DICE CHESS - Anish Giri vs Teimour Radjabov
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- čas přidán 6. 08. 2020
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Man, respect to Radja for not taking that rating thing seriously. 400 rating is a lot ! Especially at the grandmaster level !! #RadjaOP
Because he play in lichess
Thnx for the spoiler 🤷♂️😐😒😒
@@h_2577 Ouch ! Sorry XDD
@@manavjaison Spoiler! 😒
Anyways next time not gonna look comments before watching such videos. Lol 😅
@@Pranjal446 Sorry dude 😅😅😅
Let's take a moment to appreciate how Anish took the game seriously
Rather he took the math seriously
Yaar Radja is soo cute and humble. See how relaxed he was after losing so much rating points. ❤️
Radja OP
@@jaimintrivedi2207 yupp
he does not care about rating, he just wants to play good chess. His attitude towards chess is great
As a math student, this stream was too painful to watch XD
can you tell the actual probability?
@@arcadian1643 42%. Solution. Total probability of any even is 1. Now if only one dice is rolled, probability of not getting a queen is 5/6. For 3 dice, the probability of not getting queen is (5/6)×(5/6)×(5/6)= 125/216. So probability of getting atleast 1 queen is 1-125/256 = 0.42. So 42%. I am an engineer btw.
@@dhruv1863 you don't have to be an engineer to do this easy maths 😂😂😂
@@Aditya-vy2eg Not that easy tho
@@abcd-sf5ur donno bro I am just 18 and I found it pretty easy and obvious may be it is not
Samay's math teachers are looking at this video in shame
Yea!! Also anish's 😅😂
1/6 + 1/6 +1/6 = 1/18 . This is why samay's teacher would be in dismay
Kids: watching Shakuntala Devi 's insane maths calculations
Legends: 1/6 + 1/6 + 1/6 = 1/18
This is the reason you became printing engineer Samay 😂🧡
👍🏻👍🏻😂😂😂😂
Heavy
I WAS GONNA SAY THIS 😂
😂😂😂😂😂😂
🤣🤣🤣
Never do printing engineering, reason??
15:10 see this flawless maths calculation!
Haha
ROFL
See the confidence 😂😂
Hahahahahhahahah
The chess players are most humble professional players in the world
Okay so here goes (considering unbiased dice),
Probability for "EXACTLY" 1 queen: (1/6)*(5/6)*(5/6) + (5/6)*(1/6)*(5/6) + (5/6)*(5/6)*(1/6) = (1/6)*(5/6)*(5/6)*3 = (75/216)[on one of the dices queen should come but not on the other two, exactly one queen can come on any of the three dice so we add the individual probabilities]
Probability for "EXACTLY" 2 queens: (1/6)*(1/6)*(5/6) + (1/6)*(5/6)*(1/6) + (5/6)*(1/6)*(1/6) = (1/6)*(1/6)*(5/6)*3 = (15/216)[argument same as above]
Probability for "EXACTLY" 3 queens: (1/6)*(1/6)*(1/6) = (1/216)[all three dices should have a queen]
What anish was trying to calculate was Probability of "AT LEAST 1 QUEEN", which means either 1 queen OR 2 queens OR 3 queens which is the same as adding the above probabilities which is (75/216) + (15/216) + (1/216) = (91/216)
As many people correctly said, an easier way to calculate "AT LEAST 1 OF SOMETHING" is just to calculate 1 - "NONE OF THAT SOMETHING" which is
1 - [ (5/6)*(5/6)*(5/6) ] = (91/216)
Mera dimag ghom gaya 🥴🥴
U are exactly right bro , people calculated n(A u B u C) and came to 1/2 which is completely wrong. 75/216 is the right answer
I also had the same answer. But I got confused by seeing the live chat as so many people were saying its 50%. Now I'm relieved by seeing ur comment
Wow..man you did so much calculation😮😮
nice work
15:10 guys shakuntala devi would be so proud of our printing engineer!!!
I can sooo relate to what Anish was going through
Happens with me a lot of times
Something stuck in my head and I cant think properly until that worm is thrown out!!!
