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I’m sure there’s some short elegant solution here that uses beautiful vector math. Instead I went for the butt ugly coordinate geometry solution.

Let u = <a, b> and v = <c, d>

See diagram. WLOG, assume u has a smaller angle than v. We can find cos(θ) by constructing a right triangle as shown, and by finding a new vector w, in the same direction as u, which has the correct length to complete our right triangle. Once done, we will have cos(θ) = |w| / |v|.

Let’s consider each vector to be anchored at the origin. So we can say u lies on the line y = (b/a)x, and v lies on the line y = (d/c)x. To find w, let us first find z - the third side of the triangle, which we know must be perpendicular to u, and pass through (c, d).

The line perpendicular to y = (b/a)x and passing through (c, d) is y = (-a/b)(x-c) + d. So this is the line containing the third side of our constructed triangle, z. To find w, then, let’s find its point of intersection with y = (b/a)x, the line containing w.

(b/a)x = (-a/b)(x-c) + d → Setup

(b/a)x = (-a/b)x + (ca/b) + d → Distribute on right

x(b/a + a/b) = (ca/b) + d → Add (a/b)x to both sides, factor out x on left

x(a² + b²) = ca² + dab → Multiply both sides by ab

x = (ca² + dab) / (a² + b²) → Divide both sides by (a² + b²)

x = (ca² + dab) / |u|² → a² + b² is |u|²

y = (b/a)x = (b/a)(ca² + dab) / |u|² → Plug solution for x into (b/a)x, don’t bother simplifying as this form is useful in a moment

So w = <(ca² + dab) / |u|², (b/a)(ca² + dab) / |u|²>. Now we need |w|.

|w| = sqrt((ca² + dab)² / |u|⁴ + (b²/a²)(ca² + dab)² / |u|⁴) → Plugging into normal |w| formula

|w| = sqrt(((ca² + dab)² / |u|⁴) * (1 + b² / a²)) → Factor out (ca² + dab)² / |u|⁴

|w| = (ca² + dab) / |u|² * sqrt(1 + b² / a²) → Pull (ca² + dab)² / |u|⁴ out of the root

|w| = (ca² + dab) / |u|² * sqrt((a² + b²) / a²) → Combine terms in root to one fraction

|w| = (ca² + dab) / |u|² * |u| / a → Evaluate sqrt

|w| = (ca + db) / |u| → Cancel out |u| and a from numerator and denominator

So this is the length of our adjacent side in the constructed right triangle. Finally, to find cos(θ), divide it by the length of the hypotenuse - which is |v|.

cos(θ) = |w| / |v| = (ac + bd) / (|u||v|)

And note that ac + bd is u•v. So we’re done.

cos(θ) = u•v / (|u||v|), or |u||v|cos(θ) = u•v.

Note that since this formula is symmetric with respect to u and v, our assumption that u’s angle was smaller than v’s angle did not matter - so this should hold regardless of which is larger.

To be clear about your other answers, saying P(B) = 1/16 and P(C) = 1/32 are not correct - I was saying if you adjusted your formula, from 1/2^n to 1/2^(n-1), then your answers would be correct. So the probability of the game lasting at least four turns is 1/2^(4-1) = 1/2^3 = 1/8, and the probability of it lasting at least 5 turns is 1/2^(5-1) = 1/2^4 = 1/16.

But I think you were more asking about why this didn’t affect your win rate - that’s because there’s a subtle difference between “making it to the nth round” and “having the game end on the nth round” - and that difference is that once you make it to a round, you then have a 1/2 probability to end the game - which makes the probability of ending on the nth round the 1/2^n you used.

This is the correct answer, although P(A|B) should actually be 2/3 rather than 7/12 - I think you meant 1/2 + 1/8 + 1/32 + …?

The reasoning is good, either way. Since past flips won’t affect future flips, if the player has made it to turn 5, an odd turn, then their future prospects are no different than they were on turn 1, another odd turn - so A and C are independent. Similarly, A and B are dependent because your chances of winning and losing effectively flip: If you’ve made it to an even turn, then you now win if it takes an odd number of flips from there to get Heads.

So it should be an almost paradoxical-seeming situation: You win 1/3 games overall, you win 2/3 games that make it to turn 2, you win 1/3 games that make it to turn 3, you win 2/3 games that make it to turn 4, 1/3 that make it to turn 5, etc.

These are correct. It is possible to reason out which of B and C is independent of A without going into the numbers.

This is close, but you’ve got an off-by-one error where you calculate the probability of making it to the nth round as 1/2^n. Consider that that would imply you have a probability of 1/2 for the game to make it to round 1, or 1/4 to make it to round 2, etc - it should be 1/2^(n-1). Correcting this would give you the correct answers for P(B) and P(C).

As for the dependence question, I’m not sure I followed your arguments there - but saying both are dependent is not correct.

