plasmid

I get similar numbers as bonanova and prime. I did 1000 trials of playing 1000 games per trial. Average final bankroll ~ $2E10 Median final bankroll = $0.49 Number of winning trials = 465 If you're on a Windows machine with a Java JDK installed (free to download from Oracle if you don't have it already): copy the code below and paste it into a txt file called LetItRide.java, then on a command prompt in that directory type javac LetItRide.java to compile it, then type java LetItRide to run it.

I find that answer surprising if true. Let A be the slowest car in the group. Since every car behind A will eventually catch up to A and be stuck behind it, there will be one traffic jam with A and all subsequent cars, and howevermany traffic jams involving the cars in front of A. If there are M cars in front of A then there can be no more than M+1 traffic jams, and that's only in the special case where every car in front of A is traveling slower than the car in front of it (if you count a single unimpeded car to be a traffic jam). In general there will be less than M+1. So if car A is randomly distributed among the pack, M will on average be about N/2 and the total number of traffic jams should be less than that.

I was thinking along the same lines as Prime as far as the lion's strategy. But based on this calculation, that might not be a viable approach.

Should we assume the lion will always run straight for the trainer, or can the lion do whatever it feels is an optimal strategy? (I get the feeling the answer might be "calculate it both ways.") And should we consider a starting position where the trainer is at the edge of the cage and the lion is in the center, or something else? (At least with this question you can't reasonably say "calculate it for all possible starting positions." )

Oh sorry, I didn't see that you already had the spoiler in post 22 showing that there's a <50% chance of ending up with $1 or more after repeatedly letting everything ride. So that shows that the median and most likely outcome is in fact a loss.

Now's my turn to disagree with Prime. If you ride your winnings, the most likely outcome is not that you will win millions. The most likely outcome is that you will lose money. It's just that your payoff if you do win is much larger than your odds of losing, like a favorable lottery. The mean outcome of always letting everything ride is greater than your initial $1 entry wager. The median outcome of always letting everything ride is less than your initial $1. If there's doubt about this, it could be tested by simulating a bunch of runs and calculating both the mean and median. Which of those two, the mean or median, is "most important" is a philosophical argument.

I will now prove that calculating a geometric mean does not necessarily lead to an accurate conclusion, by counterexample. I offer you a game where you pay $1 to have a 99/100 chance of winning $1000 and a 1/100 chance of losing your wager. The geometric mean of the possible outcomes is zero, so it would be foolish to play in such a game if you repeatedly bet your bankroll(?) You can argue that playing an infinite number of times would guarantee that you lose your wager, but for any sensible number of plays it'd be a no-brainer. I'd even argue that after a Large number of plays, the infinitesimal chance you'll win those games times the payoff if you do win (which is very large as far as large numbers go) would make the game worthwhile.

So you're saying you wouldn't play a lottery with a 1/1,000,000 chance of winning $5,000,000 with a $1 ticket?

On the right track, but those dwarves lack a helm that you would hew yet not deliver.

Would it help to rephrase the problem by saying "You've heard of double-or-nothing games. You have a chance to play a triple-or-nothing game with 50/50 odds, but you have to bet your entire holdings each time you play. Should you take it?" It's in the same spirit as the OP but the math becomes trivially simple and doesn't call for simulations. If you start with $1 and play for N rounds, you have a 1/2N chance of winning $3N, and a (2N-1)/2N chance of losing your wager. The average outcome is 3N/2N so it's a clearly winning game from that perspective, the most likely outcome is you lose everything so it's a losing game from that perspective. I'd say play the game a few times. It's not really different from, say, rolling a six sided die and getting $10 if you roll a 1 vs lose $1 if you roll anything else.

Not any unusual scene like that, something more everyday that just requires specific creative interpretations of the clues to explain. Neither a face, nor a song, nor an instrument; it's certainly something that most people would think of as having a bridge, although "bridge" might not be the first descriptor that comes to mind.

Nice tries, but we're not there yet. A bit of direction for the efforts: the next to last line is a fairly specific clue.

Still of the "I'm not a" variety, just posted in a way to not be searchable in case JBBSen wanted to use it... I'll go back and edit it at some point and post it in text rather than jpg. But written a little too quickly, the "I" in the second line actually shouldn't be written that way because the riddle isn't asking for that thing's identity, it's asking about the entire scene. Not an engagement ring, I don't envision a pair of balconies with that answer.

Ok, I've got a relatively simple "I'm not a..." riddle that might fit the bill. I'll post it on this forum as a picture to keep the browsers from finding it.

Would condition #3 be met by my usual "I'm not a..." style of riddle like the one that's pinned in this forum? And would conditions #1-2 be violated by almost every riddle I've written (except for maybe the )? One problem I've noticed with condition #5 is that everything on this site is google-able - just search for the riddle in quotes - so anything that's been posted and solved here could be searched. I'll try to come up with a riddle that fits those criteria over the next week or so as I'm working on experiments and writing grants, and find a way to post it that can't be searched.

If a spiteful answer cannot be factually correct, then the problem is easy. If a spiteful answer can be correct as long as the person thinks they can confuse you by giving a correct answer, AND if you're not allowed to do stuff like address a question to multiple people simultaneously, or ask a question using a real name instead of a username (for example, ask "O Angelica, where art thou?" and then ask Uberkewl "Who is Cameron?" knowing that the previous question went to Angelica), then I suspect that it's not possible to determine everyone's identity. My suspicion of impossibility is based on what looks like an unbeatable strategy to conceal the truth. Suppose (without loss of generality) that Angelica is Uberkewl, Bernardo is Vaxxipaxx, and Cameron is Willywutang. If you ask a question and they are forced to tell the truth, then they will of course tell the truth. However, if they are not forced to tell the truth, then they will instead consider the alternate scenario where Angelica is still Uberkewl, but Bernardo is Willywutang and Cameron is Vaxxipaxx. If they would be forced to tell the truth if this alternate scenario were true, then they will tell what would be a truthful answer under that alternate scenario. If they would not be forced to tell the truth under either scenario, then they will just answer with "lol".

I believe that a spiteful answer could be truthful if the person you're asking thinks that telling the truth will screw up your strategy. (Otherwise the problem would be rather trivial.) I'll also assume that you can't pull shenanigans like "Ask both V and W such-and-such a question simultaneously, and then ask U whether the sky is blue, and then you can be assured that U must answer the next question truthfully." But correct me if I'm wrong about that. The best I can do is determine one person's identity. And not the person of my choosing, just some person out of the three.

I'm admittedly too much of a landlubber to come up with a more sophisticated answer.