Amazon Interview Question - Coin on a Chess Board - Probability Puzzle
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- Опубликовано: 12 сен 2024
- Amazon interview puzzle:
A coin of diameter 1 is thrown on to an infinitely large chessboard with squares of side 2. What is the chance of the coin enclosing two colours? So, you have to calculate the probability of the coin enclosing both black and white colors when it's landed on the board.
The intention of mentioning an Infinitely large chessboard is that we should not consider the edges of the board, rather we should just focus on the main board.
The difficulty level set by the composer of this puzzle is just 2 out of 5.
So, you must give it a try. Pause the video and think logically.
You can share puzzles and riddles with me on these links:
Gmail : logicreloaded@gmail.com
Facebook(message) : / mohammmedammar
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Questions are difficult, but solutions are simple, amazing....
always all ...like that
Check the effect of x and y positions of the coin independently. In both directions there's a repeated cycle of length 1+1, where the coin is over a (horizontal/vertical) line or between lines.
So the probability of being between lines in both directions is 1/2 * 1/2 = 1/4. The opposite is the coin being on a horizontal or vertical line, probability:
1 - 1/4 = 3/4.
I think your answer is incorrect coz suppose if we take the side of a square 100 cm and the coin has a diameter of only 1 cm then according to your approach the answer will be same as 3/4 but here the correct answer will be different you can think yourself that the coin has a very less chance to land on the line or you can solve it using the method shown by Amar...😊☺️ Hope u understand ✌️🤞
@@alkabindal4793 No, my solution is correct for the given problem.
For the values you gave, the length of the repeated cycle is 100, of which 99/100 the coin is between (horizontal/vertical) lines. Between both horizontal and vertical lines: (99/100)^2. On at least one line: 1 - (99/100)^2.
@@jounik4128 Ohk i understood👍👍
This is how I did it.
I got the solution a different way which was easier to calculate: To cover 2 colors the coin must be crossing a vertical vertex, a horizontal vertex, or both. Whether it's crossing a vertical or horizontal vertex are independent probabilities, if the coin is placed randomly. The coin can travel a distance of 1 while across a vertex, and a distance of 1 between vertexes, so the probability of it being on a vertex in 1 dimension is 50%. So the probability of being on neither vertex would be 50%*50%, which is 25%, so the probability of it being on at least one vertex and thus covering 2 colors is 1-25%, so 75%.
Good one Martin. Would you please do the calculation same way for a 4x4 square and coin of diameter 1. I am wondering how did you consider a probability of 50%... and what do you mean by 1 dimension.
I got your overall idea... but another example with side 4 would help everyone.
It's equivalent to the probability of at least one heads in two coin flips. if you had only vertical stripes twice the width of the coin, then, say the coin travels to the right. As the left edge of the coin leaves the left edge of one stripe, there's one coin width of travel before the right edge of the coin begins touching the next stripe. So, the coin is touching two colors half the time (ratio of coin width to stripe width). Since they're squares--essentially horizontal and vertical stripes--we must consider the probability in the vertical direction as well. That is, there's .5 chance a random X coordinate gives touching two colors, and there's .5 chance for a random Y coordinate as well. The chance of at least one head from two coins is three of the four possible outcomes. Or, consider must be both tails to fail, so it is the product of the chances of tails for each coin subtracted from one (100%). [1 - (.5 * .5)] So, considering both directions, each with .5 chance, there is .75 chance of success overall.
@@LOGICALLYYOURS Sure, by one dimension I mean just along the x or y axis. So if you're moving from left to right with the 4x4 square with diameter 1, along the x axis, you're over a vertical vertex for a distance of 1 as you move, then between vertexes for a distance of 3 as you continue to move to the right before you get to the next vertex. So you have a 1/4 probability of being on a vertical vertex and a 3/4 probability of being not on a vertical vertex. So since the same thing applies to the horizontal you have a 3/4 * 3/4 probability of being on neither vertical or horizontal vertex, so that's 9/16 probability of being on 1 color, so a 1-9/16 probability of being on 2 colors, which is 7/16 or about 44%.
Most underrated riddle channel
Do you know other good riddle channels
@@ABCXYZ-zt3dr TedEd posts really interesting riddles, though not not very often. You can check out their “can you solve this riddle?” playlist. Here’s the link: ruclips.net/p/PLJicmE8fK0EiFRt1Hm5a_7SJFaikIFW30
@@aceofspades8634 thanks☺
@minato Namikaze
PLEASE KEEP UPLOADING RIDDLES CONSTANTLY>>
YOU CAN MAKE THIS CHANNEL THE RIDDLE ENCYCLOPEDIA
This type of feedback certainly boosts my confidence... much appreciated bro :)
Riddlepedia?
