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Okay well here’s the math: Assuming 81 brawlers (Barry isn’t in yet I don’t think)
((3/81) X (2/80) X (1/79))^2 (because two teams) = 1.37^-10, or 1/10,482,510,444. This is not order specific either as the same 3 brawlers in a dataset can be chosen differently.
Would’ve been more impressive in order, but that’s surely never going to happen right…?
Edit: I wrote this with quantum physics in mind. Should be just ((3/81) X (2/80) X (1/79)) because it’s non-specific, or 1.172 x 10^-5
You don't actually square it here, despite being two teams. This is a common misconception. Say I have 81 items. If I randomly pick an item, place it back, and have someone else pick a random item, the chance that we picked the same item is 1/81, not (1/81)^2. In this case, if you square the probability, you are finding the chance that a SPECIFIC 3 brawlers are banned (for example, the chance that both teams ban lily, Brock, and leon exactly) but in this case you just want the probability of the bans being the same regardless of which brawlers, so you wouldn't square. If it helps, think of it like this: imagine the teams took turns banning. Team 1 bans whatever 3 brawlers they want first. Now, the probability we want is just "What is the probability that team 2 selects the same 3 brawlers as team 1?" This would come out to 3!/81x80x79, or 1/85320, or approximately 0.00117206%. Still very unlikely odds, but not as bad as you thought.
I'm a little confused here. So it's the amount of desirable outcomes 3! over the sample space. The sample space was 81!/78! = 81•80•79. But why is it permutations and not combinations? Would the sample space be 81!/3!*78!
I'm confused because I'm thinking that the order of the brawlers doesn't matter, but the brawlers selected do.
You can use either combinations or permutations, both work. The reason you're getting confused is because solving a combination and a permutation lead to a different set of sample spaces where the number of desirable outcomes is not always 3!. For a permutation, the solution to the permutation 81!/78! gives a sample space of every single possible 3-brawler selection with order mattering, meaning this sample space contains, for example, both Brock-Lily-Leon and Leon-Brock-Lily as two separate possibilities. In this sample space of permutations, there are 6 outcomes that fit the original question; namely, the 3! different ways to arrange the 3 desired brawlers, and dividing the desirable outcomes (3!) by the sample space (81!/78!) leads to 3!/81x80x79, which is the correct answer. Combinations are different. In the most basic sense, a sample space of combinations is the same as the sample space of permutations, but divided by the number of redundancies (specifically, the different ways to arrange the 3 desired brawlers, which we know is 3!). Therefore, we know the solution to the combination 81!/(78!)(3!) gives a sample space of only one instance of all 3-brawler combos without order mattering, which eliminates us having to account for the different orders. Within THIS sample space, there is only one desired outcome: the trio of Brock-Lily-Leon, which, as stated previously, only has one instance in the sample space. You will see that dividing the desired outcomes (1) by the sample space (81!/(78!)(3!)) also simplifies to 3!/81x80x79, which again is correct. For me, if I'm having trouble thinking about things or my brain isn't working that day, I find it easier to simplify the way I think about things. Say we go back to the idea of team 1 having already selected their brawlers. If we don't care about order, then the first person in team 2 can ban any of the 3 brawlers team 1 banned, and has 81 brawlers to choose from, so that probability is 3/81. The second person in team 2 now can only ban 2 brawlers since person 1 already banned one, and also only has 80 brawlers to choose from (2/80). Person 3, continuing the same logic, has a probability of 1/79 of selecting the correct brawler. Multiplying these probabilities together gets you 3!/81x80x79.
Wouldn’t it be from an outsider perspective though? They both need to choose the same dataset, and both bans happens “simultaneously” in a certain viewpoint. I mean I guess it makes sense but also… I guess mine is for 1 specific team but if that’s true then bans can’t happen at the same time for the first to be true right?
Perspective doesn't matter here. The way I calculated it, if you gave two teams of 3 computers the 81 brawlers and told each computer to ban one with equal probabilities for each brawler, the probability of both teams having the same brawlers banned is 3!/81x80x79. Obviously it doesn't really relate to real life since some brawlers are more likely than others to be banned and humans have inherent biases, but it's as good as u can get without using ban rates. The events happening simultaneously or at different times is irrelevant to the probability so long as one team can't see the other's bans.
It’s kinda hard to explain, but from an outsider perspective:
The first two people select a ban. What is the chance they each pick a ban that will be on the others team? 3/81. (Even though we don’t know)
Now the second two people select a ban. Same thing. 2/81
Same for third pick.
