Using a coin flip to determine the winner of a competitive sport is a HUGE L!!! It doesn’t get any worse than this. Have a shootout type end game set up in case this happens. Let the teams duel it out in game. A FRIGGIN COIN FLIP ARE YOU SERIOUS EPIC???
To me, it seems like people were calling Fortnite cringe because they quit, rather than quitting because it was "cringe". There was a time when a lot of players were falling out of love of the game until Zero Build came in because of how high the skill floor was getting.
The odds of getting the same average placement in 4 games is 1/396, or about 0.25%, which is low but isn't too low. The reason it isn't 1/100^4 is because 1/100^4 expresses the chances of guessing the exact same number in the same order. Such as: 1 - 44, 2 - 13, 3 - 95, 4 - 60 (this isn't even possible in Fortnite since 2 different people can't get the same placement in a set lobby). Whereas in this scenario where average of the 4 numbers has to be the same, it can be a variety of combinations. For example: Player 1: 51st, 51st, 50th, 50th Player 2: 1st, 1st, 100th, 100th Both these players have the same average placement of 50.5 since the sum of all their placements add up to 202. Therefore, for 2 players to have the same average placement throughout 4 games, they need to have the same sum of placements. Since the lowest possible placement is 1st, lowest sum can be 4 and the highest can be 400 (100*4). This means that the sum of placements of a player can range from 4-400, which gives the probability of 1/396 them having the same average placements.
if you roll two 1-6 dice, do you have the same probability of getting a 2 as a 6? The answer is no. While there are 396 possibilities of combined placements, that does not mean each placement has an equal chance and therefore makes the math to calculate the probability more complex. Random numbers also aren't realistic, as two players tied at a certain spot would be similar in skill level and would make their average placements much more likely to match.
@radicalaim I agree with the last part of your response that random numbers differ from actual placement of people in a game. However, I provided solution to that as that was the mathematical problem presented in the video. If you roll two 1-6 dice, the probability of getting each number will be 1/6, therefore, the odds of getting a 2 = odds of getting a 6. While it's true that not all distinct 396 sums are equally likely to come out, as values like 4 (1*4) and 400 (100*4) only have 1/100^4 chances of resulting from random generation. Whereas an average of 50 would be much higher since there are various combinations with the average of 50. However, the last paragraph was about calculating the chances of getting a specific average. The problem is to find out what are the chances that any given average (thus also sum) might occur twice. Therefore, it should be independent of the individual probabilities of getting each average.
@@eo-fikretalp1221 you're wrong. Just apply the probability to what you're trying to work out and you'll see that. You're implying that this would happen every 400ish tournaments, which is completely wrong. The answer is 1/200^4=/1/1.6 billion
@radicalaim I see your other comment that you ran a simulation, and got 1/205. While I'm confident in my answer I would still want to see how you found that value as that is also reasonable compared to other comments and you seem to actually understand the question.
To calculate the chances of a tie in percentages you'd just have to divide 100% with 100 four times so 100/(100^4)=0,000001 which is one in 100 million which is completely absurd. The number would be even more absurd if you would factor in the other tiebreakers.
You're not looking for 2 players with the same exact history, you're looking for 2 players with the same average. There are 400 choices, so at most there's a 1 in 400 chance of having a tie.
Since there are 400 choices (100 placement choices * 4 games), the maximum bound for the odds of both players having the same average placement is 1 in 400. However, it is far more likely to get an average placement towards the center, since there are more sums which add up to central numbers. For example, there is only 1 way to get an average placement of 1, but a ton of ways to get an average placement of 50 (23,48,82,47 for example). This means the distribution of average placement is not random, but instead it will follow a normal distribution. This significantly increases the chances of placements being the same. I wrote a script that runs the simulation 10 millions times and got a probability of 1 in 208.
0.05 is every 2000 tournaments There's a tournament roughly every day, for, I think, 7 regions. 2000/7≈300. So 0.05% implies that for, say, EU, every 300 tournaments this happens, which roughly once every 10 months, which is incorrect. The answer is either 1/1.6 billion or /1/100 quadrillion
@@HypaOT modern tournaments rarely have point systems that encourage ties, as reiss explains. however, this probably does happen daily among players scoring really low points. the issue is, nobody cares about lower brackets in those tournaments, so it goes unnoticed.
I feel like people forget that there’s still a lot left to cover from early Fortnite tournaments. I feel like winter, secret, and summer skirmishes are much more forgotten than fall skirmish.
Probably best way to get an approximation on likelihood of same avg. placement is: consider these players will place an avg. of at most 50. Then at 4 games multiply that by 4 and you get something like "likelihood of 2 numbers picked between 1 and 200 being the same which is 1/200 or 0.5%". Obviously, this is not mathematically correct, just a good way to gauge. Realistically, the probability is higher, since players are likely to have avg. placement between 20 and 50 and that "the number picked between 1 and 200 must be always even" so it's more close to 2%.
The average placements range from 1 to 50 (since there are 50 teams). The first team can have an average placement from 1 to 50, and the second team can also have an average placement from 1 to 50. So, the total number of possible combinations is 50 * 50 = 2500. To find the odds, divide the number of combinations where the average placements are the same by the total number of possible combinations. In this case, there is only one combination where the average placements are the same (e.g., both teams have an average placement of 25). Therefore, the odds of two teams having the same average placement over four games would be 1/2500.
