There are three fundamental equations:
1) x(distance max.) = v(in the x-direction)*t(max)
2) V = v(in y direction)-gt
3) 0 = h+v(y direction)*t-0.5g(t^2)
I'm not gonna explain these as they're simple enough to go unexplained.

In equation (3) we need to solve the quadratic for t(max). Then we need to combine(plug in) to equation(1). After doing so, the derivative dx/d(theta) needs to be taken and set to zero to maximize it.

When I did this, I got the following mess:

0=(v/g)cos(2theta)+[((v/g)sin(theta))*((v^2cos(2theta)- 2gh)/(sqrt(v^2sin^2(theta)+2gh)))]

That "mess" needs to be solved for cos(theta) so we can obtain a formula dependent upon any "h" and "v" inputs. My first inclination(reflex) was to go to Mathematica so as to not spend all night equation crunching! However, with my new computer(one week old) I haven't yet installed Mathematica(I'll do so in the next few days). Well, if that above equation can be solved for cos(theta) then I think you have your answer, right?

2) I can crunch it in Mathematica in a few days.

I'd prefer option "1", but have it your way :)

For now, I'm heading to bed(1:30am here).

You throw a ball vertically upwards in real air, so there is some air resistance. Does the ball take longer to go up or come down?

There are three mathematicians, Erdõs, Einstein and Euclid. Now being the @#$%-stirrer that he is Erdõs decides to have some fun in the way only Paul Erdõs can. He chooses two numbers and tells the product to Einstein and the sum to Euclid. Now Einstein knows that Euclid knows the sum and Euclid knows that Einstein knows the product. They are told that both the product and sum are less than 100, and that both numbers are greater than 1. They meet to discuss the problem. The conversation is as follows:

<Einstein>:I don't know the two numbers. <Euclid>:I knew you wouldn't know. <Einstein>:Now I know the two numbers. <Euclid>:Now I know the two numbers.

Given n weights each weight can occupy either the left scale, right scale or not be on the scale at all. In other words there are three possible states for each of the n weights. This gives a total of 3^n possible unique states. Now this counts no weights on the scale (zero weight) and treats weight w different from -w. So the upper bound on the number of possible consecutive unit weights is [(3^n)-1]/2.

For 1 weight the optimum weight is 3^0=1.

For 2 weights the optimum weights are 3^0,3^1 or 1,3 as in your given example.

For 3 weights the optimum weights are 3^0,3^1,3^2 or 1,3,9 allowing you to weigh any integer from 1 to 13.

For 4 weights the optimum weights are 3^0,3^1,3^2,3^3 or 1,3,9,27 allowing you to weight any integer from 1 to 40.

<zarra> criticized my analysis of the envelopes stumper saying <In your equation for expected return, <0.5(2X) + 0.5(X/2)>, you use X for two different things. In the first term, X=money in the poor envelope, whereas in the second term, X=money in the rich envelope. Shame on you!>

I responded <I don't know what you mean. "X" is not the poor envelope or the rich envelope, "X" is the unknown amount of money in the envelope that was picked.>

A casino offers a game which costs $X to play. Once you pay the $X, you get to flip a coin and try to get as many heads in a row that you can possibly get. As soon as you get tails even one time, the game is over.

There is no payoff for less than 10 heads.

If you flip 10 heads, you get $2 back.
If you flip 11 heads, you get $4 back.
If you flip 12 heads, you get $8 back.
If you flip 13 heads, you get $16 back.
etc. etc.

There is no maximum payoff.

Question - What would be a fair amount of money for the casino to charge so that this game would appeal to the players but still give them a tiny "house edge" over the gamblers?

<< ** stop reading now if you don't want to know the answer ** >>

Crazy as it sounds, this game will be unfair to the casino, regardless of how much money they charge to play. Even more crazy, it doesn't matter when the payoff starts E.g. they could start the payoff structure when you flip only 3 heads in a row, or start to pay out only when you flip 1000 heads in a row.

So think about this: the casino says "For the price of one billion dollars, you can sit down and play this game, but the payoff structure doesn't start until you flip one thousand heads in a row." You'd have to be an idiot to accept such a bet, right? According to the math, the CASINO is the idiot to offer the bet!!

If the payoff starts at 1000, the equation to tell you what the expected value of this bet is, would look something like this:

1/2^1000 * (1/2*2 + 1/4*4 + 1/8*8 + 1/16*16 ... )

Sure, 1/2^1000 is a very tiny number, but look at what's in the parentheses. It's an infinite series of ones, i.e. it's infinity! Multiply infinity by any positive constant, no matter how small, and you end up with infinity. Ergo, this game will favor the player.

This isn't really a paradox, we just think of it as a paradox since it seems to violate common sense. When you think of the game on merely human terms, there will never be enough time or money to be able to break the bank.

But think of it this way: suppose the Devil runs this casino and God decides to play. These two have nothing but time on their hands, and God has unlimited funds, so He keeps plugging away for eons on end. It might take billions of years for it to happen, but eventually God goes on a streak where he doubles so many times over that he takes away everything the Devil won since the game began.