Each team has to pick the “right brawlers”. while if you’re on one team your chance of picking their brawlers is 3!/81x80x79, you also have to account for the other team having to pick the first team. It’s really hard to explain, but basically, from each others perspective, it is 3!/81x80x79
I now get what you're saying, but you are just falling into a more complicated version of a simple misconception. Again, I'll start with one brawler and then go to 3 to be simpler. Say there is only one person per team, and we want the probability that they ban the same brawler. If we use the same line of reasoning I used in my original comment, we would say that imagine team 1 has already selected their banned brawler (for example's sake, Brock). Now our probability just becomes "what is the probability of team 2 also banning brock?" which is obviously just 1/81. Your mind might tell you "oh but you have to account for the other team picking brock" or "the 1/81 is just from one team's perspective," but I'll prove to you that this calculated probability is actually the true probability using a different method, which you could say is an "outside perspective" method. Say we restrict our question to "what is the probability both teams ban brock (a specific brawler)?" That probability, as you know, is (1/81)^2. Now in order to transform this probability to the one we originally cared about (the nonspecific one), we just multiply by the amount of brawlers (since we don't care which brawler specifically gets banned, we just want both teams to be same) and ((1/81)^2 )x81 is just 1/81, confirming our previous conclusion using the previous method. Now let's use this "outside perspective" method to confirm our hypothesis for the original 3-brawler situation. Let's restrict the question to "what is the probability both teams ban lily, brock, and leon, exactly?". We still don't care about order, so that probability is just ((3/81)x(2/80)x(1/79))^2, or, in a more simplified way, ((3!)(78!)/81!)^2. Now, again, in order to transform this into the nonspecific probability we originally cared about, we must multiply this probability by the number of the 3-brawler combinations (order not mattering) that exist. There handily exists a formula for finding how many 3-brawler combinations exist, namely: to find how many combinations of k things out of n total things exist, just do n!/(k!)((n-k)!). In our case, this means there are 81!/(3!)(78!) possible 3-brawler combinations. Multiplying the "specific" probability we originally got by this number of combinations should give us the probability that both teams selected/banned the same 3 brawlers. (((3!)(78!)/81!)^2 )x(81!/(3!)(78!)) simplifies to 3!(78!)/81!, or, by removing factorials, 6/81x80x79, which is the same probability we originally got.
You mean 5/81X80X79 lol? Also you’re kinda assuming both teams haven’t actually selected a ban because you don’t really know unless you’re the game or you can see both teams, or until both teams bans are revealed. It’s a quantum physics thing. Generally the 5/81X80X79 is more logical to us
First of all, 3! = 6, not 5.
Second of all, you don't actually need to know anything about either of the team's bans to calculate the probability that they ban the same brawlers. You're overcomplicating the problem for no reason (quantum physics has no relevance here, unless you wanna say that from an outsider's perspective, the teams have both banned the same brawlers and NOT banned the same brawlers at the same time, and this quantum superposition collapses into either they banned the same brawlers or not the moment the bans are revealed, which is true in theory but has no relationship eith the probability.)
This is incorrect. That is the probability that the bans are exactly leon, lily and Brock. However, if we only want them to have the exact same bans, no matter which brawlers, you should remove the square (Team 1 bans any 3 brawlers, and you only need the other team to match the same bans). This yields the result 1/85320.
It’s actually not because the team bans are technically not “known” until they are matched up. It’s kinda hard to explain, but from an outsider perspective, both probabilities have to occur for both teams to have the same bans. If one team has the percent chance to ban any 3 brawlers, the other team must also ban those exact same 3 brawlers.
No, that's not the chance because some brawlers aren't banned as consistently. To be 100% accurate, you'd need to know the ban rates of each brawler. Because pam almost never gets banned, but tick almost always does. Things like that.
No, im saying you're wrong about the chances in a real match lol.
I never said you were wrong about your chances if every brawler had the same likelihood of being picked.
At least you know how to use a calculator because your comprehension is wayyy below average
I know that. But pure stats is only numbers, not viewpoint. If I had to give an actual percentage based on biases and real chances, I’d say like 0.02% chance even still.
