dart board probability.

We solve a probability problem involving a square dart board using first semester calculus techniques.

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[jacemandt]

Indeed, note that the set of points equidistant from a given point and a given line is precisely one definition of a parabola (with focus at that point and directrix of that line).

[jimiwills]

I love the parts that aren't edited out :)

[kevinmartin]

This result evaluates to about 0.219, As you increase the number of sides to the polygon, the probability should approach 1/4, which is the result for a circular dart board.

[Ensivion]

great question, I'd like to add that this probability is exact for the entire square, since you assume each of those triangles are equally probable to hit and each triangle is the same, just rotated or flipped. This symmetry makes it so that the overall probability is unchanged by which triangle you hit.

[Kindiakan]

For the hexagon: apply the same technique with a triangle with vertices (0, 0), (sqrt(3), 0), (sqrt(3), 1).

[mathiest]

So, doing some messy algebra, I think I have a general solution for a regular n-gon:
let a = pi/n, where n is the number of sides (in other words, a = half a central angel)
evaluate the following:
y = csc(a) - cot(a)
then substitute that value into the following:
2cot(a)(-y^3/6 -cot(a)y^2/2 + y/2)
That will give you the probability of being closer to the center than a side. If I haven't made a mistake. Letting a=pi/4 seems to give the result shown here, so that much works at least.

[arvindsrinivasan]

I spent more time than I’m willing to admit solving this for a regular polygon with N sides. The final result is that the probability of being within the region defined by being equidistant to the perimeter and the center is given by the following expression

[DirtShaker]

Just grouching ... 1:28 ... shouldnt the parabula being the boundary of the area A intersect the orange x-axis exactly at (0.5, 0)? This point is equally spaced from center as well as the edge.

[EAdano]

It's usually good to mention the probability density in the initial statement with these kinds of geometric problems with uncountably infinite possibilities. People for some reason take "random" to mean "uniform pdf random variable, " which is what you ended up using of course. Still a good problem/cool video!

[ionutradulazar]

That inner region you calculated looks kinda cool if you extend it in the whole square

[bhavyaramakrishnan]

Add areas under parabolic arcs drawn eccentrically in x, y, - x, -y and add them up.
Then divide by the whole square area.
That way the probability looks more plausible.
The eccentricity could be 0.75 to 0.8 on all directions.
Integration of closed loop:
y^2=4×a×[x+(0.75/2)×a]
+ x^2=4×a×[y+(0.75/2)×a]
+ y^2=4×a×[-x-(0.75/2)×a]
+ x^2=4×a×[-y-(0.75/2)xa] divided by a^2
Where a = magnitude of side of square dart board

For hexagonal shaped dart board with side a:
(0.75×π×a^2) /(6×0.5×(√3/2)×a^2)

[randomperson]

This sounds like such a simple problem at the outset but has a lot more going on. Interesting.

[popkornking]

I think an interesting follow up question would be to do the same thing but with a normal distribution of darts.

[bmenrigh]

I like your use of multiple colors of chalk in this video.

[stephenhousman]

The probability will approach 0.25 as n goes to infinity. Where n is the number of sides.

[flikkie]

What about an equilateral triangle dartboard?

[michachyra]

So, apologies if the question is stupid, but I'd still like to ask. Once we have the point where y=x and the curve intersect, wouldn't it be easier to calc the triangle from 0 to -1+sqrt2 and add to it the integral from -1+sqrt2 to 1/2 of the curve?

[procerpat]

why is the distance to the cueve juat 1-x ?

[JansthcirlU]

What am I missing? I thought you could just draw an inner square inside with the edges halfway between the outer edge and the centre and then take the probability as the area of the inner square divided by the total area, which should yield 1/4?

[sandeshjain]

Nice michael sir👌👌👌
LOVE FROM INDIA SIR