Amazingly simple solution. Brilliant
Amazon interview puzzle: you have a 10 minute break, the break room is 5 minutes walk from your work station, how long is your break in the break room?
Hi Orth, that seems to be a cool riddle. Please share the exact riddle statement at logicreloaded@gmail.com.
Is it 0 ???
Well let's see... my absolute top sprinting speed is at about 18 mph. My normal walking speed is about 3 mph. So if I sprint to the break room it should take me approximately 1/6 the time it would take to walk. Given that I spend about 1.67 total minutes sprinting, I can enjoy about 8.33 minutes of actual break in the break room. XD
@@LOGICALLYYOURS👍👍 Could you tell which softwares we need to make this type videos-- software for animation, video editing.. And how you created those coins in video.
@@LOGICALLYYOURS👍👍 Could you tell which softwares we need to make this type videos-- software for animation, video editing.. And how you created those coins in video.
Hey!
I used Inclusion-exclusion principle and got the same result. (P(A): probability of enclosing vertically, P(B): probability of enclosing horizontally. P(A)+P(B)-P(AB) = 0.5+0.5-0.25 = 0.75)
Is this a valid solution or was it just blind luck?
Hi David, it's a fresh perspective.
But i see one problem here.
You don't seem to be considering the ratio of diameter and side of square.
There is an extension to this puzzle, that we should generalize the solution for diameter d and side s.
If you take different values, you'd get a nice curve showing how the probability increases with decrement in coin size.
@@LOGICALLYYOURS You can let P(A) and P(B) be d/s and P(AB) would be (d/s)^2. Then the probability is 2d/s - d^2/s^2
@@LOGICALLYYOURS logically, the probability to cover the favourable area(intersection) should increase with an increase in coin size(better chance to land on the favourable area). So what do you mean here by the satatement above that probability increase if the size decreases, isn't it wrong?
I came at if from another angle. Something I learned from other math puzzles, break it down into a simple case version, then rebuild it. Imagine it in 1 dimension where there is a single infinite line of squares and the coin only moves in 1 dimension. Moving the coin along the direction 1 square side length distance you'd notice that for half of the distance the coin is is inside the square and for half it protrudes. I used the left edge of the coin with it protruding to the right of that point to visualize. Thus you get 50% in 1 dimension of any position yielding two colors. Now you have to leap to two dimensions.
The important thing to note is that there are 4 potential outcomes; both axis double colored (the corner), one double color (an edge), the other double (the other axis' edge), and neither double colored (centered in a square). Each axis independently 50/50 split. If either or both of the 2 dimensions hits two colors the condition is satisfied, but if and only if both dimensions come up single color does it fail. Thus the fail rate (single color) is the probabilities multiplied, rather then the success (both colors). Giving you a 25% chance to see a single color, and 75% to see two.
You can also just note there are 4 states with equal probabilities, thus the chance of any single state is 1 in 4, and 3 of them satisfy the condition but this way allows for different sized squares/coins and... more complications. I'd imagine this would easily extend into three dimensions, a sphere of unit 1 tossed into a 3d grid of side unit 2, being ultimately a 7/8ths chance of occupying at least 2 boxes.
Pretty bold to assume that there is no chance of the coin to land on it's edge. =D
Super and clear cut explanation🔥 Teaching speaks 👌 Continue this to give us knowledge
Except that the perimeter of the outer squares allows for the coins that will fall along it to enclose just one color instead of falling into the square of another color. Therefore odds will be slightly decreased.
He mentioned it's an infinitely large chess board
The problem seems to be ill-posed. Nothing indicates that the coin's center should land on the central square. If we assume that it can land anywhere on the board, we get a completely different probability since squares around the seams have a smaller area adjacent to other squares. If I got this kind of interview question and the interviewer insisted that this is the correct solution I would be rightfully pissed.
How about this one? Imagine the universe is filled with alternating cubes of chocolate and vanilla ice cream with edge length 2. Suppose we can randomly extract a unit spherical scoop. What is the chance of getting only chocolate?
