bonanova

Being an engineer and not a mathematician, I built in some redundancy. And next puzzle I solve, I'll wear my glasses.

Well done all. MM, nicely metered. Pickett, I considered removing but thought a clue might be necessary. pg, lol, and in the spirit of the form, for sure.

A short one. How many lines? I'm thinking about five.

What a great Freshman physics problem that would make! Trying to think of a simple pendulum that would emulate that same motion. By analogy and reasoning, and not calculation, what might its length be? ============= Also: Suppose a New Yorker wanted to get to Los Angeles, say it's 3000 miles distant, by gravity ASAP. We know she can get there in 42+ minutes. Is that the optimal time?

By his own admission Alex spent too much time in the back corner at Morty's throwing darts. But the results could not be ignored: he cold hit the bullseye reproducibly 97% of the time. Davey, on the other hand could hit only 90% of his attempts, and Ian's accuracy was worse, at 80%. Neither if them would play Alex for that reason. Last night Alex proposed that the three of them play: Alex's single dart against the better of his opponents' two darts. Davey scribbled some 1's and 0's on a napkin, trying to work out the odds. He wasn't certain, but he figured that the two of them just might be able finally to win some beer money from their nemesis, and he persuaded Ian to give it a try. Had Alex finally made a bad bet?

Good question. Let's start withe the obvious result that the killer's chance of survival is 1/11, cuz one killer must survive. The question is a pacifist's chances with a 10-10 split. It's clearly not 1/10.

Thanks. That makes sense now.

The simulation had a counting error

I think I will resign as moderator, my brain is definitely calcifying. Good catch. Edit: There is no change to three decimal places in the two chosen scenarios. A slight increase of 0.08 in the third. i.e. it almost never is 4.

Well said. If the OP had said "It's not the case that they are both boys. What is the probability that they are both girls?" this forum would have been much shorter. Thanks to you, for posting your thoughts clearly and cogently.

I'm trying to see where my thinking has gone astray: p(other is girl) = p(other is girl if Ned spoke) x p(Ned spoke) + p(other is girl if Red spoke) x p(Red spoke) + p(other is girl if Ted spoke) x p(Ted spoke) + p(other is girl if Zed spoke) x p(Zed spoke). If the reporter had worn a name tag, we're done: we know p(other is girl if x spoke) for each x. We're only uncertain about the name tag, it seems. Jed makes each tag probability 25. Bushindo, I perceive an extra level of conditionals.

I totally get that the likelihood, or ease, of saying girl depends on the reporter's algorithm. But if Jed drew their name from a hat, the four are equally likely to have been the reporter. And then, likely or not, the OP tells us that the reporter said girl. Repeating, we are not given that the reporter said girl and then asked which reporter is the most likely to have spoken. The OP saying we don't know which one spoke was meant to be even stronger: we don't even have a clue. So I need to understand better Bushindo's analysis.

"Other" answers to "One." it does not answer to "At least one." The OP says "One," so it appropriately asks about the "other." But be careful not to identify "One" with a specific one. That leads to 1/2 unambiguously, but the OP does not support that interpretation

Did I forget to mention that uncle Jed had put their names in a hat and drew one of them to speak? More interestingly, does the "One Boy, One Girl" thread any longer have a definitive answer?

The OP intended to deny bias among reporters, since each can comment on any two-child family, Even tho some statements are denied to some reporters for some families. However ... The greater intent of this puzzle was to cool the debate somewhat, housed in the Teanchi-Beanchi thread. We don't know the algorithm used by the reporter, and now we understand that it matters. Most of us "1/3" zealots assume the reporter was like Red. But there is no basis for that [or any other] assumption.

If either is a boy. For our purposes, "One is a boy" and "At least one is a boy" are equivalent statements.

Ned, Red, Ted and Zed are identical quadruplets alike in every way except one: the way that they describe the children in families that have two children. 1. Ned says one is a boy, if that is a true statement; otherwise he says one is a girl. 2. Red says one is a girl, if that is a true statement; otherwise he says one is a boy. 3. If the older child is a boy, Ted says one is a boy; otherwise he says one is a girl. 4. Zed flips a coin and considers the taller (heads) or shorter (tails) child. If the coin-selected child is a boy, he says one is a boy; otherwise he mentions the sex of the other child. One of the four men, we don't know which, then tells us: "Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)! They have two kids, one of them is a girl." What is the probability that the other child is a girl?

Could very well be. The above curve probably debunks the catenary hypothesis. Care to cast the golden spiral inside the unit square? I'll add to the plot.

That seems to tie a bow on it. Below is a plot of A your catenary equation B my catenary equation C your final equation quoted above [plotted vs 1-x to make curves congruent] D my simulation of a thousand points using n = 1.618033989 to make cat and mouse arrive at [1,1] simultaneously. The light blue arrow is the mouse's path, at unit speed. C and D are indistinguishable, lending credence to both and casting doubt on the catenary hypothesis.