4:28 I ran a simulation, the estimated probability that two different people both pick a random number between 1 and 100 four different times each, and the averages of both people's numbers are the same, is approximately 0.4879%, or about 1 in 205.
That's very wrong. The maths says it's 1/100,000,000. Even if the maths is wrong, this would mean this happens every 205 tournaments which is just untrue
@@HypaOT no, you don't understand probability, nor the question. What was calculated was the chance that 2 players had the same average placements across 4 games given they placed randomly (which due to the similar skill level, it would be more likely for the same average, but this is truly impossible to calculate as skill level isn't really quantifiable). This only shows that once in every 205 times you compare 2 random placement averages across 4 games, the averages would be the same. Of course, in actual tournaments, this is only for the step calculating placement probabilities. The probability that they all had the same stats in the other categories is what made a coin flip very improbable. However, the probability for this happening in a real tournament isn't near 1/100,000,000. If the tournament was done with different players with the same format, this wouldn't be as unlikely to repeat itself as you would expect. As the total points and victory royales between the players would almost certainly be repeated (with a large amount of players scoring 1 point and getting no victory royales), the probability relies on the total number of eliminations and average placements. With the players being very similar skill level, I would estimate the probability that a coin flip would be needed to decide which team would move on would greater than one in a thousand. Still very improbable to happen in a big tournament but not as improbable as it would be on a glance.
@@radicalaim would be fair to nromally distribute the numbers ngl since they both aren't the best but also not the worst players so a RNG would not be the best way to calculate this
@@idontmine7215 and in addition, they didn't place in the top 10 either because they didn't get any victory points, but it is still a similar odds of them getting the same random number (0.504%). But even using random numbers, it still shows that it isn't that improbable for this to happen in a tournament.
@@radicalaim I think you can use a combinmatrics approach for this, since you're getting all the possible ways for a sum to be 4-400 and still be the same? Which then you want to square the probability since it's both people getting that number
Some players now actually compete to pay of debts and mortgages for house and bills but like imagine back then losing your house because of a coin flip smh
Wow. I like your video, but what a bad statistics in the video and the comments. Considering its a tiebraker around the 33th place there is only a narrow band of avg placements possible. Lets assume around 25-40 avg placements bandwidth. Since there is 4 rounds, there is a 0,25 difference between possible avg placements possibilities. So 15*4=60. So its actually 1/60 chance to have the same placements points. I actually think its way higher, because i think 25-40 bandwith of avg placement is pretty generous.
The other 34th place guys didnt tie the #33 guy in front of them. So it wouldnt have been unfair to them. They flipped a coin to save themselves $5,000
can you make a short on fazing into someones box every time - edit the cone to one tile and crouch walk into the corner of the cone while spraying the floor below you and you will get in every time - asking for the 16th time
@@HypaOT but If you Play two Games for examples there are 200 possible average Placements and that you geht one of them ist 100% and that the other one gets the Same ist 1 in 200 so when Theres 3 Games IT Just Doubles the possibilities of an average placemnt you can have and With 4 Games again so IT is 1 in 800 then
@@vinggee9423 keep in mind your 200 possible placements are not equally likely. the only way you get 200 as your sum (or 100 as your average) is getting 100 twice, while getting a sum of 100 can happen in 99 different ways. so, this is wrong.
its not. what youve calculated here is the chance both players specifically pick a certain number (any from 1-100) across all 4 games. first, the first player can pick a number, and the second player can just match it. 10^16 implies you have them picking the same number each game. for example, if the first player picks 60, then the second player has to match 60. they don't need to both pick 60. if the first player picks 59 its not instantly over. second, the players don't need to pick the same number each game. this is, obviously impossible, as two players cannot both be 60th place in an FN game. but also, their average is the sum of their 4 placements, and that's the only thing that needs to be equal.
That wasn't the premise though, it's way more likely because there are lots of ways to get the same average out of four numbers. e.g. 50 (50, 50, 50, 50; also 49, 50, 50, 51; 49, 49, 51, 51; ...)
4:19 it is 1 in 100.000.000 Since the chance of 2 players picking the same random number is 1/100 that happens 4 times so we do 1/100^4 which is 1/100000000 or 0,000001%
No, it's way more likely because there are lots of ways to get the same average out of four numbers. e.g. 50 (50, 50, 50, 50; also 49, 50, 50, 51; 49, 49, 51, 51; ...)
