Fortnite's HIDDEN Coin Flip Rule

[@cerealgibbon7330]

"Man, I can't believe I tied with someone, I hope I win the money!" Some random ass coin flip:

[@fujimoto909]

Hahahah

[@NSG-GAMING-REAL]

Top 400

[@Gabriel-sd1oh]

Fortnite is so real for sticking to its roots and keeping the competition as Mickey Mouse as possible to this day 🙏

[@Penini528]

the chances are 0.001421% that two players randomly selected from the 100 players will have the same average placement over four games.

[@HypaOT]

how? Jw

[@IbraAli-_-]

What is the odds of that?

[@IbraAli-_-]

Like 1 out of ___

[@jossemalmberg]

@@IbraAli-_- about 1 in 70373

[@therealpumpkinhead]

bro is smarter then us all

[@Minty.Monkey]

Reisshub is the only youtuber who can make statistics and boring competitive stuff actually engaging to watch

[@shiroonD2]

Coin flip for that is crazy 💀

[@justsomeguywithafancymusta4542]

lmfao bro didnt watch the vid

[@fujimoto909]

Net watch the vid first

[@rodrigo_tm]

always the damn skull emoji bro these kids so ignorant

[@fujimoto909]

I know always the skull like the fuck

[@Ben11sg]

@@fujimoto909 womp womp

[@lordtraxx4217]

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???

[@razorwashere5858]

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

[@Rayuzx]

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.

[@Mario285Industries]

Imagine the coin landed on the edge so they tied for that too

[@pmb4049]

Happened once in a soccer match

[@HedgieFN]

What a W format

[@fujimoto909]

Have you even watch the VID?

[@Okrawek-_-]

@@fujimoto909he’s joking 🙃

[@fujimoto909]

Oh my fault

[@eo-fikretalp1221]

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.

[@radicalaim]

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.

[@eo-fikretalp1221]

@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.

[@HypaOT]

@@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

[@eo-fikretalp1221]

@HypaOT read my original answer you will see why that isn't correct

[@eo-fikretalp1221]

@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.

[@JAL08J]

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.

[@ancellery6430]

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.

[@JAL08J]

​@@ancellery6430yeah that's my bad lol.

[@Mineoutain]

1 penny calculated 1.5 million dollars?! That's insane

[@pancake_9]

Of course this man releases this just as I’m about to sleep, guess I have to watch it.

[@ezmitich]

You cant get better edited than this. Also, a coin flip is life changing

[@bogeefn]

Yoooooo reisshub uploaded yay!!!

[@fujimoto909]

Yesss

[@HuvPlays]

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...

[@Hexxed705]

great vid keep up the work

[@fujimoto909]

First gang

[@rodrigo_tm]

you didnt watch it

[@eurofn.1]

You didnt watch it lmfao

[@pattymcfatty0]

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.

[@aaarkid]

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%.

[@Lusac11]

500 views in less than 5 minutes, don't listen to people saying you fell off you are doing great!

[@VortexZAP]

Another video!! I love you’re videos❤

[@roxinityplayz]

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

[@slapshotgaming9360]

Twist, the coin lands on its side

[@DigitYT]

4:29 i think it’s 1/1M

[@Baseballbroz3630]

This guy took me from bronze to diamond. Go reisshub!!!

[@Bob-sw3he]

If you see this I love all your videos and your my favourite Fortnite KZbinr

[@Carter.commenter]

-"So team what should we do if theres a tiebreaker in points?" "By a coin flip?" "Take my job-

[@evanwirth7881]

You know this is the main timeline if something that unlikely happened

[@aquastoneaquatics9063]

reiss i love the video man keep it up

[@fujimoto909]

Bro I love your content so much and I am not saying this for clout keep it my guy

[@Cab1664]

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.

[@Cab1664]

Just watched more of the video, released this calculation might be off. I calculated this thinking the placement could only be a integer.

[@RetroSpectral76]

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.

[@linkkuweitsi7911]

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.

[@Random1785YT]

not same placement, the sum of their 4 placements. this is the odds they had the exact same number for each round.

[@aFish17]

First and W

[@holvingar4375]

4:17 bro couldn’t do the math himself 😭

[@Patos12358]

Kreo is ol hight. Bugha with the HP advantage. It might be the shockwave play. Bugha smiling. Bugha shockwaving up. Bugha won the Worldcup.

[@radicalaim]

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. ​​

[@HypaOT]

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

[@radicalaim]

​@@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.

[@idontmine7215]

@@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

[@radicalaim]

@@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.

[@idontmine7215]

@@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

[@luvp259]

Flipping a coin to see who wins a tournament is keeping you big as hell

[@Formfv]

Who else thought this was about the metro and cooper duping coin flip?

