bonanova

Flesh that out a little ... In a line, which brother is the "last" brother? Or are you telling us that the "last" brother is the end brother who has the other four to his right? If so, does the "first" brother have anyone on his right? Or are you saying that the "first" brother has the other four brothers on his right, but the "last" brother has a sixth person, not a brother, on his right? Do the questions have to be Yes/No questions, or could they be of the type: "Please identify yourself, the king, the two liars, the unknown speaker and the copy-cat."? Sounds like an interesting puzzle, just need to know the rules. Thanks.

Tip of the hat to armcie for first suggestion of Aces and nines.

Sure! Have at it ...

HH has it. Alex, watch out .. ! Nice job.

d3k3 has it. Nice job.

The OP talks about being dealt a poker hand - that is, being dealt 5 cards. Let's remove all the variables by making the game 5-card stud poker. Each player gets 5 cards, all closed, no exchange. There are [say] five opponents, and the hands come from a single deck.

Nope, No jokers. No pun intended by saying nothing is out of the box here - except the cards. I won't comment on the two posted answers yet, except to say they aren't [quite] correct. By the way, Doyle gave dms's answer.

You're drawing a trend line through a single datum, which randomly happened to exceed 1/4 - by an amount commensurate with finite sampling. I don't simply assert the deviation is random. Repeated simulations included results less than 25%. Sampling noise swamps comparison tolerance noise, [of the order of 10-13], which itself swamps the effect of including or excluding the Lesbegue-nonexistent cases of the center-of-square landing on the triangle's perimeter. The decision to include or exclude points on the line has zero effect. OK. That's the discussion of the simulation. On to the OP problem analysis... I'm thinking that you do disagree with the probability method. No discussion points there. But the inconvenient problem with your objection is that you also disagree - every bit as much - with the perimeter method. All you've said is that the correct answer for a specific case is not the correct answer for the OP. My preference, evident also in my roll-of-the-die analysis, is to find the shortest path to final answer. Given the OP, .5 x .5 = 1/4 has a certain charm for me. Your preference, there and here, is to analyze particular cases and sum, or average, afterward. That's how you think. The world needs us both.

What if you and all your friends had even numbers? What if everyone would be executed if any of the last nineteen gave the wrong color?

I grabbed an online dictionary and looked up net as an adjective... Quick definitions (net) ▸ adjective: remaining after all deductions ("Net profit") ▸ adjective: conclusive in a process or progression ("The net result") Not meant as a clue, just a clarification. And ... I just love it when people show their work. It's more fun to comment on that.

It was poker night at Morty's, and true to form, Alex arrived early to start out at his lucky table. As the night wore on, the number of players dwindled. And when it finally came down to Alex against Doyle the new kid in town, Alex lay down his cards and issued a challenge. I'll bet my stack of chips here against yours, right now, that you can't answer one of the simplest poker questions: a puzzle that any poker player worth his spit should be able to solve. These weren't the first words Alex had spoken. All thru the tournament he had been verbally sizing up the new kid, looking for a goad that would make this moment a reality. Doyle, as it happened, was a sharp player. Better perhaps in raw talent than even Alex himself. But he had a weakness. Doyle couldn't let a pi**ing contest go by without getting involved. And when Alex suggested that he couldn't answer even a basic question about poker, he jumped at the bait. You're on! he shouted, laying his cards down as well. What's this question you think I can't answer? Easy now, Alex began to milk the moment, let's not get all excited about it. Like I said, it's a simple question. All poker players know the precedence of suits and hands. All I'm asking you is to tell me the best Full House you could possibly be dealt. Doyle thought for no more than a moment before he answered. And Alex then began to explain simply and clearly why he now owned all of Doyle's chips, and had become the tourney champion. What answer would you have given?

Let's get rid of any caveats. Prime is correct that the OP is weirdly stated. My bad. Let these statements replace the corresponding ones in the OP Prime Robot is placed at the center of the rail. Twelve robots are then placed at random on one side, and twelve more are placed at random on the other side of Prime. etc... d3k3,

Chuck, That's a basically zero-probability case; but if it happened I included it. I included the cases where the triangle areas having the center as one vertex equaled* the area of the triangle that didn't. *To within a comparison tolerance of 10-13. Prime, Simulations are finite samples, like all measurements. If you see someone's data points lying exactly on the theory curve, take it to the bank the data were doctored. My purpose was to establish 1/4 as the answer, as opposed say to 1/2 or some other simple fraction. I was satisfied that it did that. The deviation is not systematic. That would indicate a poorly designed experiment. Equal areas computed to a finite number of decimal places have an unbiased even chance of meeting the criterion. Saying "on the line" points are exterior is no more valid than saying they are interior. The number of "on the line" cases has Lesbegue integral measure of zero; the point is moot.

Given that ... Twenty-five Robots, all named after BrainDenizens, are placed at random on a rail 1 mile long. The robot named Prime begins as the thirteenth robot from the North end of the rail at its midpoint. Each robot faces North or South with equal probability, and travels at 1 mile/hour in the direction it faces. When two robots meet or reach the end of the rail, they reverse direction. On average, what is the net distance Prime has traveled after one hour?

That leaves 3. 7,8,5,3,4

Nice one.

but that's only valid if you want it equal to zero, if it is asking a different question that i have no clue.

Right. There is no N, not matter how large, for which all of the 2N distinct strings of flips will produce all possible triples. For example, one of the distinct strings is N consecutive Ts, and another is N consecutive Hs. As N increases, these comprise an increasingly small fraction of the distinct cases, but in these strings only the triples TTT and HHH, respectively, appear. Well, we can say there is no value of N that will ensure a triple will occur after N flips. One thing we can ask, though, is the expected number of coin flips for a triple to occur: the expected "wait time."

Prime asks: So are we looking for the exact probability of precedence in 17 tosses for each triplet? Better, probability of precedence in a string of tosses that only needs to be long enough for the first one to appear. Chuck Rampart notes three pairs, with one being likely to appear first. Correct on the first two; and I think you can calculate their likelihood: you almost have it for the first, most glaring, case. But symmetry can be invoked on the first two tosses for your third pair: THT and HTT are equally likely to appear first. Can you calculate the likelihoods for HTT before TTT and for HHT before HTT? If we want to ignore pairs [all 14 nonredundant pairs can directly be calculated] it's possible also just to think of how many tosses on average are needed for each of the eight triples occur individually. I don't know that this can be directly calculated; but it can be simulated. For starters, since p=1/8, you might expect 8 tosses for each. But would some actually occur later on average? And why does the p=1/8 fact seem to be a red herring in this puzzle?