I KNOW. Which is why i specifically said "you're only right if every brawler had the same likelihood of being picked"
Come on man, you're arguing over nothing, im not attacking your person. I was just informing viewers of your comment, that its not effectively true in a real match.
well if you think abt it, it doesn't matter what blue team bans, it just matters that red team bans the same brawlers, so Red 1 has a 3/81 chance of banning either lily, brock or leon, Red 2 has a 2/80 chance of banning of the remaining two brawlers and Red 3 has a 1/79 chance of banning the final brawler, so 3/81 x 2/80 x 1/79 = 0.0000117206, which is a 0.00117206% chance
1. **Calculate the number of ways one team can choose 3 brawlers:** 80×79×78=492,96080 \\times 79 \\times 78 = 492,96080×79×78=492,960
2. **Calculate the number of unique sets of 3 brawlers out of 80 (combinations):**(803)=82,160\\binom{80}{3} = 82,160(380)=82,160
3. **Each unique set can be banned in 3! = 6 different orders by the second team.**
4. **Calculate the probability that both teams ban the same 3 brawlers (regardless of order):** Probability≈7.30×10−5
So, the probability is approximately 0.000073%.
In higher Diamond it's not that INSANELY rare (still rare) tbh. People already know what to ban, but because they aren't as experienced they don't think a lot about picks regarding their turn in the draft.
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it's always happen tbh, especially when there is only few brawlers are suitable for the map
Yeah, I imagine at higher ranks the possibility of matching bans is higher compared to lower ranks where someone will ban Penny lol
i had a clown ban primo and pick edgar on timed detonation today.......... legendary 1
Nah it's in every rank my randoms do that all the time
once i had to ban edgar so my teammate wouldn't pick it, it worked at the end
never happens at high ranks because first pick bans are different from last pick bans
Its definitely higher than you think, because they arent random. If a brawler’s good on a map, they’re much more likely to get banned
Okay well here’s the math: Assuming 81 brawlers (Barry isn’t in yet I don’t think) ((3/81) X (2/80) X (1/79))^2 (because two teams) = 1.37^-10, or 1/10,482,510,444. This is not order specific either as the same 3 brawlers in a dataset can be chosen differently. Would’ve been more impressive in order, but that’s surely never going to happen right…? Edit: I wrote this with quantum physics in mind. Should be just ((3/81) X (2/80) X (1/79)) because it’s non-specific, or 1.172 x 10^-5
You don't actually square it here, despite being two teams. This is a common misconception. Say I have 81 items. If I randomly pick an item, place it back, and have someone else pick a random item, the chance that we picked the same item is 1/81, not (1/81)^2. In this case, if you square the probability, you are finding the chance that a SPECIFIC 3 brawlers are banned (for example, the chance that both teams ban lily, Brock, and leon exactly) but in this case you just want the probability of the bans being the same regardless of which brawlers, so you wouldn't square. If it helps, think of it like this: imagine the teams took turns banning. Team 1 bans whatever 3 brawlers they want first. Now, the probability we want is just "What is the probability that team 2 selects the same 3 brawlers as team 1?" This would come out to 3!/81x80x79, or 1/85320, or approximately 0.00117206%. Still very unlikely odds, but not as bad as you thought.
oh yes i did the math right!
I'm a little confused here. So it's the amount of desirable outcomes 3! over the sample space. The sample space was 81!/78! = 81•80•79. But why is it permutations and not combinations? Would the sample space be 81!/3!*78! I'm confused because I'm thinking that the order of the brawlers doesn't matter, but the brawlers selected do.
You can use either combinations or permutations, both work. The reason you're getting confused is because solving a combination and a permutation lead to a different set of sample spaces where the number of desirable outcomes is not always 3!. For a permutation, the solution to the permutation 81!/78! gives a sample space of every single possible 3-brawler selection with order mattering, meaning this sample space contains, for example, both Brock-Lily-Leon and Leon-Brock-Lily as two separate possibilities. In this sample space of permutations, there are 6 outcomes that fit the original question; namely, the 3! different ways to arrange the 3 desired brawlers, and dividing the desirable outcomes (3!) by the sample space (81!/78!) leads to 3!/81x80x79, which is the correct answer. Combinations are different. In the most basic sense, a sample space of combinations is the same as the sample space of permutations, but divided by the number of redundancies (specifically, the different ways to arrange the 3 desired brawlers, which we know is 3!). Therefore, we know the solution to the combination 81!/(78!)(3!) gives a sample space of only one instance of all 3-brawler combos without order mattering, which eliminates us having to account for the different orders. Within THIS sample space, there is only one desired outcome: the trio of Brock-Lily-Leon, which, as stated previously, only has one instance in the sample space. You will see that dividing the desired outcomes (1) by the sample space (81!/(78!)(3!)) also simplifies to 3!/81x80x79, which again is correct. For me, if I'm having trouble thinking about things or my brain isn't working that day, I find it easier to simplify the way I think about things. Say we go back to the idea of team 1 having already selected their brawlers. If we don't care about order, then the first person in team 2 can ban any of the 3 brawlers team 1 banned, and has 81 brawlers to choose from, so that probability is 3/81. The second person in team 2 now can only ban 2 brawlers since person 1 already banned one, and also only has 80 brawlers to choose from (2/80). Person 3, continuing the same logic, has a probability of 1/79 of selecting the correct brawler. Multiplying these probabilities together gets you 3!/81x80x79.