promt for ChatGPT: User what are the changes that if you take the sum 4 random numbers from the range 1-100 twice, that the sum is the same answer:To calculate the probability that the sum of 4 random numbers chosen from the range 1-100 is the same when done twice, we can use a combinatorial approach. First, let's calculate the total number of ways to choose 4 numbers from the range 1-100. This can be done using the combination formula: (100 4)=100!4!(100−4)!(4100)=4!(100−4)!100! Then, we need to determine the number of ways in which the sums of the two sets of 4 numbers can be equal. Since the order doesn't matter and the numbers are unique, we can simplify this problem by considering the sums' averages. The average of a group of numbers is the sum divided by the number of elements. If the sums of two groups are equal, their averages will also be equal. Let's denote the sum of the first set of 4 numbers as S1S1 and the sum of the second set of 4 numbers as S2S2. The probability that S1=S2S1=S2 is equivalent to the probability that their averages are equal. The average of a group of 4 numbers from 1-100 would be the sum of those numbers divided by 4. Now, we want to find the number of pairs of sets where their averages are equal. Let's denote the average of the first set as A1A1 and the average of the second set as A2A2. We need to find the number of pairs (A1,A2)(A1,A2) where A1=A2A1=A2. After calculating the total number of ways to select 4 numbers twice and the number of pairs of averages that are equal, we divide the latter by the former to get the probability. First, let's calculate the total number of ways to choose 4 numbers from the range 1-100: (100 4)=100!4!(100−4)!(4100)=4!(100−4)!100! =100!4!96!=4!96!100! =100×99×98×974×3×2×1=4×3×2×1100×99×98×97 =3921225=3921225 So, there are 3,921,2253,921,225 ways to choose 4 numbers from the range 1-100. Now, let's find the number of pairs of averages that are equal. The average of 4 numbers chosen from the range 1-100 would be the sum of those numbers divided by 4. Since the numbers are unique and the order doesn't matter, we can focus on the average sums. The possible range of the sum of 4 numbers chosen from 1-100 is from 4 (1+1+1+1) to 400 (100+100+100+100). There are 397 possible values for the sum. We need to find pairs of sums (averages) that are equal. Let's denote SS as the sum of 4 numbers, and AA as the average of 4 numbers. So, A=S4A=4S. We need to find pairs (S1,S2)(S1,S2) where S1=S2S1=S2. Let's calculate the number of pairs of sums (averages) that are equal: Number of pairs=Number of unique sums−Number of unique averagesNumber of pairs=Number of unique sums−Number of unique averages =397−96=397−96 =301=301 Now, we divide the number of pairs by the total number of ways to choose 4 numbers twice to get the probability: Probability=Number of pairsTotal number of ways to choose 4 numbers twiceProbability=Total number of ways to choose 4 numbers twiceNumber of pairs =3013921225=3921225301 ≈0.0000768≈0.0000768 So, the probability that the sum of 4 random numbers chosen from the range 1-100 is the same when done twice is approximately 0.00768%0.00768%.
@@Piggenwings chatgpt is not always correct. in fact, its like wrong most of the time. below you can see it churning out an answer completely different from what you go
probably should explain lol the average is the mean, but how you calculate the sum is incorrect. i assume you made the most common mistake of most people here, which is you assume both players get the same ranking every game instead of just across all 4 games.
To find the probability that two players randomly pick the same number 4 rounds in a row, we can break it down step by step. First, the probability of both players picking the same number in one round is 1 out of 100, because there are 100 possible numbers to choose from and they want to pick the same one. So, the probability of this happening once is 1 100 100 1 . Since they need to do this four times in a row, we multiply the probabilities together: ( 1 100 ) 4 = 1 10 0 4 = 1 10000 ( 100 1 ) 4 = 100 4 1 = 10000 1 So, the odds that they pick the same number 4 rounds in a row are 1 in 10,000.
no, they dont need to pick the same number every round. first this is impossible as both players cant both be 33rd, but second, we want their average to equal, or the sum across their 4 games to be equal.
That's wrong that's the amount of different calculations to get to all the different answers when you put it on a bell curve the answer in 0.00169 chance
I tried calculating it and I got 1 over 100 quadrillion (16 zeros). Theoretically the chances of a player getting a certain placement is 1/100. Now let’s say we want TWO players to somehow get that same placement. To calculate the probability of two events happening in succession, we multiply them, meaning it’s 1/100 x 1/100 which is 1/10000. This is the probability of getting the same placement in their FIRST round, which means there are still 3 more rounds. 1/10000 to the power of 4 (represents all 4 rounds) will result in a final answer of: *1 in 100 Quadrillion (100,000,000,000,000,000) Keep in mind this is only a THEORETICAL probability and could be wrong. If this also included the chances of all other stats being the same, the number would get so big it would require the use of e.
The problem with your calculations are that you're calculating the chance that 2 players get the same placement 4 times in a row. Whereas in this scenario it's how likely is it that the 2 players' averages from the sum of their placements are the same. You should also take into account that 2 people can't get the same placement in a match. The actual answer is along the lines of 0.4% that I got from running a simulation, hence approximate. Also I don't think you did, but using ChatGPT for these kind of problems is useless, because it literally can't do this level of math consistently.
The odds of two people getting the same total placement given that all possibilities are equally likely from 1-100 after 4 games is 0.479% or 1 in 209 or 479387303143/10^16 *This assumes that the two players can get the same placement in the same game cause else the math is too hard.
