bonanova

Exactly. Random generally implies uniform distribution. One authority offers this guidance: . A random number is a number chosen as if by chance from some specified distribution such that selection of a large set of these numbers reproduces the underlying distribution. .It is impossible to produce an arbitrarily long string of random digits and prove it is random.. If [2] is true, then random triangles and numbers only exist in bounded regions. A clue, perhaps, to the difficulty of this discussion.

Can we come up with a function f(x) with these properties for 0 <= x < = inf? . f(x) is defined, continuous and infinitely differentiable [no piecewise linear portions] f(x) > 0 [strictly positive] Integral [x=0, inf] exists and is finite.But still it is not the case that lim [x->inf] f(x) = 0.

1/4 for covering the center of a circle when the triangle fits within the circle. That's independent of radius, so just let the radius increase without bound. Voila! The answer for the plane is 1/4. I've come full circle. Now the center of the circle need not be at the origin to allow its increasing radius to swallow up the plane. So this result applies to any point - not just to (0, 0) Thus a "random" triangle covers every point in the plane 1/4 of the time! This means the average area of a random triangle is 1/4 the area of the plane. An interesting result, and one that negates any notion of a random triangle having a finite area. The OP was formulated naively I think, as it probably anticipated that random triangle had finite area. But perhaps not. Only if it's finite is the covering probability zero. So I now raise the question I kept silent on while the debate brewed: Is it possible to select at random a real number from the interval [-inf, inf]? This process is required, to specify the vertices of - and even to discuss - a random triangle in the plane. If you feel that it is possible, can you describe the process? And, especially if we conclude a random triangle in the plane cannot be constructed, how surprising is it that we nevertheless know its average size!

I received this answer as an email from a new member Mathephobe who had trouble posting it. It arrived about 2:00 pm EDT July 31 2009.

DeeGee [with a very nice proof] and others have it for part 1. Pieater is correct that parts 1 and 2 differ. Basically, part 2 boils down to finding the average size of a triangle that fits in a unit circle. Call that area <A>. The probability of a randomly chosen point [e.g. not necessarily the center] being covered by a random triangle is then just the ratio of areas: pcover = <A>/pi. BrightRocks, sparx1 has it right. Happily there's no ambiguity in this case. Select the triangle's vertices from a uniform grid of points within the circle. If you were going to program this, you could select x and y at random from the interval [-1, 1] and discard any selection for which x2 + y2 > 1. The seemingly equivalent method of selecting r, theta uniformly from the intervals [0, 1] and [0, 360) won't work, for reasons that are evident after a little thought.

Since finite realms do have midpoints, and thus are intuitive and programmable I wrote for the unit circle.

Caveat: not for the faint of heart. Several have noted that for an infinite series to converge, the sequence must converge to zero. That is, the terms of the series must have zero as a limit. For example the series 1/2 + 1/4 + 1/8 + 1/16 + ... + = 1 would not converge if the terms 1/2n did not approach 0 as n-> infinity. For those familiar with the integral calculus, where the integral can be treated as the limit of a series, a corollary seems to be that, for the improper integral of a function to exist [converge], the function, if it's continuous, must have a limit of 0. So we ask: is that true? That is, if . f(x) is defined, continuous, and differentiable for 0<= x <= infinity, andIntegral [x=0, infinity] f(x) dx exists and is finite, . then must f(x) have a limit as x-> infinity, and must that limit be zero? Clearly if it's not true then there must be a ... counter example.

Good one, and I agree there must be others. This one, owing to the preponderance of points far from the circle, will create a lot of near-diameters. Nonetheless, it's random in the sense the generating point lies at a random point in the plane.

To complement DeeGee's thought provoking puzzle of . A triangle is drawn at random in a circle centered at the origin. What is the probability the origin lies inside the triangle? . A triangle is drawn at random in a circle that contains the origin. What is the probability the origin lies inside the triangle?. I could comment on whether the answers are the same, but I won't. Enjoy!

Choose two real numbers at random. If the probability of their falling on opposite sides of the number zero is 1/2, with what probability do they fall on either side of the number . onetwo pione million10100ten raised to the power of the previous number?. It is tempting to treat the continuum as if it were bounded, with elements that fall in equal measure about a single, unique point.

Let's compare zero probability of covering the origin with some related statements. A finite triangle is placed at random on the infinite plane. Consider the following statements: There are no points inside the triangleThe probability of any point in the plane falling inside the triangle is zero.The probability of any point in the plane, that was chosen before the triangle was placed, falling inside the triangle is zero. . Now consider: If a finite, randomly placed triangle covers the origin 25% of the time, with what probability does such a triangle cover the point (0, 1)?Does a finite, randomly placed triangle cover some points in the plane more often than it covers other points?Can a finite, randomly placed triangle cover every point in the plane 25% of the time? Finally consider: An event with zero probability can nonetheless happen. A deck has an uncountably infinite number of cards, all different, that includes the Ace of Spades. Is it possible to cut the Ace of Spades?Does the probability of cutting the Ace of Spades differ from that of cutting any other card in the deck? What is the probability of cutting the Ace of Spades?

The OP asks to draw a triangle on the plane. If one or more of the vertices may be placed at infinity then yes, you get a different [undetermined] probability. But by the time you drew that triangle, none of us would be here to discuss it. \

Sure it does. The odds of a random point being near [whatever near means on the infinite plane] the origin are vanishingly small. If we think of three points each individually appearing in quadrants 1-4 with equal likelihood, we're making a non-random assumption about the location of the origin. Imagine displacing the origin by 1021 units in some arbitrary direction. The equal likelihood for the 4 quadrants just went out the window. Nothing in the process of creating three random points on the infinite plane puts us "near" the origin as a starting point. The answer has to be the same if you consider covering any particular point in the plane - e.g. the point [1021, 0] On the other hand, limit the three points to a finite area, for example a circle centered on - or covering - the origin, and a non-zero probability is easily calculated. It will again involve the ratio of areas, but this time both areas will be finite.

Not really. Specify any time t<1; the OP will describe the state with unconditional clarity.

By my analysis, the prime had to be even, thus one unique answer.

After further review ...

It could be.

Must the antidote be drunk after the poison to be efficacious?

Pedantic or not, it would be incorrect. But let's leave the discussion of infinite series for another time. For the present purpose, make the assumption that it does in fact equal 1. If it helps your approach, change "after 1 minute" to "after 1.0000000000001 minutes".

Think of them as math/logic questions; some of them have physics ramifications that should be ignored.