[@bluebeanie-Boy]

Great video reiss

[@gruffy686]

I can't be the only one that thought this was the mero, cooper coin flip😭

[@mrrotteneggs8178]

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

[@N3GATiiVE]

the likelihood is 1 in 16 million

[@katherinephilippakis1270]

Can you do an updated version of the perfect fortnite lootpool video.

[@MiksonMacloud]

I remember a DH Finals where Verox and znappy tied on everything

[@bryce56]

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

[@Random1785YT]

thats honesty really true

[@rl.kageyama]

God how much I love your videos

[@fujimoto909]

Me too

[@HogneeDZN]

No idea how I missed the coin flip back than and only now I found out, but that's crazy...

[@neptune030]

Yippee another reisshub vid

[@survivalsearcher]

Biggest takeaway: LAN event Teams couldnt dominate like normal 🤔🤔🤔

[@jaswoj6810]

I was hoping you’ll mention Kami And Setty dreamhack tiebreaker

[@MyNameIsGrape_]

In the Lachlan pickaxe cup I tied 500th place with another team, top 500 is what you needed to win. They won the skin.

[@fujimoto909]

Mmm mmm.

[@MyNameIsGrape_]

@@fujimoto909 mmmm mmmm

[@Reflex_ed]

womp womp

[@kaktus4162]

W

[@fujimoto909]

Hi

[@Redwarior59]

What if they tied on rank as well

[@glowninja]

?

[@Piggenwings]

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%

[@jossemalmberg]

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; ...)

[@Piggenwings]

@@jossemalmberg that’s true but in the question given to chat gpt it’s correct

[@-horoshilov-8272]

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%.

[@Random1785YT]

@@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

[@Piggenwings]

@@Random1785YTI didn’t use chargpt, I calculated it myself

[@IKeyOK]

First To see how smart reiss is

[@HyugaJeremy]

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

[@KwikBR]

how? both are 50/50

[@scorchy9447]

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.

[@Random1785YT]

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.

[@kjhcfad]

Here after the video's name changed 👇

[@thespecialkid1384]

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

[@Random1785YT]

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.

[@Jwmd1]

The possibility is around 0.000001% chance or 1/100000, crazy odds

[@Random1785YT]

how did you get this number? its not correct btw anyways

[@Warsaw_Pact]

A 1v1 would have been more fun to watch

[@Prodigyyhimself]

Mero x Cooper was also a coin flip 🐐

[@somebody_JB]

Fortnite is so high skilled

[@playboicraze]

I thought this was another Cooper Mero video

[@rachelridgill6499]

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

[@Random1785YT]

wrong

[@Random1785YT]

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.

[@4Cone]

courage was commentating and probably eat the points for 11th place and higher

[@tod3034]

Can’t wait for a team to tie on the exact same victory Royale eliminations average placement and rank 😂

[@Doctor_Maulwurf]

Sometimes its elims before wins with tefus world cup qualifyer

[@djorimooofc8281]

This is crazy 😂❤️

[@chillinedits.]

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

[@bradderrs6221]

The coin then lands vertically 💀

[@Lendogg6124]

Hi nice vid

[@OPTIXZZZZZZ]

Best fn youtuber

[@Fireisnotred-if7wx]

How on earth did some of yall watch a 9 minute video in 4 minutes.

[@neilA-gr7cd]

Reisshub a w fr

[@eurofn.1]

People be sayin great vid after its been out only for 1 minute and the video is 8😂

[@Wolfjam01]

First😮

[@Mikael9781]

Imagine that the coin fell on its middle

[@macchiplayz]

IM pretty early

[@360ninja2yt]

How can a coin do that

[@F1Forlife-hn1tj]

Good video

[@Domancave]

Just awful by Epic. Glad this is over with

[@jamason1235]

Guys 1 view in 55 seconds, i think its time to accept it...

[@Subupop]

Controller or KBM?

[@dannymendelsohn6400]

the chance of the placement i exactly 1 in one million

[@ChessKnight-1]

Imagine it landed on its side...

[@TheOnlyBlinx]

I would take the chances

[@MeneerVerspiller]

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.

[@been_block]

Fell

[@JacobBake999]

A coin flip was unfair just half half the money

[@Flynewt544]

Pretty cool info

[@fujimoto909]

I know bravo

[@ilikegalaxies]

should have done a 2v2 or something

[@Sanjifv]

w upload

[@fujimoto909]

Did you watch it

[@fang3562]

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.

[@fang3562]

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)

[@dadanl2843]

@@fang3562bro it’s not that serious /j Nah but fr good job

[@Random1785YT]

@@fang3562 upload it as a pdf file using overleaf !!!!

[@Kinggold2011]

How do u get the replay mode footage

[@bluisok]

first

[@fujimoto909]

No me

[@WateRoblox]

Did the thumbnail change?

[@BilalKhalfi-wm8bi]

3:00 150 thousend dollars?????

[@fujimoto909]

Here before 2minutes

[@fujimoto909]

Ressihub have you won a tournament yet?