On the scale of Radjabov to Anish how lucky are you? 😛
- Radjabov
Chess
@@netraparihar867 So lucky you are. Because -ve * -ve becomes positive. Sorry😂🙌 But was a Maths Stream so did a Maths joke😂🤣
@@MohitKumar-gz2rt ohh God 😑😑what a joke😂😂
i am probably RCB
Here before Samay makes it Members Only content
This won't be made members only
1/6+1/6+1/6=1/18- Samay Raina(Printing engineer, Stanford)
Probability of getting at least 1 queen(assuming unbiased dice) is 1-(5/6)^3=91/216
Probability of getting exactly 1 Queen-(3C1)(1/6)^1(5/6)^2=75/216(binomial distribution)
Me after answering the most standard JEE question-JEE advanced AIR 1
shivansh the probability of getting at least one queen can also be solved using binomial distribution. [3c1(1/6)(5/6)^2+3c2(1/6)^2(5/6)+3c3(1/6)^3 ]=91/216.
@@varuntangtur9708 ik but other method is easier and shorter to write
And that ladies and gentlemen is 42%, the answer to everything in life.
I can't believe samay laughed at the person who said 91/216.
1/216 fool 😂😂😂😂
@@vipulbhardwaj9772 would it not be 36/216
‘’1/6+1/6+1/6 is actually 1/12 no no 1/18’’ Printing Engineering zindabad 😂😂. See 15:10 if you don’t get it
Lol
Bro 1/12 nahi 1/2 lol😂😂
@@lost2932 yep xD
Well this is awkward
Maybe NEP was drafted because of Samay’ mathematics
15:10 SAMAY- THE LEGENDARY MATHEMATICIAN😂😂😂😂😂😂😂🤣🤣🤣🤣
Man ! This stream had more maths than my online classes in this lockdown !!!
There can be two approaches
1 by considering cases with no queen at all , which means
Cases where no queens comes is 5/6*5/6*5/6 = 125/216
Thus cases for queen= 1- 125/216
= 91/216
2. Its long method we have to consider three possibilities with one queen , two queens and three queens
For 1 queen let us take queen at dice 1 then there will be 25 cases as for each other element and fir 3 pos total= 75
When 2 queen's queen can change 3 pos like d1 d2 d1 d3 d2 d3 dices then 5+5+5 cases= 15
3rd case us all 3 queens = 1
So total = 75 + 15 + 1 = 91 casee
Thus probability is 91/216
Which is roughly equals to 42 percent
These can be two approaches to solve the problem.
Engineer hu yehi aata hai bs😅
Radja is a great sport for not appearing to be bothered by the decline in his rating. Samay, you should ask Rakesh bhai to retrospectively change this game to unrated, or reverse back their ratings. That seems only fair :)
Best part :- Radjabov about relentless calculations of anish- "IT IS UNNECESSARY"
1/6+1/6+1/6=1/18 WOOW samay your maths just blew my mind away🤯🤯🤯
😂😂😂😂😂😂
ohh bhai, maro mujhe maro
Probability of Getting a Queen on any 1 of the 3 dices :
P(Not Getting a Queen in any of the dices) = (5/6)*(5/6)*(5/6) = 125/216
P(At least one Queen) = 1-(125/216) = 91/216
=42.13%
The title should be RIP maths by Anish Giri!
Respect Radjabov for showing concern for Air India crash today at Kerala despite being Foreigner while Samay was still unaware. Huge respect man and condolences to family 🙏
Binge Watched Comicstaan again today (for the third time), because why not! And damn, it just reminded me why I liked you since the very first (audition) episode itself! I remember watching a clip of your audition bit in the teaser/ trailer that was shown on TV, and though I hadn't watched much of Comicstaan Season 1, I was really excited for Season 2! During the decision making, when Zakir said that all the other contestants are to be judged by considering you as the gold standard, I felt that! 🤗 The respect for you increased multiple folds, since it came from my favourite Stand Up Comedian! 🙌 So please please please upload some of your old standup comedy videos! That needs to get out there! It's an honest request! Much love! You are doing great currently! ❤️ (I shall post this comment on a few of your videos so that it gets noticed and you read this! Sorry for the spam in advance)
Being a Jee aspirant this probability thing looks super easy to me. So let's calculate the probability of never getting a queen:x=5*5*5/216, so probability of getting atleast one queen= 1-x= 91/216.
Being a jee aspirant, you should be not watching these streams.
Your sincerely, a senior!
@@vikash126 Hahahahaha
@@vikash126 there's some entertainment too...not much but little bit
@@vikash126 being a Jee aspirant I'm tired waiting for the F#### exam, and pissed by people who are still filing petitions for postponing it😭. Na sahi s padh sakate h na sahi s maze karsakte h
@@mainwonahihun256 haha, don't worry man. You're pissed at the delay now, once you get the college of your dreams, you'll be more pissed about your choices in life 😛
" There are a total of 6 x 6 x 6 = 216 possible outcomes when rolling three dice with six sides each.
