Grid Probability
The grid image below is not drawn to scale ... consider a grid like the one pictured, in which each space is a perfect square. If you toss a coin ( either penny, nickel, dime or quarter... the coin should be smaller than a square ) randomly onto the grid, what is the probability of the coin landing on a corner of one of the squares?
posted by Zaux at
10:47 AM
26 Comments:
Can't be done using the info given. Try it with a coin of 0.00000001" diameter. You'll almost never touch a corner.
Does the coin have to fall entirely within the grid or can it overlap the corners around the perimeter?
There are several different possible solutions depending on the answer to this.
We also need the coin and grid dimensions.
there is a general answer without knowing specifics ... but if, you must, assume that if the coin is position directly in the center of a square, at the 4 closest points the edge of the coin will be 1/8" from the edge of the square. Does that help?
I got it - very sneaky but now I see the correct meaning of the question.
Given that no measurements were given I guessed this was more than a mere maths question, but I couldn't see why until now.
The probability is zero.
Nice one Zaux.
My above answer assumes that the coin cannot fall outside of, or across, the outer lines which border the grid and into the blank white space around the edge.
I believe this is the correct interpretation given the "trick" in the question.
Just off the top of my head - given r is the radius of the coin and s is the length of the side of one square - r < s/2 and the coin cannot protrude outside the edges of the grid:
9*pi*r^2 / (4*s^2 + 8*(s*(s-r)) + 4*(s-r)^2)
Not taking into account the fact that there is a slightly probability of the coin landing closer to any edge of the grid in "real life".
-h
I like this one because it is perplexing ... Hagar has the right idea .... but the published solution is different.
The published version states that the probability of the coin landing on a corner is about 50%. They go on to say that the probability can be calculated by dividing the area of the coin by the area of a single square on the grid.
So ... what ... πr²/s².
The reason I divulged the published solution so soon is that I am hoping one of you math wizards (which I am not), will be able to explain why that is the probability of the coin landing on a corner. In other words, why is the ratio, of the area of the coin to the area of the square, the desired probability? Help!
Grid Probablity
Assume coin must be entirely within grid
Calling bottom left corner (0,0) and top right corner (4s,4s)
The centre of the coin may then fall within (r,r)and(4s-r,4s-r)
A=(4s-2r)^2=total area centre of coin may fall within
Area for centre to fall within causing collision with corners
Assumption: The side length is large enough that these areas do not overlap
A=Pi*r^2 for each corner
9 corners
Acorners=9*Pi*r^2
P=Acorners/Atotal=9*Pi*r^2/(4s-2r)^2
P=9*Pi*(r/(4s-2r))^2
Assume instead a xby x grid
we have (x-1)^2 internal corners
for the x^2 Area square
Atotal=(x*s-2r)^2
Acorners=(x-1)^2*Pi*r^2
P=(x-1)^2*Pi*r^2/(x*s-2r)^2
As x-> inf
P=x^2*Pi*r^2/(x*s)^2=Pi*r^2/s^2
P=Pi*r^2/s^2
Answer:
for a finite grid the expression is:
P=(x-1)^2*Pi*r^2/(x*s-2r)^2
for the 4x4 grid the expression is:
P=9*Pi*(r/(4s-2r))^2
for a infinitely large grid the expression is:
P=Pi*r^2/s^2
Cam
OK I see that we are to assume a particular set of dimensions and
an infinitely large grid.
For the coin to overlap a corner, its centre must be within it's
own radius of one of the corners of the square. As the probability
density function for where the centre of the coin lands is uniform
(with respect to area), the probability of the coin overlapping a
corner is p = 4(1/4)πr²/s², where s is the length of the side
of each square and r is the radius of the coin. So p = πr²/s².
I can only see how this can be true when r ≤ s/2. After that,
some of the areas would be multiply counted.
I've no idea why the source said about 50% probability. For 50%
need r ≈ 0.4s
Intuitively, and mathemeatically the probability is dependent on coin size to square size, so the published ~50% solution must make some assumptions regarding the ratio. And it looks like it assumes the grid is infinitely large to come up with the Pi*r^2/s^2 expression.
Cam
Hi Cam. I see our solutions overlapped. I interpreted infinite grid because of the πr²/s² expression.