I did this using generating functions start with the generating function f(x) = x/(1-x ) - x^101/(1-x ) this is the same as the function x + x^2 + x^3 ... x^100 so if we raise it to the power of 4 we get the amount of ways that each total placement can be made so the function g(x) = (x/(1-x ) - x^101/(1-x )) ^ 4 has coefficients such that the coefficient of x^n is equal to the total number of ways that a total placement of n can occur. g(x) can then be expanded to get g(x) = x^404/((1-x )^4) - 4 * x^304/((1-x )^4) + 6 * 2^404/((1-x )^4) - 4 * x^104/((1-x )^4) + x^4/((1-x )^4) we can write the 1/((1-x)^4) as the sum from n=0 to infinity of (n+1)(n+2)(n+3)x^n / 6 (not going to derive this but u can differentiate 1/(1-x) 3 times to get it) substituting this into g(x) we have sum from 404 to infinity of (n+1-404)(n+2-404)(n+3-404)x^n / 6 - 4 * the sum from 304 to infinity of (n+1-304)(n+2-304)(n+3-304)x^n / 6 + 6 * the sum from 204 to infinity of (n+1-204)(n+2-204)(n+3-204)x^n / 6 - 4 * the sum from 104 to infinity of (n+1-104)(n+2-104)(n+3-104)x^n / 6 + the sum from 4 to infinity of (n+1-4)(n+2-4)(n+3-4)x^n / 6. So when n is in the range of 1-100 the coefficient of x^n = (n+1-4)(n+2-4)(n+3-4)/ 6, when in the range of 101-200 it is (n+1-4)(n+2-4)(n+3-4)/ 6 - 4(n+1-104)(n+2-104)(n+3-104) / 6, when in the range of 201-300 it is (n+1-4)(n+2-4)(n+3-4)/ 6 - 4(n+1-104)(n+2-104)(n+3-104) / 6 + (n+1-204)(n+2-204)(n+3-204), and when in the range of 300-400 it is (n+1-4)(n+2-4)(n+3-4)/ 6 - 4(n+1-104)(n+2-104)(n+3-104) / 6 + (n+1-204)(n+2-204)(n+3-204) - (n+1-304)(n+2-304)(n+3-304)x^n / 6. And when it is greater than 400 it will be 0, we can call this function h(n). We can take the sum if h(n) for all values of n to find there are a total of 10^8 ways, which makes sense as for each game you can get from 1-100 so it should be 100^4 which it is. We can then find the probability of each average placement by doing h(n) / 10^8. So we can find the chance of two players getting the same score k by doing (h(k) / 10^8)^2 summing this value for all values of n we get 479387303143/10^16 meaning this is our probability of two players getting the same score. This is roughly 1 in 209 which makes some sense as we would expect it to be less than 1 in 400 (the total number of different scores) as we have a distribution of score that is more clustered. (The formatting is really bad on this so sorry)
"Man, I can't believe I tied with someone, I hope I win the money!"
Some random ass coin flip:
Hahahah
Top 400
Fortnite is so real for sticking to its roots and keeping the competition as Mickey Mouse as possible to this day 🙏
the chances are 0.001421% that two players randomly selected from the 100 players will have the same average placement over four games.
how? Jw
What is the odds of that?
Like 1 out of ___
@@IbraAli-_- about 1 in 70373
bro is smarter then us all
Coin flip for that is crazy 💀
lmfao bro didnt watch the vid
Net watch the vid first
always the damn skull emoji bro these kids so ignorant
I know always the skull like the fuck
@@fujimoto909 womp womp
Using a coin flip to determine the winner of a competitive sport is a HUGE L!!! It doesn’t get any worse than this. Have a shootout type end game set up in case this happens. Let the teams duel it out in game. A FRIGGIN COIN FLIP ARE YOU SERIOUS EPIC???
Reisshub is the only youtuber who can make statistics and boring competitive stuff actually engaging to watch
7:41 What a coincidence that Colton quit at the same time it became popular to call Fortnite “cringe.” Colton totally didn’t hop on a bandwagon at all
To me, it seems like people were calling Fortnite cringe because they quit, rather than quitting because it was "cringe". There was a time when a lot of players were falling out of love of the game until Zero Build came in because of how high the skill floor was getting.
Imagine the coin landed on the edge so they tied for that too
Happened once in a soccer match
The odds of getting the same average placement in 4 games is 1/396, or about 0.25%, which is low but isn't too low.
The reason it isn't 1/100^4 is because 1/100^4 expresses the chances of guessing the exact same number in the same order. Such as: 1 - 44, 2 - 13, 3 - 95, 4 - 60 (this isn't even possible in Fortnite since 2 different people can't get the same placement in a set lobby).
Whereas in this scenario where average of the 4 numbers has to be the same, it can be a variety of combinations.
For example:
Player 1: 51st, 51st, 50th, 50th
Player 2: 1st, 1st, 100th, 100th
Both these players have the same average placement of 50.5 since the sum of all their placements add up to 202.
Therefore, for 2 players to have the same average placement throughout 4 games, they need to have the same sum of placements.
Since the lowest possible placement is 1st, lowest sum can be 4 and the highest can be 400 (100*4).
This means that the sum of placements of a player can range from 4-400, which gives the probability of 1/396 them having the same average placements.
if you roll two 1-6 dice, do you have the same probability of getting a 2 as a 6? The answer is no.
While there are 396 possibilities of combined placements, that does not mean each placement has an equal chance and therefore makes the math to calculate the probability more complex.
Random numbers also aren't realistic, as two players tied at a certain spot would be similar in skill level and would make their average placements much more likely to match.
@radicalaim I agree with the last part of your response that random numbers differ from actual placement of people in a game. However, I provided solution to that as that was the mathematical problem presented in the video.
If you roll two 1-6 dice, the probability of getting each number will be 1/6, therefore, the odds of getting a 2 = odds of getting a 6.
While it's true that not all distinct 396 sums are equally likely to come out, as values like 4 (1*4) and 400 (100*4) only have 1/100^4 chances of resulting from random generation. Whereas an average of 50 would be much higher since there are various combinations with the average of 50.
However, the last paragraph was about calculating the chances of getting a specific average. The problem is to find out what are the chances that any given average (thus also sum) might occur twice. Therefore, it should be independent of the individual probabilities of getting each average.