This series can be rewritten as:

2 * lim (n → ∞) Σ (k = 0 to n) k / 2^k

The final term of this series will of course be n / 2^n. So let’s try to write each term with 2^n as its denominator:

2 * lim (n → ∞) Σ (k = 0 to n) k * 2^(n-k) / 2^n → Multiply by 2^(n-k) / 2^(n-k)

= lim (n → ∞) 1/2^(n-1) Σ (k = 0 to n) k * 2^(n-k) → Factor the denominator out of the sum, combine it with the 2 from in front of the limit

What will this sum look like, though? Its first few terms are:

0*2^n + 1*2^(n-1) + 2*2^(n-2) + … + (n-1)*2^1 + n*2^0 (my construction below works backwards, from the last term here to the first)

This can be written as a double sum: Σ (k = 0 to n-1) Σ (j = 0 to k) 2^j (which is 2^0 + 2^0 + 2^1 + 2^0 + 2^1 + 2^2 + 2^0 + 2^1 + 2^2 + 2^3 + …, so we have n copies of 2^0, n-1 copies of 2^1, etc)

So we have:

lim (n → ∞) 1/2^(n-1) Σ (k = 0 to n-1) Σ (j = 0 to k) 2^k

= lim (n → ∞) 1/2^(n-1) Σ (k = 0 to n-1) (2^(k+1) - 1) → Evaluate innermost sum, just sum of powers of 2 from 0 to k

= lim (n → ∞) 1/2^(n-1) * (2^(n+1) - 2 - n) → Evaluate next sum, which is powers of 2 from 1 to n, and don’t forget the -1 part

= lim (n → ∞) (2^(n+1) - n - 2) / 2^(n-1) → Combine leftover terms into one fraction in the limit

= lim (n → ∞) 4 - n/2^(n-1) - 1/2^(n-2) → Distribute denominator to each term

= 4 - 0 - 0 → First term is just 4, other terms go to 0

= 4

I’ve been checking and thinking about this at least a bit each day, and today’s the first day a solution came to me.

Edit: Checking your solution now, it’s obviously much nicer than this - but at this point I’m just happy I solved it at all. Obviously I knew it was 4 from just calculating partial sums from day 1, but proving it proved challenging.

For both of these, it’s sufficient to consider only sin(x)^sin(x). It stands to reason that the maximums will happen when sin(x)^sin(x) is at its maximum value, and its minimums will happen when sin(x)^sin(x) is at its minimum value. The x^ part really just makes the function prettier to look at. The function lacks any continuity when sin(x) is negative, so we can consider just the regions where sin(x) is positive.

For the maximums, this is very easy: The largest sin(x) can be is 1, and so the largest value for sin(x)^sin(x) is also 1. So the monotonically increasing function which passes through each local maximum is y = x^1, or just y = x.

For the minimums it’s a bit trickier: The smallest sin(x) can be is 0, and while 0^0 is indeterminate, it happens to approach 1 in this case - our minimum is somewhere else (side note: sin(x)^sin(x) approaching 1 for these values is why our max function also exactly bounds the flailing arms of the function). So we turn to derivatives. We can do this using either sin(x)^sin(x) or x^x - I chose to use sin(x)^sin(x) here and x^x (sort of) in the extra credit. Using sin(x)^sin(x) here has the added benefit of also showing the maximums are when sin(x) = 1.

y = sin(x)^sin(x)

y = e^(ln(sin(x)) * sin(x))

y’ = sin(x)^sin(x) * (cos(x)/sin(x) * sin(x) + ln(sin(x)) * cos(x)) → Chain rule, product rule, and chain rule again

y’ = sin(x)^sin(x) * (cos(x) + ln(sin(x)) * cos(x)) → Simplify

y’ = sin(x)^sin(x) * cos(x) * (1 + ln(sin(x))) → Factor

We can set each of these factors to 0.

sin(x)^sin(x) will never be 0.

cos(x) will be 0 when x = π/2 + 2πn (2πn instead of πn because we’re skipping over the solutions where sin(x) is negative). These are the maximums, because when x = π/2 + 2πn we have sin(x) = 1.

1 + ln(sin(x)) = 0

ln(sin(x)) = -1

sin(x) = e^-1 = 1/e

And that’s actually sufficient for our purposes - we could of course say x = arcsin(1/e), but our goal is to calcualte the value of sin(x)^sin(x). Since sin(x) = 1/e, we have sin(x)^sin(x) = (1/e)^(1/e). So our monotonically increasing function which hits all the minimums is y = x^((1/e)^(1/e)), or: y = (e-th root of e)-th root of x.

Graphed solution visible here.

For the extra credit, we can do the same basic idea: Find the maximum of |sin(x)|^sin(x). Since we know we’re in a region where sin(x) is negative, we can model this as x^-x, and the solution will be valid as long as x is something between 0 and 1, within sin(x)'s positive range.

y = x^-x = e^(ln(x) * -x)

y’ = x^-x * (-1 - ln(x))

x^-x won’t ever be 0, and -1 - ln(x) will be 0 when ln(x) = -1, or x = 1/e again, same as before. So our maximum value happens at (1/e)^(-1/e), which simplifies to e^(1/e). In other words, our new maximum function is x^(e^(1/e)), or x to the power of the e-th root of e. Graph of all three solution functions

And, finally - I mentioned that we should also be able to expand the function to its less-defined regions using y = x^-|sin(x)|^sin(x). I don’t care enough to post the full work here, but the reciprocal of the three answer solutions above are what works here: y = 1/x, y = 1/x^((1/e)^(1/e)), and y = 1/x^(e^(1/e)). Graph. Note that the upper function here only applies in regions where the original function is already defined, and so the other parts of this extension are invalid - you can toggle the main function at the top off to see only the parts that truly apply.

Nothing particularly interesting happens if you extend things to values of x less than 0 - the graphs (more or less) just reflect across the y-axis.

Bonus graph with everything + phase shifts and pretty colors