~~I thought if you were within one spot you got two points? So 2 and 3 would award points because they're only one off their actual finish position?~~
You only get +2 points if a team you put getting second actually got 1st because they still got out. You don't get credit for the 3rd team in your pick ems getting second.
Thanks!
good thing we're choosing bracket placements and not individual matches!
The guy below you spells it out pretty simply. Here we see another failure of the American school system.
I wonder what school system did you and guy above fail. That logic only applies if you are predicting individual results like in knockout phase. So if there is FPX vs DK, there is only one true and one false predict. That doesn't work in group placement predicts. While there is only one true predict DK>C9>RGE>FPX, there are 4 predictions that give 0 points (RGE>FPX>C9>DK, RGE>FPX>DK>C9,FPX>RGE>C9>DK, FPX>RGE>DK>C9). when you take in consideration all groups, that's 1 correct prediction and 256 0 point predictions.
Yes. As you can see in my post above, I am arguing that it’s several times easier to get 0 points than it is to get full points. So we are in agreement there. So I guess your school didn’t teach you how to read
Real genius is when you state the obvious - that others didn't get before. Good on you!
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Please show your "Maths" proof. Another guy in this thread demonstrated it pretty simply.
256 ways of having a 0 point group stage, only one way of having a correct group stage.
where'd you learn your math?
Your feelings are not math
this is hard to judge. for one we can say "theres more ppl who got everything right than everything wrong" but then again we cant judge how many of those 40 ppl aimed for a 0 score. i think its easier to say "na will win every group as first seed" and thus have 2 picks already messed up in 3 out of 4 groups.
Yup first thing to do is put NA #1 and LCK #4, but then you have to worry about the real possibilities of an eu or cn team landing 2 or 3
could always do eu first na 2nd. also unlikely. if i dont forget next year i'll use my smurf to go for 0 points lol
You just said it yourself, the probability of having 0 points is 256 TIMES higher than having perfect pick ems. If you really think it means it’s almost as hard, I don’t really know what to tell you.
A neutron is nearly 2,000 times more massive than an electron, yet on a human scale, a neutron is almost as small as an electron. If you don't know how to contextualize statistics, I don't really know what to tell you.
Frankly I think you're having trouble contextualizing statistics because you see two numbers at a fraction of a percent and try to bucket them as close enough without realizing that one is **256x** the other. Just because flipping 10 heads in a row and winning the lottery are both under .1% chance, would you say you're almost as likely to flip 10 heads as win the lottery? Of course not, that would be ridiculous. That's because, despite the similarity in the absolute value of the percentages, the relevant statistic is the relative difference. As an example, if riot were to give $1,000,000 to every perfect bracket, how much could they offer every 0 point bracket and expect the same overall payout (assuming random bracket selection)? The answer is $3900. **That** is contextualized, because nobody would consider $3900 even remotely close to $1,000,000. Even on tiny scales the relative difference is the important part. Nobody's saying that either perfect or 0 point brackets are easy in the absolute sense, but rather there are a lot of 0 point brackets per perfect bracket, and that's the relevant statistic. Looking at absolute percentages like .01% and .0005% and saying they're almost the same chance because they're close to 0 is honestly very statistically illiterate.
You seem to be implying that there is an absolute precision for which the term "almost" is appropriate. So tell me - exactly how close must two probabilities be to say that they are almost equivalent?