@@eo-fikretalp1221 you're wrong. Just apply the probability to what you're trying to work out and you'll see that. You're implying that this would happen every 400ish tournaments, which is completely wrong. The answer is 1/200^4=/1/1.6 billion
@HypaOT read my original answer you will see why that isn't correct
@radicalaim I see your other comment that you ran a simulation, and got 1/205. While I'm confident in my answer I would still want to see how you found that value as that is also reasonable compared to other comments and you seem to actually understand the question.
1 penny calculated 1.5 million dollars?! That's insane
What a W format
Have you even watch the VID?
@@fujimoto909he’s joking 🙃
Oh my fault
To calculate the chances of a tie in percentages you'd just have to divide 100% with 100 four times so 100/(100^4)=0,000001 which is one in 100 million which is completely absurd. The number would be even more absurd if you would factor in the other tiebreakers.
You're not looking for 2 players with the same exact history, you're looking for 2 players with the same average. There are 400 choices, so at most there's a 1 in 400 chance of having a tie.
@@ancellery6430yeah that's my bad lol.
You cant get better edited than this. Also, a coin flip is life changing
It is always a good day when this legend uploads🎉
Of course this man releases this just as I’m about to sleep, guess I have to watch it.
Since there are 400 choices (100 placement choices * 4 games), the maximum bound for the odds of both players having the same average placement is 1 in 400. However, it is far more likely to get an average placement towards the center, since there are more sums which add up to central numbers. For example, there is only 1 way to get an average placement of 1, but a ton of ways to get an average placement of 50 (23,48,82,47 for example). This means the distribution of average placement is not random, but instead it will follow a normal distribution. This significantly increases the chances of placements being the same. I wrote a script that runs the simulation 10 millions times and got a probability of 1 in 208.
only correct answer in this comment section lmao
0.05% chance of having same placement happening
Hope this helps 😀
i believe its alot more rare than this as they are both independent of eachother
0.05 is every 2000 tournaments
There's a tournament roughly every day, for, I think, 7 regions. 2000/7≈300. So 0.05% implies that for, say, EU, every 300 tournaments this happens, which roughly once every 10 months, which is incorrect. The answer is either 1/1.6 billion or /1/100 quadrillion
@@HypaOT modern tournaments rarely have point systems that encourage ties, as reiss explains.
however, this probably does happen daily among players scoring really low points. the issue is, nobody cares about lower brackets in those tournaments, so it goes unnoticed.
I feel like people forget that there’s still a lot left to cover from early Fortnite tournaments. I feel like winter, secret, and summer skirmishes are much more forgotten than fall skirmish.
highest rank in battle royale decided who moves on is crazy 💀
knowing my "luck" I'd be the guy who would tie on
1. Total Vic Roys
2. Eliminations
3. Avg Placements
4. Total Seconds Alive...
You know this is the main timeline if something that unlikely happened
4:29 i think it’s 1/1M
Flipping a coin to see who wins a tournament is keeping you big as hell
Another video!! I love you’re videos❤
Twist,
the coin lands on its side
Yoooooo reisshub uploaded yay!!!
Yesss
Can you do an updated version of the perfect fortnite lootpool video.
Probably best way to get an approximation on likelihood of same avg. placement is: consider these players will place an avg. of at most 50. Then at 4 games multiply that by 4 and you get something like "likelihood of 2 numbers picked between 1 and 200 being the same which is 1/200 or 0.5%". Obviously, this is not mathematically correct, just a good way to gauge. Realistically, the probability is higher, since players are likely to have avg. placement between 20 and 50 and that "the number picked between 1 and 200 must be always even" so it's more close to 2%.
-"So team what should we do if theres a tiebreaker in points?"
"By a coin flip?"
"Take my job-
Imagine losing a tournament to a coinflip, they should just do a 1v1 instead
Imagine the deciding factor between you and $1.5 milli in your bank account is just a random gamble on heads or tails 😭
I remember a DH Finals where Verox and znappy tied on everything
Btw the chance that 2 players have the same average placement isn't as low as you think if they are of similar skill level
I can't be the only one that thought this was the mero, cooper coin flip😭
Kreo is ol hight. Bugha with the HP advantage. It might be the shockwave play. Bugha smiling. Bugha shockwaving up. Bugha won the Worldcup.
What if they tied on rank as well
?
Does anybody remember the experimental boxfight tournament? What if they brought that back for tie breakers and did a pg a zw or a bxf
3:17 1 day after 4/20 is crazy
Who else thought this was about the metro and cooper duping coin flip?
great vid keep up the work
First gang
you didnt watch it
You didnt watch it lmfao
This guy took me from bronze to diamond. Go reisshub!!!
The average placements range from 1 to 50 (since there are 50 teams). The first team can have an average placement from 1 to 50, and the second team can also have an average placement from 1 to 50. So, the total number of possible combinations is 50 * 50 = 2500.
To find the odds, divide the number of combinations where the average placements are the same by the total number of possible combinations. In this case, there is only one combination where the average placements are the same (e.g., both teams have an average placement of 25).
Therefore, the odds of two teams having the same average placement over four games would be 1/2500.
Just watched more of the video, released this calculation might be off. I calculated this thinking the placement could only be a integer.
500 views in less than 5 minutes, don't listen to people saying you fell off you are doing great!
4:17 bro couldn’t do the math himself 😭
4:28 I ran a simulation, the estimated probability that two different people both pick a random number between 1 and 100 four different times each, and the averages of both people's numbers are the same, is approximately 0.4879%, or about 1 in 205.