I'm figuring within a given flavor cube, there's a unit cube where the center point of the scoop can be and get only that flavor. So, to get only chocolate, you have .5 [prevalence of chocolate] * (1/8) [volume of target cube/volume of flavor cube] = 1/16 chance of only chocolate.
infinately large chessboard? By definition a chessboard is 8x8 squares. so the maximum area size is 16 x 16 = 256. By using the term chessboard you cancel out the word infinately so the chessboard has edges too where the coin can land on. The correct calculation should include the side areas where the centre of the coin could be outside the board. For all side squares together the possible areas would be 4x corner area of 0.5 x 0.5 plus 32 x side area of 0.5 x 1 plus 56 (all side areas x 2 minus 8 corner parts) quarter circles where the coin could be touching one colour. The correct chance should therefore be ( 1 + 16 + ( 14 x 0.785 == 14 full circles with a radius of 0.5 = 11 ) + 64) = 92 : 256 ==> 35.9375 % chance of hitting 1 colour and 64.0625 % of hitting 2 colours. For pi i used 3.14 only.
There is a simpler way to see. The 2x2 square can accomodate the coin in 4 places evenly. Totally there are 16 chances of coin falling into either inside the color or on the line. So 4/16 = 25% is the chance it can fall in the color. Therefore (1-.25) 75% on the line.
Or
The coin can be accomodated evenly in 4 places within the square. So 1/4 is the chance it will fall within square. So 3/4 or 75% is the chance it can fall on the line.
Answer: Get me the coin and the infinite chessboard. If you can't let's toss one, there is a higher probability of me winning than this stupid question.
Also it's not 50/50 less than 50, but way higher when dealing with infinities without a frame of reference or derivation.
Always play to your advantage. This is fun.. i subscribed bro.
Use this formula
d(2x -d) / x^2
Where,
d: coin diameter
x: side of square
Can you explain why this works?
I got the Answer with Perfect Reasoning!!!!!
Nice puzzle
Awesome puzzle
My ans is also 75% but different approach
My approach= largest circle that can come inside that square " r=1"
"Area= π"
Area of that coin =0.25π
Probability that coin come inside the square= Area of coin/area of big circle
=0.25
Probability that coin does not come inside square= 1-0.25=0.75
"75%"
Please leat me know weather my approach is correct or not
Largest Circle that can come inside the circle have r=4
@@clonemrk4163 no square side is 2
That's way circle diameter is 2 , so r=1
@@ABCXYZ-zt3dr what is r here ?
@@clonemrk4163 r is radius
@@clonemrk4163 now I realised that my approach is wrong, when you change the size of the coin my approach will not work
As I opened the VDO eventually it's speaks out from my mouth
What's up logical people this is Ammar .........😂😀😀😂🤣🤣
All you have to do is to realize if the coin is totally inside a square, the center of the coin can be no closer than a half inch from the edge of any square. So a half inch on each edge it cannot be in. That leaves only an inch square in the middle of each square that it could be in and touch only one color.
So only one square inch out of the four total square inches of each square could it be in and touch only one color. So that means it's a 75% chance of it touching two colors.
Perfect David !!
One more excellent video from the factory or AMMAR'S LOGIC INDUSTRIES.
thank you ammar for a very nice riddle, not very hard, but very nice.
btw, what is the chance of this coin to cover only 1 white area , and 1 black area?
(I mean not, for example, on the cross middle of 4 squers, forming 2 white and 2 black areas, like the logo of BMW, and not near the corner, covering 2 black areas and 1 white)
What is the chance to cover ONLY 1 white and 1 black areas? That problem is turning the riddle from Amazon interview question, into NASA interview question...
Can you please, make part 2 to this video, which in you show the answer to my challenge?
Thank you Ammar for very good video, and very simple explanation!
It looks really simple after doing the original question. The answer to your question is 1/2 or 50%.
Just imagine that you divide the single square in 16 equal parts :
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
Each part will be of dimensions 0.5×0.5 and radius of coin is also 0.5 .
Coin's centre should not be in parts 22,23,32,33 as the coin will lie in a single square as explained in the video.
The coin's centre cannot be in parts 11,14,41 and 44 as it will cover 3 or 4 different squares of two different colours. As the part's sides and coin's radius are equal so,if the coin is in those parts the radius of coin will surely exceed all of the sides of the parts and cover 2 or 3 other squares sharing their perimeter with that part.
We will get desired result on parts 12,13,21,24,31,34,42,43 as they share there perimetre with only 1 other square so the coin can only cover total of 2 squares.
So out of 16 parts 8 parts give us desired result.
So the probability = 8/16 = 1/2 = 50%
@@alfezkhan591
Thank you Alfez Khan for your time and answer!! Looks like a beautiful solutions! I will read it carefully. I, myself, didn't solevd it yet, even though I suggest the new variant.