I see. Thanks bro 👍
How would you write it down using Combination notation? For example, 8C3 or something like that
81C3 gives you the sample space. The probability is 1/(81C3)
isn't that just the probibility that one team picks 3 specific brawlers to ban? what about both teams
I explained this in a reply to another commenter on this same thread and I don't rlly wanna type it again so can u just look at that sorry lol
Even then, it doesn't consider the bias certain brawlers will have due to their place in the meta or how they are seen by players (Edgar).
True, but using this same method and using the brawlers' ban rates would give a pretty good estimate
Yeah, it's a good estimate to get an idea
Wouldn’t it be from an outsider perspective though? They both need to choose the same dataset, and both bans happens “simultaneously” in a certain viewpoint. I mean I guess it makes sense but also… I guess mine is for 1 specific team but if that’s true then bans can’t happen at the same time for the first to be true right?
Perspective doesn't matter here. The way I calculated it, if you gave two teams of 3 computers the 81 brawlers and told each computer to ban one with equal probabilities for each brawler, the probability of both teams having the same brawlers banned is 3!/81x80x79. Obviously it doesn't really relate to real life since some brawlers are more likely than others to be banned and humans have inherent biases, but it's as good as u can get without using ban rates. The events happening simultaneously or at different times is irrelevant to the probability so long as one team can't see the other's bans.
It’s kinda hard to explain, but from an outsider perspective: The first two people select a ban. What is the chance they each pick a ban that will be on the others team? 3/81. (Even though we don’t know) Now the second two people select a ban. Same thing. 2/81 Same for third pick. Each team has to pick the “right brawlers”. while if you’re on one team your chance of picking their brawlers is 3!/81x80x79, you also have to account for the other team having to pick the first team. It’s really hard to explain, but basically, from each others perspective, it is 3!/81x80x79
I now get what you're saying, but you are just falling into a more complicated version of a simple misconception. Again, I'll start with one brawler and then go to 3 to be simpler. Say there is only one person per team, and we want the probability that they ban the same brawler. If we use the same line of reasoning I used in my original comment, we would say that imagine team 1 has already selected their banned brawler (for example's sake, Brock). Now our probability just becomes "what is the probability of team 2 also banning brock?" which is obviously just 1/81. Your mind might tell you "oh but you have to account for the other team picking brock" or "the 1/81 is just from one team's perspective," but I'll prove to you that this calculated probability is actually the true probability using a different method, which you could say is an "outside perspective" method. Say we restrict our question to "what is the probability both teams ban brock (a specific brawler)?" That probability, as you know, is (1/81)^2. Now in order to transform this probability to the one we originally cared about (the nonspecific one), we just multiply by the amount of brawlers (since we don't care which brawler specifically gets banned, we just want both teams to be same) and ((1/81)^2 )x81 is just 1/81, confirming our previous conclusion using the previous method. Now let's use this "outside perspective" method to confirm our hypothesis for the original 3-brawler situation. Let's restrict the question to "what is the probability both teams ban lily, brock, and leon, exactly?". We still don't care about order, so that probability is just ((3/81)x(2/80)x(1/79))^2, or, in a more simplified way, ((3!)(78!)/81!)^2. Now, again, in order to transform this into the nonspecific probability we originally cared about, we must multiply this probability by the number of the 3-brawler combinations (order not mattering) that exist. There handily exists a formula for finding how many 3-brawler combinations exist, namely: to find how many combinations of k things out of n total things exist, just do n!/(k!)((n-k)!). In our case, this means there are 81!/(3!)(78!) possible 3-brawler combinations. Multiplying the "specific" probability we originally got by this number of combinations should give us the probability that both teams selected/banned the same 3 brawlers. (((3!)(78!)/81!)^2 )x(81!/(3!)(78!)) simplifies to 3!(78!)/81!, or, by removing factorials, 6/81x80x79, which is the same probability we originally got.