I cannot see how you can assign a probability to a finite grid without introducing some assumptions. I might be a mile away from the grid when tossing the coin - the probability will then be indistinguishable from zero.
Ooops. I see you did make an assumption. I don't believe that the finite grid was intentional - because the problem is then poorly defined.
Oooooh! "intuitively and ...", Cam, I'm surprised at you. That was a pseudonym for the dreaded "obviously". ;)
That the centre of the coin is equally likely to land anywhere (uniform PDF WRTT area) is very clear.
hey there!
I'm not really familiar with the actual size of USA coins. But, considering the grid as a set of imaginary lines in a definite area, then the chances of the coin landing on one of the corners is 4/16 or 25%
This is, of course, assuming discrete values only (disregarding throws in which the coin lands between two or more squares).
Chris,
Intuition is merely our gut feel on the answer and can often lead us astray, (especially when dealing with puzzles or quantum mechanics).
When a worked out solution matches our intuitve response, it provides some measure of reassurance that the answer is correct. When it doesn't match our intuition it should force a recheck of the solution, to ensure something hasn't gone astray.
As far as, "obviously" goes, I try to use it sparingly. If things were obvious we would have a unified field theory, and we wouldn't need schools.
Cam
This post has been removed by the author.
Deleted last as forgot to delete a paragraph when editing. Also to acknowledge Cam's post. I had written the following before Cam's last post.
Divide big square into a n identical sub-squares, then it is
equally likely that the centre of the coin ends up in any one of
the sub-squares. So the probability that the centre of the coin
ends up in any given sub-square is 1/n. The probability that the
centre of the coin ends up in any one of m arbitrarily chosen
sub-squares = m/n. But m/n is the ratio of the areas of the m
sub-squares to the big square. This is true now matter how large
we choose n. So in the limit that n tends to infinity, the
probability that the centre of the coin lands in a given region
is equal to the area of the region/area of the containing square.
If the coins radius is over half the side of a square we start to
get overlapping areas. So we'd be counting the overlapped areas
twice, because that's fairly tedious messy to calculate, the
formula p = πr²/s² is only true if r ≤ s/2.
Hi Cam. My point was that because you and I are obviously (;) are familiar with this stuff, it doesn't mean it's second nature to everyone here. I didn't want them to be left out of this classic probability puzzle. Zaux obviously (;)) hasn't got our intuition that's partly come about by experience. No offence intended to anyone, I'm just trying to be helpful in the best way I know to be.
Hi Julio. I've just written some background material onwhy the area of the coin is very important. An extremely small coin is very unlikely to overlap a corner, the probability would be close to 0, not 1/4.
This post has been removed by the author.
Deleted last as got suck, some fixing typo at same time. There is an extension to this problem: Instead of using a coin, choose a complex shape, like a doodle with a thick pen. Then, provided the "shape" isn't too large, the probability of it overlapping a corner point, is equal to it's area/area of the square.
Trivia, when r = s/2, p = π/4 ≈ 0.7854.
Nostalgia. 0.7854 used to be engraved on slide rules. It was to help calculate the area of a circle using A = πd²/4, where d is the diameter.
More nostalgia. A slide rule does multiplication, but not addition. To add on a slide rule you used:
x + y = x(1 + y/x)
I still maintain that if the coin cannot leave the boundary of the grid (which the published solution assumes) then the probability is zero.
The questions specifically states what is the probability of the coin landing on the corner of one of the squares.
It cannot possibly land on the corner of one of the squares, it can only land on the corners of four of the squares simultaneously.
Given that Zaux has a solution of 50% this is clearly not the intended intgerpretation of the question as I thought, but I still like it as an answer.
DualAspect, I agree with you: the phrasing does suggest that it is a trick question (and so 0 is a possible answer, depending on what you think the trick is). But Zaux's comment suggests that that was unintentional. It may be that he rephrased the question to avoid possible copyright issues. I also suspect that drawing a finite grid was unintentional. But he's not clarified that. If it was intentional, then more information should have been given. I can think of several valid ways to tackle that, but they give very different results. Cam made the ad-hoc decision to erect a wall around the grid and only consider coins that landed flat on the grid.
I would purposely flip the coin so that it would land in the square tails up in the second column, first row. Therefore 100% chance of success.
From The Godson
I'd like to see you try that!
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