plasmid

I may be missing something, but by playing on the opposite side of the center square as player B, player A seems to be able to control the outcomes of four rows (both of the diagonals and the horizontal and vertical rows through the center), not six. The topmost and bottommost horizontal rows, and the leftmost and rightmost vertical rows, are not so easily determined. Player A can control four row outcomes, which is enough to prevent B from being able to win if B goes for either an even or an odd outcome, but it's not clear if A can prevent B from achieving a neutral game outcome using that strategy.

Ah, right you are fabpig. Your post two above mine does seem to say what I was saying, I just hadn't completely followed it.

But this is not the approach with the minimum number of weights, in case anyone wants to go after that problem.

I can't find one where I can prove that the wait time for A is longer than B while A is more likely to happen first, but I can prove a case where the wait time for A is longer than B and they are equally likely to happen first.

Do you mean equilateral instead of isosceles?

Well solved, Witch, and nice to see you back.

Neither of those. The clue about helms that are hewn but not delivered is an important one referring to something that's simply cut where it is on the dwarves.

How do you partition the flips of a coin with p(tails)=0.4 and p(heads)=0.6 in the scenario where it comes up with ten consecutive tails?

Stabs at two more These multi-celled hosts hold the key to life's rejections Puts them down for all to see, but can they make corrections? They're commonly portrayed in these parts its safe to say As all they're meant to be in performing the same way

"Young sir or madam, if it were not for your actions, would the merman whose remains are before you still be alive?" *Realizes after saying this that it might or might not even be relevant* *Searches for book on Atlantean law to find out how they define murder* *Realizes that underwater critters have no feathers and therefore the Atlanteans never developed the quill and paper, and therefore never went on to write law books* *Acts like I knew that all along*

We're writing both the accusing and defending arguments? If we get to decide what the prosecutor's arguments will be, this should be an easy defense.

A web among the tree branches could fit most of the clues. But I have something in mind that always (for all practical purposes) has multiple dwarven guardians with helms that need hewing.

Here's the tricky part: The OP doesn't ask whether the first player who answers can be sure that he's correct, it asks whether you can be sure that you're correct if you see two brunettes. If you wait a while and then both of the other guys correctly answer brunette, then are you safe or in danger? Remember, you didn't get a chance to discuss strategy with the others before this all started.

In case 2 I think Alice is not traveling at the median speed, if you count the median of the markers rather than the median of the conveyor belts. Edit: But this starts to touch on an important point. The way the OP is phrased, for each car there are an infinite number of identical 1-mile segments that repeat as if on a conveyor belt. A car gets counted for each time it's placed within a 1-mile segment, which is not necessarily the same as the number of times Alice sees one of those cars passing or getting passed. Suppose there are cars moving at 45 MPH, 50 MPH, 55 MPH, and c/2 where c = speed of light. Neglecting relitivistic effects, even though there's only 1 warpspeeder per 1-mile segment, Alice will be passed by a ton of repeating 1-mile segments if she's going between 50 and 55.

Getting things started, for Packs quite a wallop, when one really sings Motions to give the shaft, certain body rings I'll have to take it up a notch to make a decent try at the others.

If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter. Would this be the same as considering the function y = x2 and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all?

#5 is factually true but I disagree with its implications, mainly because the OP doesn't say you have to play a Large number of times. Simulations show that the odds of winning after 1000 plays are not too far from 50/50 (roughly 45%), adding another couple of lines to the code showed the odds of winning >$1000 are roughly 17%, and you have at least a 1/1000 chance of winning a bankroll that has to be expressed in scientific notation. I'd consider that favorable for a $1 wager.

The good news is that using equations for a more efficient approach strategy by the lion as the trainer runs around the perimeter doesn't require trig functions. The bad news is that I nonetheless end up with a differential equation that I can't see how to solve. But I can show that the lion is able to catch the tamer, albeit with a much less optimal approach.