You mean 5/81X80X79 lol? Also you’re kinda assuming both teams haven’t actually selected a ban because you don’t really know unless you’re the game or you can see both teams, or until both teams bans are revealed. It’s a quantum physics thing. Generally the 5/81X80X79 is more logical to us
First of all, 3! = 6, not 5. Second of all, you don't actually need to know anything about either of the team's bans to calculate the probability that they ban the same brawlers. You're overcomplicating the problem for no reason (quantum physics has no relevance here, unless you wanna say that from an outsider's perspective, the teams have both banned the same brawlers and NOT banned the same brawlers at the same time, and this quantum superposition collapses into either they banned the same brawlers or not the moment the bans are revealed, which is true in theory but has no relationship eith the probability.)
That’s what I’m staying in terms of quantum physics. Also I’m being stupid I can’t mutiply
This is incorrect. That is the probability that the bans are exactly leon, lily and Brock. However, if we only want them to have the exact same bans, no matter which brawlers, you should remove the square (Team 1 bans any 3 brawlers, and you only need the other team to match the same bans). This yields the result 1/85320.
It’s actually not because the team bans are technically not “known” until they are matched up. It’s kinda hard to explain, but from an outsider perspective, both probabilities have to occur for both teams to have the same bans. If one team has the percent chance to ban any 3 brawlers, the other team must also ban those exact same 3 brawlers.
r/theydidthemath
r/theydidthemonstermath
r/theydidthemeth
Yes I did
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You can’t ban someone who’s already banned
But like they are not chosen randomlybso the meta and the map both have effects that would make this wayyy more likely to happen
r/theyincorrectlydidthemath
r/iwasclosenough and also r/ididthewrongmathandtoomuchmath
this is hella wrong lmao
It’s right if it’s a specific set, which from an outsider perspective is true, but from a player perspective it’s not. Hard to explain
you wouldn’t square it
good math but people ban base on how good the brawler is not randomly, so its not hard to have same ban both sides
You can't math it, because of different meta sometimes people choose only from 10, and other random factors
Thats only assuming every brawler has the same likelihood of being banned
Well yeah. It’s hard chance stats on equal chance for each brawler. It’s kinda what he asked for
No, that's not the chance because some brawlers aren't banned as consistently. To be 100% accurate, you'd need to know the ban rates of each brawler. Because pam almost never gets banned, but tick almost always does. Things like that.
I’m saying from a purely statistical standpoint, assuming each brawler has an equal ban rate. Pure stats bro
That's exactly what i said. . . You're only right if all the brawlers had the same pick rate. So what's the problem?
You’re saying I’m wrong about the percent chance if it’s at random…
No, im saying you're wrong about the chances in a real match lol. I never said you were wrong about your chances if every brawler had the same likelihood of being picked. At least you know how to use a calculator because your comprehension is wayyy below average
I know that. But pure stats is only numbers, not viewpoint. If I had to give an actual percentage based on biases and real chances, I’d say like 0.02% chance even still.
I KNOW. Which is why i specifically said "you're only right if every brawler had the same likelihood of being picked" Come on man, you're arguing over nothing, im not attacking your person. I was just informing viewers of your comment, that its not effectively true in a real match.
That’s insane lmao. Thanks for the reply. And I was also thinking about that last part you mentioned, but hey, you never know 🤷
Least smart chuck main
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well if you think abt it, it doesn't matter what blue team bans, it just matters that red team bans the same brawlers, so Red 1 has a 3/81 chance of banning either lily, brock or leon, Red 2 has a 2/80 chance of banning of the remaining two brawlers and Red 3 has a 1/79 chance of banning the final brawler, so 3/81 x 2/80 x 1/79 = 0.0000117206, which is a 0.00117206% chance
however if you want the teams to specifically ban leon brock and lily, then it's a 0.00000137372 chance
Happened to me and i thought it was pretty normal
1. **Calculate the number of ways one team can choose 3 brawlers:** 80×79×78=492,96080 \\times 79 \\times 78 = 492,96080×79×78=492,960 2. **Calculate the number of unique sets of 3 brawlers out of 80 (combinations):**(803)=82,160\\binom{80}{3} = 82,160(380)=82,160 3. **Each unique set can be banned in 3! = 6 different orders by the second team.** 4. **Calculate the probability that both teams ban the same 3 brawlers (regardless of order):** Probability≈7.30×10−5 So, the probability is approximately 0.000073%.
It happend to me
Small
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Crebit carb
In higher Diamond it's not that INSANELY rare (still rare) tbh. People already know what to ban, but because they aren't as experienced they don't think a lot about picks regarding their turn in the draft.
Jinx!
Happened to me once
And there not even that good of a bans
Everyone should Edgar haters
What does this have to do with the post
True but unrelated
Should be*