bonanova

If OP is taken to identify "one" of the two, and to say that "one" is a girl, the probability is indeed 1/2 that the "other" is, too. If OP is taken to say it is not the case that they are both boys, then one can deduce a probability of 1/3, although it is not necessary to conclude 1/3 since our fact-reporter's algorithm has not been disclosed. It is incorrect to say there are 3 combinations and one is removed so p=1/2. That reasoning ignores the equal-likelihood requirement: I buy a lottery ticket, and there are only two possibilities: winning and losing. Therefore my chances of winning are ... Before adding to this thread, read through some of the posts first. Additional, thoughtful comments are always appropriate.

The group 382 mentioned was his group of three friends. I meant that to be a given, but you also correctly deduced it.

What, no side pockets? Calling Allison Fisher .... !

Assumptions? Distance (1000 m) traveled is with respect to a stationary coordinate system, not that of the star? Elastic collisions with the walls, with equal angles of incidence and reflection? Since the ball is not a point, what bounce angle is assumed if the ball hits an interior point of the star?

They wear numbers in the land of Truthtellers and Liars, and it was a great help. I had to find Bob, and I knew only that he was across the room talking with two of his friends. I decided on a direct approach. I asked the group, "Which of you is Bob?" The person wearing 576 replied, "I am." The person on his left, wearing 238, disagreed: "Not at all. It is I who am Bob." Since number 382 remained silent I prodded him: "You're not helping all that much by your silence. Might I trouble you for at least a clue?" He smiled slyly and ventured, "We're a strange group, we are, I would not expect to receive all that much help: only one of us will ever tell you the truth." "Thanks," I said. Next, I fixed my attention on Bob and told him: "I really need to find Charlie." I knew he was in the room elsewhere, and it held only two other small groups: 3117 was talking with 1137, and 4741 was standing next to 2305. Bob helpfully(?) motioned to one of the groups. Hmmm... should I believe him? Not only did I want to meet Charlie, I also wanted to know whether he would tell me the truth. I decided on the second pair, and blurted out my question "Does Charlie tell the truth?" Number 2305 replied (with a yes or no;) and with that I was able to deduce Charlie's number. But ... could I trust him? What do we know about Bob's number and type, and about Charlie's number and type?

Nice diagram. I actually get Bayes, and conditional probability (I think of it as "conditioned" probability.) And I had the right thinking and I had visualized the right numbers. And then I just wrote 5/6 when I should have written 4/5. My Bayesian aversion comes when the numbers are too large or the steps too complicated to visualize. And you just plug, plug, plug, and turn the crank. As an engineer I should love that. That's when the tool really shines. But it also robs me of the opportunity for the puzzle to shape my intuition. and that's part of my love for probability puzzles.

Given four points A, B, C, D, and four directly similar quadrilaterals AP1P2B, I'm struggling with picturing these similar quadrilaterals. Proportional sides are needed, No? If AB is common, which sides are proportional?

Two to go. Clue at post 10. OTIHIS, LSTH. IGOLAG, BTTS. OWCBTS, FFTS. SHWW, BTTS. (18)CTIT, TN. ANOGSYFTBATS. YKIT, TN. YFFYLIAK, TT. (8)WATC, MF. AWKOFTTE. WATC. WATC. NTFL, CWATC ... OTW. (17)SLY, Y-Y-Y. SLY, Y-Y-Y. WALLT, YKYSBG. (11)IIB. V. IIB. V. IIB. V. IIB. V. (10)TMBTTAIKY, KMTYDAISY ATMLJ. IGTMLJ. IGTMLJ. (15)IOTEOGAIHOAMOT. OOTEOGAIHOAMWY. IOTE, TE, TE, TE, TE, TE, TE, IOTEOGAIHOAMWY. IOTEWY. (14)IDGMTRTFT. IF, IATTAL. NIKATWTTSAFBMH. AWWIBTD, TIFSYLF. NISI, E. AIK... TIA, IA IATL. (11)

One to go. Clue at post #5. CMI. (8)TBITL! (7)IATGOCP, STS. LUM! (15)IWTBOT, IWTWOT. (16)AAAE, BSAAMETO. (10)

Agree with Bushindo and note it converges rather quickly. After only a few days it's a fraction of a percent from the limit values.

Does the wording " ... let the entrant grab his share of the market ..." say there is a fixed market? I.e. Is there a zero-sum assumption made? i.e. One party's loss is the other party's gain?

Looks like convergence to me....

Yes. I think the three of us agree that I made a mistake in the second case. I should say the four of us.

I'm saying it seems that is so. 216 is a counter example against the definitions being equivalent. I read some more about congrua having to correspond to primitive Pythagorean triples, which includes 24 but excludes 9x24 = 216, but the reason for that requirement escaped me.

Calling Dr. Bayes ... Where is Bushindo when you need him?

A proof would be interesting to see. A post on red hot pawn, where this problem was discussed a few years back, makes this claim: See: http://www.redhotpawn.com/board/showthread.php?threadid=141670&page=&page=3 I agree with this reasoning for some sufficiently large number of points in the plane. However, I am not sure how this translates to, say, the initial problem of 6 points in the plane. I do not really know the answer yet. The best I have come up with is the same as what Palynka [the pentagon 1.902 answer] proposed, but I do not have a proof that it is the minimum. I do have a proof that, for example, the answer cannot be less than y/x = sqrt(3). But that does not help that much. He does not post the proof itself, although I believe it. But note his proof is "cannot be less than" which is different from "can be equal to."

Rob, I think this would work, because I haven't quickly found a counterexample, and it seems intuitive. BobbyGo's answer is a proof, and got there first.

I won't spoiler this because it's not a solution but a refutation. Consider a regular pentagon with side t and points at center and vertices. The radius r of the circumcircle is the minimum distance among the six points. The diagonal d is the maximum distance among the six points. t = 2r sin 36o = 1.1756 r d = PHI t where PHI is the golden ratio = 1.618... M/n = d/r = 2 PHI sin 36o = 1.902 = sqrt (3.618) > sqrt (3). This is the answer from my where I conjectured a more dense arrangement of points. I don't believe there is a more dense configuration of six points.

I am exploring one more possibility.

I had read a while back about congrua. This puzzle gave me a reason to look at them again. As unakade notes, 216 is not listed by Wolfram as a congruum. But a sub-multiple (24) is, and it does satisfy both definitions: 25 + 24 = 49 25 - 24 = 1 where 25, 49 and 1 are squares. 24 = 4ab(a+b)(a-b) where (a,b) = (2,1). I wasn't able to see where scaled-up versions of a lowest-term congruum are excluded, but it does seem that 4ab(a+b)(a-b) is the more fundamental definition, from which x2 + c and x2 - c being squares can be derived. Putting another way, x2 + c and x2 - c being squares seems to be a necessary but not sufficient condition for c to be a congruum.