That's very wrong. The maths says it's 1/100,000,000. Even if the maths is wrong, this would mean this happens every 205 tournaments which is just untrue
@@HypaOT no, you don't understand probability, nor the question. What was calculated was the chance that 2 players had the same average placements across 4 games given they placed randomly (which due to the similar skill level, it would be more likely for the same average, but this is truly impossible to calculate as skill level isn't really quantifiable).
This only shows that once in every 205 times you compare 2 random placement averages across 4 games, the averages would be the same. Of course, in actual tournaments, this is only for the step calculating placement probabilities. The probability that they all had the same stats in the other categories is what made a coin flip very improbable.
However, the probability for this happening in a real tournament isn't near 1/100,000,000. If the tournament was done with different players with the same format, this wouldn't be as unlikely to repeat itself as you would expect. As the total points and victory royales between the players would almost certainly be repeated (with a large amount of players scoring 1 point and getting no victory royales), the probability relies on the total number of eliminations and average placements. With the players being very similar skill level, I would estimate the probability that a coin flip would be needed to decide which team would move on would greater than one in a thousand. Still very improbable to happen in a big tournament but not as improbable as it would be on a glance.
@@radicalaim would be fair to nromally distribute the numbers ngl since they both aren't the best but also not the worst players so a RNG would not be the best way to calculate this
@@idontmine7215 and in addition, they didn't place in the top 10 either because they didn't get any victory points, but it is still a similar odds of them getting the same random number (0.504%).
But even using random numbers, it still shows that it isn't that improbable for this to happen in a tournament.
@@radicalaim I think you can use a combinmatrics approach for this, since you're getting all the possible ways for a sum to be 4-400 and still be the same? Which then you want to square the probability since it's both people getting that number
the likelihood is 1 in 16 million
In the Lachlan pickaxe cup I tied 500th place with another team, top 500 is what you needed to win. They won the skin.
Mmm mmm.
@@fujimoto909 mmmm mmmm
womp womp
God how much I love your videos
Me too
Some players now actually compete to pay of debts and mortgages for house and bills but like imagine back then losing your house because of a coin flip smh
No idea how I missed the coin flip back than and only now I found out, but that's crazy...
I was hoping you’ll mention Kami And Setty dreamhack tiebreaker
Wow. I like your video, but what a bad statistics in the video and the comments. Considering its a tiebraker around the 33th place there is only a narrow band of avg placements possible. Lets assume around 25-40 avg placements bandwidth. Since there is 4 rounds, there is a 0,25 difference between possible avg placements possibilities. So 15*4=60. So its actually 1/60 chance to have the same placements points. I actually think its way higher, because i think 25-40 bandwith of avg placement is pretty generous.
I always smile when I see that ricehub posts
Bro I love your content so much and I am not saying this for clout keep it my guy
If you see this I love all your videos and your my favourite Fortnite UA-camr
Biggest takeaway:
LAN event
Teams couldnt dominate like normal
🤔🤔🤔
The other 34th place guys didnt tie the #33 guy in front of them. So it wouldnt have been unfair to them. They flipped a coin to save themselves $5,000
thats honesty really true
can you make a short on fazing into someones box every time - edit the cone to one tile and crouch walk into the corner of the cone while spraying the floor below you and you will get in every time - asking for the 16th time
Fortnite is so high skilled
People be sayin great vid after its been out only for 1 minute and the video is 8😂
Sometimes its elims before wins with tefus world cup qualifyer
reiss i love the video man keep it up
same avg placement in 4 rounds is 1/800 but onnly in theorie because nobody is gonna get an average placement off 100 etc
It's either 1/200^4=1/1.6 billion or 1/100 quadrillion btw
@@HypaOT but If you Play two Games for examples there are 200 possible average Placements and that you geht one of them ist 100% and that the other one gets the Same ist 1 in 200 so when Theres 3 Games IT Just Doubles the possibilities of an average placemnt you can have and With 4 Games again so IT is 1 in 800 then
@@vinggee9423 keep in mind your 200 possible placements are not equally likely. the only way you get 200 as your sum (or 100 as your average) is getting 100 twice, while getting a sum of 100 can happen in 99 different ways. so, this is wrong.
Here after the video's name changed
👇
im pretty sure if my maths is correct the odds of 2 players placing the same average placement in 4 matches is 1*10^-16
its not. what youve calculated here is the chance both players specifically pick a certain number (any from 1-100) across all 4 games.
first, the first player can pick a number, and the second player can just match it. 10^16 implies you have them picking the same number each game. for example, if the first player picks 60, then the second player has to match 60. they don't need to both pick 60. if the first player picks 59 its not instantly over.
second, the players don't need to pick the same number each game. this is, obviously impossible, as two players cannot both be 60th place in an FN game. but also, their average is the sum of their 4 placements, and that's the only thing that needs to be equal.
A 1v1 would have been more fun to watch
W upload 🎉
W
Hi
the chance for 2 players to have identical placements across 4 games is (maybe) 1/10^16, aka 1 quadrillion squared. DAS A BIG NUMBER
Yep, whice is same as what i said 1/ 100^4 (100 is much easier to calucate because the normal and most optimal game is 100 players)
1/1 quadrillion ^2 would be 1/1 nonillion, which isn't anywhere near 1/10^16
That wasn't the premise though, it's way more likely because there are lots of ways to get the same average out of four numbers. e.g. 50 (50, 50, 50, 50; also 49, 50, 50, 51; 49, 49, 51, 51; ...)