First I will try to solve it, then I read your solution.
Ty !!
(Edit: Part 2 of my comment)
I soleved it now with dividing the chess square into 16 parts:
A B C D
E F G H
I J K L
M N O P
When the center of the coin is on area:
F,G,J,K = 1 color.
A,D,M,P = 3 or 4 colors.
B,C,E,H,I,L,N,O = 2 colors only!
Our solutions are the same (didn't understand why 11, 12,13, etc. but good as A,B,C,... as well)
You solved it correctly!!
Thank you.
I think Ammar will like it😃
Thanks Tamir for the appreciation and importantly for the fresh aspect of this problem. I just saw your email and searched for the comment. I'll give it a try at the earliest and will also verify the answer from Alfez khan.
Of course one more puzzle would be a greater idea.
Hi Ammar, can you put some riddles from Smullian plz?
Here we have an another amazing video 👌👌👌 Keep it up. But please reduce your uploading period.
This one's a beauty 👌
Awesome awesome. Mind blowing
Thanks Kumar :)
And to make the probability 50%, the squares would need a side length of around 2.414
No, the Chess squares need a side length of 3.414 to get 50% probability
@@clonemrk4163 which is 2+root 2, I’m guessing?
I'm getting x = 2 + root2 for P=.5
For x = 1 + root2, I'm getting P ~ .657
Have one doubt, what if the squre's side is of length 1 ? then according to same formula shaded area = main square area - point = main sqaure area that means 100%. But we know that if coin lands on center of square it will not cover two colors and since board is infinite, there are infinite such points where coin can land. so what about the probability now? formula is giving 100% but we have cases where it covers only one color.😁😁
I got the answer without seeing the answer 🤩
🤓
Or you could get the probability that it doesn't enclose two squares horizontally (1/2), multiply that by the probability that it doesn't enclose the squares vertically (1/2), and subtract that from 1.
Extra credit: What if the coin diameter is 2? There is exactly one point in each square where the coin will be on one color only. Everywhere else the coin will be overlap at least 2 squares. If the area of a single point is zero, the calculated probability is 100% that the coin will overlap two colors, yet we know there is a non-zero probability that the coin will fall exactly in the center of a square. This appears to be a contradiction or paradox based on the assumption of infinite divisibility.
That's where mathematical limits come in. The chance that the coin will be on two colors is not exactly 1, but 0.99999... , which approaches 1.
Actually, the probability that the coin lands at the center IS 0, I'm not sure what the problem is.
Ok. Suppose a side of squares is 1, is the chance of the coin landing on only one color 0 %? The coin can never land on the center of a square?
The coin should land exactly on the centre of the square to enclose just a single color, but the problem is that a single point does not have any 'area'.
In fact I had the same thought while trying for different values.
Theoretically (as per my understanding), the area of a point is negligible in comparison with the remaining area of the square. So, chances of enclosing two colors would be 100%, or the other way, chances of enclosing single colour would be 0%.
I will still give a thought on this.
Readers, please share your thoughts.
@@LOGICALLYYOURS It will tend to 100% but will be slightly lesser than it (concept of limits). The probability of coin covering 2 colours would be 99.999999.......%.
@@rahilsanghavi9347 Suppose the chance landing on the center of squares is 0% because a single point doesn’t have any area, every other single point than center doesn’t have any area either. It means, the coin never lands on the chessboard! lol
@@tommynickymicky That is the beauty of an uncountable sample space. The probability associated with any single 'point' tends to zero but that associated with the entire board is 1.
Meeting point Corner of 4 square if coin cover that point than it is one throw enclose 4 square in 1 throw.
1st can be is 1, 2nd can be 2 & 3rd can be 4, total in 3 throw best maximum enclosement will be 7 square.
I heard the problem and then figured it out on my morning walk.
Looke at it as a '1 more before going to sleep kinda deal', instantly shut down my PC and solved it while trying to sleep lol
nope, it's ambiguous
the solution depends on HOW you throw randomly
which random distribution? (uniform was not stated and even if it was it doesn't exist on infinite domain)
I think the answer is :
=(3n+1)*(n-1)/(4n*n) for a chessboard of length L=n*2 and n is the number of squares vertically and horizontally
Amazing Sir . I could have solved this, due to my curiosity I saw the answer.
So, we have four possibilities.
The coin can be completely in one square
The coin can enclose 2square
The coin can enclose 3squares
The coin can enclose 4squares
The last 3 cases are favorible. So 3/4
Isn't that also right?
