As my body building friend has repeatedly yelled in the gym, training is about mind and body. So to keep our minds sharp and healthy to stave of dementia, here is my puzzle thread where I will post a number of brain teasers. Here is the first and second and third and nth (from a post I made on FB): Here is the probability of survival I've calculated for the n prisoners: 0.5+(1/(n^2)) If you can get higher, please post your working and I'll post mine. THERE ARE NO LOOP HOLES, THIS IS A LOGIC PUZZLE. Also if you answered this on FB correctly, please do not post your answer here unless you have calculated the nth. UPDATE TO PUZZLE: the warden assures you that there is always at least 1 red hat and 1 blue hat. My friend assures me that this is now solvable.
Isn't this a question of the difference between one coin flip and a set of coin flips? Although each sequential coin flip only has 1/2 chance of being heads or tails, the distribution of heads and tails among say, 10 coin flips is likely to be around 50/50 such that if a prisoner saw two red hats he would have a good chance of guessing correctly that he has a blue hat based on a 50/50 distribution. Probability of having 1 coin flip wind up heads is 50%. Probability of having three coins flip heads is 12.5%. Any prisoner that sees a majority of red hats has a better than 50% chance of surviving if he guesses that he has a blue hat. I don't want to work out the equation because I am very, very tired.
Hey! Mascarenhas! HEY! Sorry, nothing useful to add, but I've been wanting to say that for ages, carry on Mitch, hey!
My friend points out that this is a gamblers fallacy because each event (coin flip or hat) is independent (it is not necessarily tending back toward a mean). Dependent events that tend toward their base distribution such as drawing a high or low card in black jack.
Oops, yeah good call. Going to go sleep now. Maybe brain work better later. Edit: Wait, that only works if you're applying that to one viewpoint. There is only one configuration of hats that results in all three prisoners having red hats. There are three configurations in which one prisoner has a blue hat and the other two have red hats. He or she can assume that all other prisoners realize this fact. He/she can assume that because they have not spoken up, they do not see a majority of red hats. Therefore it is more likely that he/she is wearing a blue hat and the other players have a 50/50 distribution and don't know how to guess.
I didnt figure it out till he explained it to me just now. We assume things are going to tend back to 50/50 but it you think about throwing a coin, it doesnt matter if you throw it for the 1st time or the 1000th time - the probability is still 50/50. UPDATE TO PUZZLE: the warden assures you that there is always at least 1 red hat and 1 blue hat. My friend assures me that this is now solvable.
I'd look at the reflection of my hat in the pupil of another prisoner. If the hat distribution is random, and no communication is allowed, I don't see how you could enact a strategy. I'm probably being too linear. It is my way. But, on the subject of coin flips and the gambler's fallacy; how does radioactive decay fit into that? An individual atom decays in a completely unpredictable amount of time, but a whole bunch of them decay in an extremely predictable amount of time: http://demonstrations.wolfram.com/RadioactiveDecayAsAProbabilityDistribution/ http://en.wikipedia.org/wiki/Radioactive_decay
I'm terrible at this kind of thing, but doesn't the distribution of colours for the first three prisoners change the odds once you start introducing more? As in, you have to know what colour hat you're wearing - if you see two hats of the same colour on the other prisoner's heads you're golden, if not it's 50/50. This is with the no communication proviso. Does the strategy talk between the four prisoners include being able to talk about what colour hat they're wearing?
the strategy talk beforehand is before they are given hats. and hats are redestributed between rounds. only 1 prisoner needs to guess his hat correctly for a stay of execution for the lot, others can stay silent/not guess. if a prisoner guesses incorrectly, they all die. hence its best to stay silent.
So one prisoner at least knows the color of his hat (in a party of three), because at least one prisoner will see either two red hats or two blue hats.
Can we assume each prisoner is a rational and intelligent being? Such that if there is a solution they each are able to reach it...
I'll tell you what I can't work out: who are you, Mascarenhas? In the sense of presumably having had a different MAP username in the past? Cheers.
The prisoner has the certainty of guessing the right hat in 2 cases (that is, all the other hats being red or all the other hats being blue), and he has a 0.5 probability in the other n-2 cases (the other hats being of different colours). I'm having trouble putting this into one equation though, perhaps someone more mathematically versed can help me out EDIT: doh! I misunderstood it. I thought the prisoner having to guess was chosen beforehands. Sorry, I haven't had any coffee yet.
I'd say it's clear when there's three of them. When there are 4 prisoners, and more then one hat is a different color then the others (in which case it's the same as before), they can wait a little. If nobody proclaims right away the answer, it means nobody can see three same colors. Therefore on of them will make an assumption there are 2 reds and 2 blues and can guess his own correcly. When more of the prisoners are introduced, I'll have to think a bit more about that, but I think when discussing strategy, they can set something like rounds. For example: first ten seconds will be reserved for an answer of the one who sees that every prisoner (except for him) has the same color. Next ten seconds will be for those, who see for example all reds and 1 blue, in which case, they're blue. Next ten seconds for those who see all reds/blues and 2 other colors.... So the first one to answer will be on of those who have the minority color.... and when there are the same number of reds and blues, everyone should be able to answer, then. Could it possibly be like that?