Now, we can calculate the probability that one of the dice will have queen by finding the probability of the complementary event, i.e., the probability that none of the dice will have queen, and then subtracting this probability from 1.
The probability that the first die will not have queen is 5/6, and the same goes for the second and third dice. So, the probability that none of the dice will have queen is:
(5/6) x (5/6) x (5/6) = 125/216
Therefore, the probability that one of the dice will have queen is:
1 - 125/216 = 91/216 ≈ 0.4213, or about 42.13%" - 2023 Chatgpt.
Radja really won our hearts😁 Anish was a sweetheart too!
20:25 , yes the probability of getting a desired piece atleast on one dice is 42.13% because lets see.. the probability of not getting a desired piece on one dice is 5/6 , so the probability of not getting a desired piece on all the three dice is 5/6 * 5/6 * 5/6 . So to get the probability of getting a desired piece atleast on one die is = 1-( probability of not getting a desired piece on any die ) = 1-[(5/6)^3] = 0.42
Anish question can be modelled by binomial distribution with p = 1/6. The answer would be summation i = 1 to 3 (3 choose i) p^i(1-p)^i = 1 - p^0 x (1-p)^3 = 1- 125/216 ~ 42%
I was just studying probability, (preparing for JEE), and it was hard to sit through them trying to calculate the probability 😂
Please rename the title to -
*Anish Giri solving Math Problems for an hour*
title change to Online math class ft Anish Giri 🥴😂😂
we should prepare our math problems for next time anish streams 🤭🤣🤣
I didn't know Saif's kid was so good at chess.
LOL.
Legends say anish is still calculating
Probability of getting 'atleast a queen'( IT can also be considered the piece you are expecting to come)=>
We have to consider 3 scenarios:- 1 queen, 2 queens and all 3 queens.
Now, total possible outcomes= 6*6*6=216.
1 queen=> Queen can be any of the 3 dices( ORDER SHOULD BE CONSIERED AS We have included Q-Q-R and Q-R-Q as different outcomes in 216). Now once you fix the queen in one of the dices the other two can be anything but queens So, 5 possibilites for each of the other two i.e, 5*5=25. Now, since three possible places for the queen. So total outcomes with only 1 queen=3*25=75.
2 Queens=> Again 3 possible arrrangements for the 2 queens are possible. The remaining position can be any of the other 5 pieces but queen. So, total outcome with 2 queens=3*5=15
3 queens=> 1
Probability= Favourable outcomes/ Total outcomes= (75+15+1)/216=91/216.
Inverse:-
1- probability that none of the pieces are queen=1-( 5/6*5/6*5/6)=91/216
I'm liking this whole concept
On each face there is knight, bishop, rook , queen , king , pawn. So probability of each piece is 1/6.
Probability of not getting a piece in one dice is 5/6 . Probability that the piece will not come in all three dice is 125/ 216.
So probability of getting a piece on atleast one dice is
1- 125/216
=91/216
Just curious to know howany inaccuracies, nistakes, blunders and missed wins these guys got
Binomial distribution with parameters n=3,p=1/6
Your maths blew my mind 🤣😅
Aww Anish chill❤
And Samay, please buy NCERT Class 6th Maths textbook.😂❤
LOTS OF LOVE FROM NEPAL
01) If the event/s occur simultaneously then you multiply
02) If the event/s occur at different time then you add
SIMPLE ;)
For getting some other piece than the queen probability is 5/6. So, for getting some other piece on 3 tosses of die, probability is 125/216. So, getting the queen on one of them is 1-125/216=91/216
25:04 I never knew anybody who watched Detective Conan growing up...brings back old memories!
Anish does look a bit like kiteretsu and conan xD
Dude all props to Raja for being such a huge sport. Respect!
This kind of chess is a cage for players. God decides your moves no matter how well you plan. Sounds like life
1/6 +1/6 +1/6 =1/18 - GM, SAMAY RAINA
Now bm samay raina!
Fields Medal Winner
1/6+1/6+1/6 ≠ 1/18 because they have common denominator so just add the numerator (15:20)
15:10 this is going to be a meme material..mark my words
How to add time stamp
@@jaymakwana8636 just write the numbers no prefix no suffix..just 4 numbers with a colon in between
4:02 when life gives you lemon make it ayesha takiya
Radja you should write a book titled : "How to be like Radja" ❤
Brace yourselves, comments with absurdest probability solutions are coming
All of these answers are wrong, the probablility is 43/216= 19.907%
1/6+ 1+36 + 1/216
@ABHISHEK GAUTAM
Bhai mujhse behes mat kar mere Jee Mains mai poore 42 number the 😁😁 haha
answer toh tera bhi sahi lag rha hai bro, but meri reasoning kyub galat hai please batana?