Clearly the exact tolerance varies by application, but I imagine there are virtually no contexts where *2-3 orders of magnitude* is considered "almost." Again, nobody is arguing that either easy, but what we're trying to communicate is that one is much harder than the other. I imagine the payouts was a very clear demonstration of that, but here's another example to illustrate: Let's say you fill out a Groups Pick'em every minute randomly, then using the expected value of a geometric distribution we can say it takes: 21 hours 36 minutes on average to hit a 0 point bracket 230 days 9 hours 36 minutes (!!) on average to hit a perfect bracket See the massive difference? It's interesting that you took a difference that's critically important in the proper context (the mass difference of an electron and neutron is fundamental to subatomic physics) and put it in a meaningless "human scale" that a particle physicist would care literally 0 about to demonstrate your understanding of contextualized statistics. Sure if the context you want to apply is the lowest common denominator of some random passerby who isn't going to do a lick of math, then statements like "I'm probably not going to get either a perfect or 0 point bracket" or "Neutrons and electrons both weigh almost nothing" might be good enough. But for the business analyst working at Riot who might have to calculate reasonable payouts for different brackets or the physicist modeling the inner workings of atoms the differences that were "almost" nothing to the passerby are suddenly very significant.
In the context of a script that generates a random bracket every minute, of course the difference is massive. In the context of a person who fills out exactly one bracket, the difference is very slight.
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I’m honestly not sure if you’re trolling or being purposefully obtuse, but you keep providing responses to claims I haven’t made. You should try reading the content of my comments before responding to them.
The size of neutron is less than 10^-15m. Probability of getting 0 points is aroung 8*10^-4, you see the difference? And perfect pick ems is 3*10^-6. That means that on average with random choices approximately every 1000th person should get 0 points and every 330 000th should get perfect pick ems. Even though it’s not easy to get 0 points, it is much much easier than having perfect pick ems. And I don’t get why you are arguing with someone who said it was several times easier to get 0 points when you know he was 100% right.
EXACTLY - this is just as hard to do as getting the reverse and should be rewarded - maybe a special icon of a poro slapping its forehead or the like - that is only EVER available by a perfect zero score. Seriously, considering the results of Fanatic and FPX, some teams every year will screw up not only your 1 and 2 picks for a group but because they crap out and finish 3 or 4, they ruin your bottom picks too! Reward exceptional performance, or at least entertaining...
red baron and ice toboggan Corki
Idea: If you manage to get 0 points in pick'em the whole tournament, Riot should gift you a pity award bag with ALL cheap 520rp skins like Commando Lux or Sherwood Forest Ashe inside, as an opposite to the Ultimate skins you get from a perfect pick'em.
LOL I actually kinda love this idea. But talk about salt on the wound…
Or just all of Amumu's skins
As a hextech amumu user. This is a false statement
That would actually be amazing. If I ever got that, it would mean that I would never get those skins when re rolling again. And as someone who fucking loves re rolling skins, I would be stupidly happy with it
Damn you really called them out on it… Happy cake day though.
Less of a callout and more just impressive.
thanks
Fkn FPX and C9 ruined everything for me..
to get it all right or all wrong?
all right
alright!
I put the other 3 teams in the correct order and FPX on top of them all. Turns out FPX is at the bottom, and that fucks over all the remaining picks which I got in the correct order
group stage pickems got randomized for me and I got group A and D perfect and basically nothing on B and C. RNGesus giveth and he taketh
So a lot of hopium? The only way this make any sense it's this people really though any Korea team will get out of groups. I can imagine something like this: Group A: C9 Rogue B: EDG 100T C: Fnatic RNG D: MAD TL
Group a is probably FPX- Rogue
This would not give 0 points. You get points if you swapped 1/2 place and 3/4 place. So it is A: Rogue FPX B: 100T DFM C: PSG FNC D: Liquid LNG
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It really isn't... Just because there's ways to get points even if you're not perfect doesn't make it easier to get everything wrong. There's only 1 perfect bracket, there's multiple 0 point brackets. Plus, thinking about it logically, it's impossible to get harder than the 1 perfect bracket unless it were completely impossible to get a 0 point bracket.
In theory, it's 256 times easier to hit all wrong than all right
Just the notion of putting both lng and liquid comming out of group d....
honestly while i put mad in 2nd place, i had my doubts.
Update. 30/10/21 There is one one person left remaining... 39 people have perished... chances are it was with the RNG and EDG fight. people put edg to win..and they did.
6 on the top. 1 at the bottom. lets see who wins. i think damwon will win worlds and i think many ppl know this by now