"The coin can elclose 3squares"...how?👀
@@beatbox9794 If the edge was on a point in the middle of 4 squares the sides on two adjacents would be overlapped by the coin. This means as long as the center of the coin is not on one of the four edges of the squares around the central point it would be on three different squares
@@beatbox9794 If the edge of the coin*
And this isn't actually a valid way to approach this problem because different outcomes don't necessarily have the same possibilities. If I buy a lottery ticket, I can either win or lose. But that doesn't mean I have a one half chance of winning just because there are two different possible outcomes. The chance of me winning could be 1/800,000,000 which would mean the chance of me loosing would be 799,999,999/800,000,000. You cannot universally treat possibilities as equally probable
Same probability of happening*
nice one. but there can be a corner case. the probability is valid for middle 6*6 = 36 squares but not for outermost squares. There, if we have to consider if coin can be on the outer edge or not. Based on which answer will change.
Question states infinite chessboard. We won’t have any corners
Param has mentioned it correctly. In fact, i wanted to give a clarification in the video... but the beauty of a puzzle lies in fewer information.
Probability theory is magic.
Let's just consider the square the center of the coin lands in. If the center of the coin is distance r (radius) or less from one of the edges, it'll overlap the adjacent squares. In this case, it's 0.5 away. So there is a 1 by 1 square in the center of each 2 by 2 square thats safe for the coin center to land. 1*1/(2*2) gives the percent of safe area, or 25%
That's perfect!
bro your videos are very helpful for for me thanks
Nice easy prob
I always wait for your vedio sir 👍👍
I doubt this puzzle has to do anything with any interview. Nice puzzle on its own.
what if the diameter of a coin is same as side of square, what is the probability that it falls exactly in a square? 1/infinity?
Can Jeff Bezos solve this?
I think Jeff Bezos is done with any puzzles now & thats why he created Amazon for others to solve such puzzles.
@@nitinagrawal6637 ya bro ,his head only tells how much puzzles he has solved.
🤣
I solved a little bit different,
I considered 4 squares, 2 blacks and 2 whites,
I needed to calculate the area around the axes, which was 12 over the total area 16, it gave me the same output as yours 3/4.
I have a doubt with my approach...
Can you please help..
So I consider that if circle lie inside square then it will enclose only one colour then probability of enclosing one colour is equal to (2*2 - .25π)/4 = 45/56
Now probability of enclosing two colours= 1 - 45/56
Hi Ammar. I love your channel and your puzzles. Keep them coming! I know a very good puzzle that I learned a long time ago and it is one of my favorite puzzles. I would like to share it with you so you can post it on your channel. How can I send it to you?
Hi Mark, really glad to see your comment. I'd love to see your puzzle and post it on my channel. Please share it at logicreloaded@gmail.com.
When video starts mind is like this must be solved, after listening the problem statements my mind is like ok see the solution, from next problem I will solve that.
Easy one. Nice.
Bro I have a doubt with your answer...
75% probability is satisfying with all the squares but not with the 4 edges right 🙄
*One of the best channel*
elegant
I have a puzzle for you
We have a tank of 50ltr and its filled with 50ltrs of water. Without using any object ,only presence of you and tank ,how can you remove exact half of tank ie is 25 ltr
This question is asked in my interview please can yoy solve this
I will rate It 3/5
Because it took almost 3-4 min to solve.
Who else was able to solve????
I solved in less 30 seconds.
@@arjunannamalai.m3447 great bro. 👍
The propability of the coin landing in a square is the area of the coin devided by the area of the square: pi/4pi which is 1/4 which is 25%. Anything else it means the coin lands outside and that is 75% chance
This is like when people look up answers to coding problems and then can't explain it lol
next task is to calculate the chance of the coin enclosing 1/2/3/4 different cells
the answer is below :)
25%/50%/5,365%/19,635%
Fantastic
Great video
This is bothering me if someone can give enlightement, so if the coin falls in the lines between the shaded area and the non shaded area, it wont be able to hit 2 colors, in other words the probability of the coin hitting 2 colors should be less than 75% not equal to 75%???
Why did I have to scroll this far to see someone make this point smh
The riddle states it's an infinitely large chessboard. To be a chessboard, it must have 64 (8x8) squares. If it has more squares than that it is not a chessboard. So the 25% must be decreased by the probability of landing on an outside square, overhanging the edge.