19.907% is the actual probability of getting AT LEAST one desired peice in the 3 dye.
In the first dice, 1/6. Simple to understand.
In the second dice, 1/36.
In the third 1/216.
1/6= 16.67% chances you get Queen on first dice.
1/36=2.77% chances you get Queen on 2nd dice.
1/216= 0.4% on the third dice.
Notice how the chances keep getting lower? This is why we dont win the 777 jackpot that often.
Another way to understand this is,
Hoe many total outcomes? 6*6*6=216
How many favourable outcomes on first dye?
1
How many on second?
6. (Why? Because These ARE DEPENDENT events.)
Let me explain this with an illustration of all favourable outcomes on the second dye.
Pawn Queen
Bishop Queen
King Queen
Knight Queen
Rook Queen
Queen Queen(AT LEAST ONE QUEEN, not ONLY ONE queen)
Total 6 favourable outcomes.
Similarly, corresponding to EACH one of these outcomes, there will be 1 Queen outcome on the 3rd dye.
Meaning, on 3rd dye,
6*6=36 outcomes.
SO,
Total favourable outcomes =1+6+36 =43.
43/216 = 19.907%
If you would add the fractions:
1/6 + 1/36 + 1/216
You would get,
(36 + 6 + 1) /216
= 43/216.
=19.907%
@ABHISHEK GAUTAM Bhai mai 4th mai year pahoncha hun abhi just.
Third tier college hai no surprise there. 7 cgpa. No projects, lund bhi nahi aata programming ka aur placements start ho chuki hain
Government ko faaltu hi gaali dete hain sab, mere paas sab kuch hote huye bhi mai unemployable hun, and this is the reality of most of the undergrads.
Waise maine actually jee aur boards dono mai probability aisa iklauta chapter prepare kia tha jiske saare answers correctly diye the.
Abhi 4 saal baad sab bhool gaya aur upar se 1st aur 2nd year mai
random discrete variables, probability density function, advanced stats
ye sab padhke dimaag ka bhosdha ho gaya, saare concepts hil gaye haha
@ABHISHEK GAUTAM aur bhai, mai 42 total jee ke bata raha tha, not only maths xD
42/360. lmaoo
@Ashutosh Bhai probability aise solve nahi kr skte ye wali kyonki har dice pe aane vale pieces interdependent nahi h. For each piece on a individual die , the probability is 1/6. So we can have 3 cases , that queen on 1 soace5 , queen on 2 spaces and queen on all 3 spaces.The calculation is - 3*(1/6)*(5/6)*(5/6)+3*(1/6)*(1/6)*(5/6)+(1/6)*(1/6)*(1/6)
= 91/216
Ps I understand where you are coming from , but all I'm gonna say is that dont add probabilities non interdependent events like this.
13:46- when Samay makes the biggest mistake of triggering AG, "Let's calculate it". Rest is history
Probability of atleast 1 queen is 1 - (not getting any queens)
1-5/6*5/6*5/6
All the math teachers in India would be so proud of this comment section.
Thanks Anish 😄
Guys, Just saw the stream, I think Anish is right, probability of getting a piece in a 3 dice role event should be 1/6+1/6+1/6=1/2 ('+' because piece on the first dice or on the second or on the third). If we calculate the probability of getting the piece at least once them the probability is 91/216.
There will be case of overcounting if you simply do 1/6 + 1/6 + 1/6
The addition rule works for mutually exclusive cases but here there are overlaps so you need to subtract possibilities off using the inclusion-exclusion principle.
Radja respect op man for being such a sport
Next stream should be Understanding Maths ft Anish😂
91/216 is the probability of coming a piece
I can watch these people all day
It would be hard for me to appreciate Anish Giri from now on even he still a legend in chess.
isnt it amazing that you have got these gms playing dice chess :D
At 10:10 samay ask Anish how much is the probability and later than he said Anish to stop calculation 😂😂
Thank god! None of them tried to become a mathematician
Those calculations made me think of my mathematics classes!!!!
...............audience = teachers trying hard to teach ;
................samay, anish : students who ignore the correct answer to find the wrong one perfectly!
😂😂😂🤐
Samay making me feel good about my maths skills
Probability of atleast one queen = 1 - Probability of No queen.
So 1 - (5*5*5/6*6*6) is 91/216.