Mero x Cooper was also a coin flip 🐐
How can a coin do that
Can’t wait for a team to tie on the exact same victory Royale eliminations average placement and rank 😂
courage was commentating and probably eat the points for 11th place and higher
I thought this was another Cooper Mero video
5 million views in 43 milliseconds????? Reisshub popping off fr
4:19 it is 1 in 100.000.000
Since the chance of 2 players picking the same random number is 1/100 that happens 4 times so we do 1/100^4 which is 1/100000000 or 0,000001%
No, it's way more likely because there are lots of ways to get the same average out of four numbers. e.g. 50 (50, 50, 50, 50; also 49, 50, 50, 51; 49, 49, 51, 51; ...)
@@jossemalmberg that’s true but in the question given to chat gpt it’s correct
promt for ChatGPT: User
what are the changes that if you take the sum 4 random numbers from the range 1-100 twice, that the sum is the same
answer:To calculate the probability that the sum of 4 random numbers chosen from the range 1-100 is the same when done twice, we can use a combinatorial approach.
First, let's calculate the total number of ways to choose 4 numbers from the range 1-100. This can be done using the combination formula:
(100 4)=100!4!(100−4)!(4100)=4!(100−4)!100!
Then, we need to determine the number of ways in which the sums of the two sets of 4 numbers can be equal.
Since the order doesn't matter and the numbers are unique, we can simplify this problem by considering the sums' averages.
The average of a group of numbers is the sum divided by the number of elements. If the sums of two groups are equal, their averages will also be equal.
Let's denote the sum of the first set of 4 numbers as S1S1 and the sum of the second set of 4 numbers as S2S2. The probability that S1=S2S1=S2 is equivalent to the probability that their averages are equal.
The average of a group of 4 numbers from 1-100 would be the sum of those numbers divided by 4.
Now, we want to find the number of pairs of sets where their averages are equal.
Let's denote the average of the first set as A1A1 and the average of the second set as A2A2. We need to find the number of pairs (A1,A2)(A1,A2) where A1=A2A1=A2.
After calculating the total number of ways to select 4 numbers twice and the number of pairs of averages that are equal, we divide the latter by the former to get the probability.
First, let's calculate the total number of ways to choose 4 numbers from the range 1-100:
(100 4)=100!4!(100−4)!(4100)=4!(100−4)!100!
=100!4!96!=4!96!100!
=100×99×98×974×3×2×1=4×3×2×1100×99×98×97
=3921225=3921225
So, there are 3,921,2253,921,225 ways to choose 4 numbers from the range 1-100.
Now, let's find the number of pairs of averages that are equal.
The average of 4 numbers chosen from the range 1-100 would be the sum of those numbers divided by 4. Since the numbers are unique and the order doesn't matter, we can focus on the average sums.
The possible range of the sum of 4 numbers chosen from 1-100 is from 4 (1+1+1+1) to 400 (100+100+100+100). There are 397 possible values for the sum.
We need to find pairs of sums (averages) that are equal.
Let's denote SS as the sum of 4 numbers, and AA as the average of 4 numbers. So, A=S4A=4S.
We need to find pairs (S1,S2)(S1,S2) where S1=S2S1=S2.
Let's calculate the number of pairs of sums (averages) that are equal:
Number of pairs=Number of unique sums−Number of unique averagesNumber of pairs=Number of unique sums−Number of unique averages
=397−96=397−96
=301=301
Now, we divide the number of pairs by the total number of ways to choose 4 numbers twice to get the probability:
Probability=Number of pairsTotal number of ways to choose 4 numbers twiceProbability=Total number of ways to choose 4 numbers twiceNumber of pairs
=3013921225=3921225301
≈0.0000768≈0.0000768
So, the probability that the sum of 4 random numbers chosen from the range 1-100 is the same when done twice is approximately 0.00768%0.00768%.
@@Piggenwings chatgpt is not always correct. in fact, its like wrong most of the time. below you can see it churning out an answer completely different from what you go
@@Random1785YTI didn’t use chargpt, I calculated it myself
Why couldn't they just write names on small papers and add it to a box, shake and pick on random? Its way better and perhaps fair than a coin flip imo
how? both are 50/50
The possibility is around 0.000001% chance or 1/100000, crazy odds
how did you get this number? its not correct btw anyways
The coin then lands vertically 💀
How on earth did some of yall watch a 9 minute video in 4 minutes.
Controller or KBM?
4:38
I believe it to be a 1 in a million chance assuming the average used is the mean. Sorry if I’m wrong
wrong
probably should explain lol
the average is the mean, but how you calculate the sum is incorrect. i assume you made the most common mistake of most people here, which is you assume both players get the same ranking every game instead of just across all 4 games.
Imagine that the coin fell on its middle
To find the probability that two players randomly pick the same number 4 rounds in a row, we can break it down step by step. First, the probability of both players picking the same number in one round is 1 out of 100, because there are 100 possible numbers to choose from and they want to pick the same one. So, the probability of this happening once is 1 100 100 1 . Since they need to do this four times in a row, we multiply the probabilities together: ( 1 100 ) 4 = 1 10 0 4 = 1 10000 ( 100 1 ) 4 = 100 4 1 = 10000 1 So, the odds that they pick the same number 4 rounds in a row are 1 in 10,000.
no, they dont need to pick the same number every round. first this is impossible as both players cant both be 33rd, but second, we want their average to equal, or the sum across their 4 games to be equal.