And if the board is truly infinite in size then the sides of each squares must also be infinite
Too good
Provided the entire coin lands completely within the board, the same logic for the centre square cannot be applied to the squares on the sides as the squares on the sides are bounded on 2 sides
But since its an infinite chessboard there is no "outside" of the board.
@@AleyinMica You are right. 👍 I forgot that part.
@@AleyinMica A chessboard is 8 x 8. So an infinite chessboard is impossible, even theoratically.
What if falls near the corner and touches three or four squares
Osm logic application 👍👍
Since the outer square is 2 and the inner layer is 1 and shaded also adds up to 1, if the shaded 1 is substracted from 4 then does the probability shift to 75% in favour of the other ? Maybe the answer is 50% ?
the shaded area does NOT add up to one, it adds up to 3 because it's just the remainder of outer square - inner square i.e. 2²-1²=3
@@TigruArdavi instead of giving the dimensions, let me know the size of the shaded area. If that's 0.5 + 0.5 on each side then it becomes 1 and so subtracting this area from the bigger / outer square would give 3 which would be the inner square.
1000th veiw
The correct answer is "I don't know. How about you hire me to do some engineering while you HR guys fuck around with more games."
👍wow
Whoa, I actually got this one right.
Clever, but how will this help in real life?
What if chess board was not infinitey large?
In that case you'd need to consider the squares on the edges in a different way.
@@LOGICALLYYOURS 👍
@@LOGICALLYYOURS chess match?
What if the coin land in other square rather than the middle one but still touching both sides ?
It's an infinite length chess board, probability of coin landing on any given square is equal, hence the condition of circle falling between two square will remain same and so will be probability (3/4)
The Answer is wrong bro, who says the center has to land on the center square bro? it can land on other square too calculate the others too
Why are we focusing on only one square? Seems incomplete question
Because the chess board is infinite and of equal squares so the coin landing on any of the square independently remains constant only factor which changes is the coin center landing position on any particular square
👍
SUPER PUZZLE
The question should be rephrased, if the chessboard is infinite in size then the chance is of the coin landing on both colors is also infinite, because 75% of infinity is still infinity
The intention of mentioning an Infinitely large chessboard is the we should not consider the possibility of landing the coin on the edges of the board.
With infinite number of squares, the ratio of probabilities still remains the same, i.e. 75:25 (both colors: single colors)
You are surely right that 75% of infinite is infinite but the question is not about finding the actual value of the area where the coin can lie but just the probability of the coin lying in that area or in other words the percentage of the area in which the coin can lie, which we got 75%.
I get this approach but there is a paradox on this kind of puzzles
Beacaus we dont calculated the probabilitie of the coins landing exactly on the sides of the small inside square !!!!
Good point Tony, but the inner or outer square is made up of points which are negligible in size. So the squares are indeed invisible lines.
@@LOGICALLYYOURS thank u for clarifiying !!!
can i give the answer as 0.75
Works fine
Old one.
Q- What would happen if the coin has a diameter of 2
A- We assume that the center of the coin is zero, but is actually the area of the center, because the shortest length in physics is the length of the plank. For this reason an area of the center of the coin will also be formed, but very little.
Your answer is wrong. The outside squares have a different probability.
sry, they said infinite chess board, so there are no outside squares
¾
75%
Yo Amar
pro-ba-BIL-i-ty
1 solved by it with a different method and the answer is same.
What was your method.
it's 50/50 - it did or it didn't. ;)
Easy puzzel. I will give 1 star for it, because I able to sole this very easily
so r u the next CEO of amazon
@@Sahil-15 probably
Big fan bro play chrss with me
I will like to play with u
@@alexcruiser5277 lichess?
@@chandraprakashgupta61 yes iam part-time chess player.recently crossed level 6 at chess free App
0.75
this one was quite easy. Actually, a RD sharma problem of 10th class
No not at all similar to that question
The answer is wrong as it is not a valid case for corners
There are no corners... The board is infinite...
@@minecrafting_il No because chessboards only have 8x8 grids in them
@@alaydeonrono3355 but he talked about a theoretical infinite board...
@@minecrafting_il The condition that it's infinitely large is paradoxical with there being a finite number of squares on a chess board with their lengths given
@@alaydeonrono3355 but the riddle is not about a normal chess board.
It is about a theoretical infinitely large board with infinite squares
What if I tell you that you don't need any math to solve this puzzle? In any chance, the coin can only cover areas from 4 squares at the most (75%) and 1 square at the least (25%). So chance of the coin enclosing two colours is 75%.
Amazing question... But don't forget to give units for area... It annoys me😐😐😐