Total number of possibilities are 6*6*6=216
Total number of ways we could not get a specific piece are 5*5*5=125
So, total number of ways we could get a specific piece are 216-125=91
So, chance of getting a specific piece is 91/216.
In percentage it is 91*100/216=42.129%
Oh boy, Radja got completely screwed over by the dice
Probability will not add but the total number of outcomes will increase , right now it is 216(6*6*6) , and probality of getting at least one queen is 1- probality of no queen.
Probability of no queen 5/6*5/6*5/6. So probality of at least 1 queen is 91/216 and if dices are increased probability of at least 1queen will keep on increasing.
Which app is he using for rolling the dices??
that was some serious maths
RADJA IS SUCH A SWEETHEART. MAN I LOVE HIM NO MATTER WHAT HAPPENS IN FUTURE.
Samay I can create dices which is also stop rolling pieces which are not in board
Probability you don't get a desired piece on any of the dice
= (Probability you dont get on the first die)* (Probability you dont get on the second die)* (Probability you dont get on the third die)
= 5/6 * 5/6 * 5/6 = 125/216 (This is the probability you won't get the desired piece on any of the dice)
So probability you will get your desired piece on atleast one of the dice = 1 - 125/216 = 91/216 (42.16%)
Cheers :)
FInally a comment with the right answer!
The probability of any piece coming up atleast once is 50%. And here's how the probability adds up (which is the only thing Anish couldn't figure out) : Assume you get three different pieces (which is a clear possibility and has happened often in the game), let's say a pawn, a bishop and a knight, on the three dices. You've now got three out of the possible six pieces, in other words 50% (3/6) of the pieces. Hope you can be at peace now, Anish.
We want Radjabov's ratings back. Samay do something !!!!
Please make them play unranked very sincere request,always worried about their points being ruined
Anish also has another channel in which he had done ice bucket challenge
case 1: 1 queen on three dices-1/6*5/6*5/6=25/216
case 2: 2 queen on three dices-1/6*1/6*5/6=5/216
case 3: 3 queens on three dices-1/6*1/6*1/6=1/216
total probability=25/216+5/216+1/6=31/216
prob of queen 1-(5/6)^3
the chance for one piece appearing at least once on three dices is approximately 0.421296 %
So the secret is revealed.... Samay is deleting his secondary CZcams channel with 10 subscribers.🤣🤣🤣🤣🤣😂😂😂😂
Name of the channel
Btw anish also has anothr channel in which he did ice bucket challenge
You were not supposed to spoil the fun! ** facepalm **
@@abhishekgadhia1669 I am Anish Giri of CZcams Comments section 😅
MANNNN WHY DO YOU HAVE TO SPOIL IT? ......PLEASE DELETE THIS COMMENT........SAMAY WILL NOT LIKE THIS...
1:38 Chikna is gali in Nepali Anish reaction .
Bring that Harddisk from sumit saurav Where All your Stand up Is In it bro And Put your comic Videos man on Your channel And when that stream Happen 1600 cross Or delete Account.
Lol😂😂👏
42:38 "a pawn a lover"
Never knew chess players are the coolest people ever.
People spend lifetime in getting the rating u lost today...
it will be 3/216 for one queen and 7/216 for atleast one queen
its 21/56 to get any piece( not considering the sequence of pieces.)
"You know what a dice is?"
Should've said no
Annis is wright chans of get one peace 1/2
If we have 7 dice their have chance 7/6
You thing how it would be possible because probability never cross 1
But in this example it should cross 1 because there is 6 pieces and we have 7 dice so always we have 7 pieces after rolling dice but there is 6
Was this the first stream with Radjabhov
Btw, the probabiltiy of getting atleast one queen is 1-((5/6)^3).
P.S: i know this is correct. Please dont come and argue
Does anyone have a link for the dice chess
want to play it with some friends and can't find any link
Ok Specially for Anish Again, if you Want to Come to 42% differently then it's (3*(5^2)+3*5+1)/6^3 = 91/216 ~ 42%
Kids: 25/216+75/216+1/216=91/216
Intuitive: 1/6+5/36+25/216=91/216
Nerds: 1-125/216=91/216
Legends: 1/6+1/6+1/6=1/2
Ultra legend: 1/6+1/6+1/6=1/12
Ultra legend in beast mode: 1/6+1/6+1/6=1/18
Bruh no it's all dice are identical you're not supposed to multiply with 3 and 2. I think at least one is 25/216 + at least 2 5/216 and all 3 1/216 adding 31/216
legends say Anish is still calculating