It’s a 100 to the power of four likely hood that two people would draw with average placement without rounding 4:47
1 in 10million
That's wrong that's the amount of different calculations to get to all the different answers when you put it on a bell curve the answer in 0.00169 chance
1 in 500 games not factoring the other criteria in and rounding
@@seb8073 can you show how you got this? might be interesting
i thought this video would be about the coin mero flipped that made him duo with cooper and win globals based on the title
Guys 1 view in 55 seconds, i think its time to accept it...
Can i ask a question?
Is the fncs comes once in year of every season have its own fncs 😅
Can you put like to make hime replay for my question??😅😊
Just awful by Epic. Glad this is over with
Imagine it landed on its side...
kreo still mad at courage
I tried calculating it and I got 1 over 100 quadrillion (16 zeros). Theoretically the chances of a player getting a certain placement is 1/100. Now let’s say we want TWO players to somehow get that same placement. To calculate the probability of two events happening in succession, we multiply them, meaning it’s 1/100 x 1/100 which is 1/10000.
This is the probability of getting the same placement in their FIRST round, which means there are still 3 more rounds.
1/10000 to the power of 4 (represents all 4 rounds) will result in a final answer of:
*1 in 100 Quadrillion (100,000,000,000,000,000)
Keep in mind this is only a THEORETICAL probability and could be wrong. If this also included the chances of all other stats being the same, the number would get so big it would require the use of e.
The problem with your calculations are that you're calculating the chance that 2 players get the same placement 4 times in a row. Whereas in this scenario it's how likely is it that the 2 players' averages from the sum of their placements are the same. You should also take into account that 2 people can't get the same placement in a match. The actual answer is along the lines of 0.4% that I got from running a simulation, hence approximate. Also I don't think you did, but using ChatGPT for these kind of problems is useless, because it literally can't do this level of math consistently.
not same placement, the sum of their 4 placements. this is the odds they had the exact same number for each round.
Reisshub a w fr
The odds of two people getting the same total placement given that all possibilities are equally likely from 1-100 after 4 games is 0.479% or 1 in 209 or 479387303143/10^16
*This assumes that the two players can get the same placement in the same game cause else the math is too hard.
I did this using generating functions start with the generating function f(x) = x/(1-x ) - x^101/(1-x ) this is the same as the function x + x^2 + x^3 ... x^100 so if we raise it to the power of 4 we get the amount of ways that each total placement can be made so the function g(x) = (x/(1-x ) - x^101/(1-x )) ^ 4 has coefficients such that the coefficient of x^n is equal to the total number of ways that a total placement of n can occur. g(x) can then be expanded to get g(x) = x^404/((1-x )^4) - 4 * x^304/((1-x )^4) + 6 * 2^404/((1-x )^4) - 4 * x^104/((1-x )^4) + x^4/((1-x )^4) we can write the 1/((1-x)^4) as the sum from n=0 to infinity of (n+1)(n+2)(n+3)x^n / 6 (not going to derive this but u can differentiate 1/(1-x) 3 times to get it) substituting this into g(x) we have sum from 404 to infinity of (n+1-404)(n+2-404)(n+3-404)x^n / 6 - 4 * the sum from 304 to infinity of (n+1-304)(n+2-304)(n+3-304)x^n / 6 + 6 * the sum from 204 to infinity of (n+1-204)(n+2-204)(n+3-204)x^n / 6 - 4 * the sum from 104 to infinity of (n+1-104)(n+2-104)(n+3-104)x^n / 6 + the sum from 4 to infinity of (n+1-4)(n+2-4)(n+3-4)x^n / 6. So when n is in the range of 1-100 the coefficient of x^n = (n+1-4)(n+2-4)(n+3-4)/ 6, when in the range of 101-200 it is (n+1-4)(n+2-4)(n+3-4)/ 6 - 4(n+1-104)(n+2-104)(n+3-104) / 6, when in the range of 201-300 it is (n+1-4)(n+2-4)(n+3-4)/ 6 - 4(n+1-104)(n+2-104)(n+3-104) / 6 + (n+1-204)(n+2-204)(n+3-204), and when in the range of 300-400 it is (n+1-4)(n+2-4)(n+3-4)/ 6 - 4(n+1-104)(n+2-104)(n+3-104) / 6 + (n+1-204)(n+2-204)(n+3-204) - (n+1-304)(n+2-304)(n+3-304)x^n / 6. And when it is greater than 400 it will be 0, we can call this function h(n). We can take the sum if h(n) for all values of n to find there are a total of 10^8 ways, which makes sense as for each game you can get from 1-100 so it should be 100^4 which it is. We can then find the probability of each average placement by doing h(n) / 10^8. So we can find the chance of two players getting the same score k by doing (h(k) / 10^8)^2 summing this value for all values of n we get 479387303143/10^16 meaning this is our probability of two players getting the same score. This is roughly 1 in 209 which makes some sense as we would expect it to be less than 1 in 400 (the total number of different scores) as we have a distribution of score that is more clustered. (The formatting is really bad on this so sorry)
@@fang3562bro it’s not that serious /j
Nah but fr good job
@@fang3562 upload it as a pdf file using overleaf !!!!
the chance of the placement i exactly 1 in one million
Yippee another reisshub vid
How do u get the replay mode footage
everyone who has massively improved after watching this man
👇
Me
What if the coin landed on its side🤣😂🤣😂🤣 tibreaker number 7😂
Fell